A Recreational Application of Two Integer Sequences and the Generalized Repetitious Number Puzzle
aa r X i v : . [ m a t h . HO ] M a r J.R.M. Antalan Research PaperMay 28, 2019
A Recreational Application of Two Integer Sequencesand the Generalized Repetitious Number Puzzle
Abstract
In this article, we give a particular recreational application of the sequence A000533and A261544 in “The On-line Encyclopedia of Integer Sequences” (OEIS). The recre-ational application provides a direct extension to “The Repetitious Number” puzzle ofMartin Gardner contained in The Second Scientific American Book of MathematicalPuzzles and Diversions published in 1961. We then provide a generalization to therepetitious number puzzle and give a related puzzle as an illustrative example. Finally,as a consequence of the generalization, we define a family of sequence in which thesequences A000533 and A261544 belong.
Keywords:
Integer Sequence · Repetitious Number · Generator · Length · Replication Number · ( l , r ) co-divisor · ( l , r ) co-divisor number · ( l , r ) co-divisor sequence. AMS Classification Numbers:
Author Information:
John Rafael M. Antalan(Assistant Professor, Department of Mathematics and Physics, College of Arts and Sciences, Central LuzonState University, 3120), Science City of Mu˜noz, Nueva Ecija, Philippines.e-mail: [email protected]
J.R.M.Antalan
We begin this section by giving a brief introductory information on the
On-line Encyclope-dia of Integer Sequences or OEIS. We then discuss the two integer sequences under study.Finally, we present the
Repetitious Number Puzzle of Gardner that will serve as the “source”of the recreational application.
The OEIS (available at https://oeis.org/ ) is an on-line collection of over quarter-millionnumber sequences initiated by Neil J.A. Sloane in early 1964 [1].OEIS aims are (based on [2]):1. To allow mathematicians or other scientists to find out if some sequence that turns upin their research has ever been seen before. If it has, they may find that the problemthey’re working on has already been solved, or partially solved, by someone else. Orthey may find that the sequence showed up in some other situation, which may showthem an unexpected relationship between their problem and something else.2. To have an easily accessible database of important, but difficult to compute, se-quences.We illustrate the first aim using the paper of Rabago and Tagle in [3]. Their paper aimsto find the integral dimensions of a rectangular prism (i.e. length, width and height) inwhich the surface area and the volume are numerically equal. The solution to their problemwritten in lexicographic order is surprisingly the sequence A229941 [4] in the OEIS whichgives a way for three regular polygons to snugly meet at a point.For the second aim, a particular example of important but difficult to compute sequencein the OEIS is the sequence of
Mersenne primes [5]. The existence of which is equivalentto the existence of an even perfect number and the largest known prime number [6].
We now turn our attention to two particular sequence in the OEIS. They are the sequencesA000533 and A261544.The sequence A000533 [7] in the OEIS is the sequence defined by a ( ) = a ( n ) = n + , n ≥ . Its first 15 terms are:1, 11, 101, 1001, 10001, 100001, 1000001, 10000001, 100000001, 1000000001,10000000001, 100000000001, 1000000000001, 10000000000001, 100000000000001, . . . . ecreationalApplicationofTwoIntegerSequences... a ( ) =
11 and a ( ) =
101 are the only prime terms of thesequence up to n =
100 000” [7]. Also, it is unknown whether there are other prime termsin the sequence.On the otherhand, the sequence A261544 [8] is the sequence defined by b ( n ) = n ∑ k = k . Its first 10 terms are:1, 1001, 1001001, 1001001001, 1001001001001, 1001001001001001,1001001001001001001, 1001001001001001001001, 1001001001001001001001001,1001001001001001001001001001, . . . .It can be verified that unlike the first sequence, the terms of this sequence are all com-posite except for the zeroth term “1”. A complete solution to this claim may be viewed in[9]. With the two sequences in the OEIS already introduced, we are now ready to considerthe
Repetitious Number Puzzle . In [10], Martin Gardner presented the puzzle given below:“
The Repetitious Number . An unusual parlor trick is performed as follows.Ask spectator A to jot down any three-digit number, and then to repeat thedigits in the same order to make a six-digit number (e.g., 394 394). With yourback turned so that you cannot see the number, ask A to pass the sheet of paperto spectator B, who is requested to divide the number by 7.Dont worry about the remainder, you tell him, because there won’t be any. B issurprised to discover that you are right (e.g., 394 394 divided by 7 is 56 342).Without telling you the result, he passes it on to spectator C, who is told todivide it by 11. Once again you state that there will be no remainder, and thisalso proves correct (56 342 divided by 11 is 5 122).With your back still turned, and no knowledge whatever of the figures obtainedby these computations, you direct a fourth spectator D, to divide the last resultby 13. Again the division comes out even (5 122 divided by 13 is 394). Thisfinal result is written on a slip of paper which is folded and handed to you.Without opening it you pass it on to spectator A.Open this, you tell him, and you will find your original three-digit number.Prove that the trick cannot fail to work regardless of the digits chosen by thefirst spectator.”
J.R.M.Antalan
This puzzle was originally written by Yakov Perelman in [11].In section 3, we discuss the solution of the puzzle and state some important questionsnecessary for its extension. At the moment, we discuss some important notations and math-ematical concepts needed in understanding the solution of the puzzle and its generalizationin the next section.
The following terms will be encountered in the succeeding sections of this article.
Definition 2.1.
Let n = d d . . . d k d d . . . d k . . . d d . . . d k be a positive repetitive integer.We say that the positive integer g = d d . . . d k is a generator of n if g is a positive integersuch that replicating g a finite number of times generates n. Definition 2.2.
Let g = d d . . . d k be a generator of n. Then the length of g denoted by l ( g ) is the number of digits in g. Definition 2.3.
Let g = d d . . . d k be a generator of n. The replication number of g denotedby r ( g ) is the number of replication performed in g in order to generate n. To fully understand the concepts being discussed, we consider some examples.
Example 2.4.
Consider the positive repetitive integer n =
394 394 in the Repetitious Num-ber Puzzle. The positive integer g = is a generator for n with length l ( g ) = andreplication number r ( g ) = . Example 2.5.
The positive repetitive integer n =
111 111 is generated by g = withlength l ( g ) = and replication number r ( g ) = . The integers 11, 111, and 111 111 arethe other generators of n . Example 2.6.
The positive integer n =
223 344 generates itself with length and replica-tion number . Remark 2.7.
We emphasize that a generator is not unique. Also, for any positive integer n,n is a generator of itself. Moreover, if n is not repetitive, then its generator is unique.
Remark 2.8.
If g generates n with replication number r, we write n = g r . For completeness, we recall some essential concepts in elementary Number Theory.
Definition 2.9.
Let a and b be two positive integers such that a ≤ b. We say that a divides b written in symbol by a | b if there is a positive integer c such thatb = ac . (1) ecreationalApplicationofTwoIntegerSequences... If there is no positive integer c that satisfies equation (1), then we say that a does not divides b and this situation is denoted by a ∤ b. If a | b, we can also say the following:(i) b is a multiple of a, (ii) a is a divisor of b and (iii) a is a factor of b. Example 2.10.
Let us consider the positive integer
394 394 . Note that divides
394 394 since
394 394 = ×
56 342 . However, does not divides
394 394 since we cannot find any positive integer c that cansatisfy the equation
394 394 = × c . The property of divisibility given below is important.
Lemma 2.11.
Let a , b and D be positive integers such that D ≤ a and D ≤ b. If D | a andD | b, then D | ( ax + by ) for any positive integers x and y. The proof of Lemma 2.11 follows directly from the definition of divisibility and is stan-dard in any elementary Number Theory textbooks. Letting x = y = Corollary 2.12.
Let a , b and D be positive integers such that D ≤ a and D ≤ b. If D | a andD | b, then D | ( a + b ) . Remark 2.13.
The property of divisibility stated in Corollary 2.12 can be easily extendedinto a finite number of multiples. Given D | a and D | b, by Corollary 2.12 we have D | ( a + b ) . Now, if D | c given D | ( a + b ) applying Corollary 2.12 once more gives D | (( a + b ) + c ) or D | ( a + b + c ) . If b is divided by a then either a | b or a ∤ b . In both cases however, we may write b interms of a ; this is guaranteed by the next theorem. Theorem 2.14 ( Division Algorithm).
Given integers a and b with a > , there are uniqueintegers q and r satisfying b = qa + r , ≤ r < a . Remark 2.15.
The integers q and r are respectively called the quotient and the remainder of b upon division by a. Note also that a | b if and only if r = , and that a ∤ b if and only ifotherwise. Definition 2.16.
A positive integer p > is said to be a prime number if its only positivedivisors are and p itself. Example 2.17.
The positive integers , and are all prime numbers; since their onlypositive divisors are and their selves.While the positive integer
394 394 is not a primenumber; since from Example 2.10, we know that not only the positive integers and
394 394 divides
394 394 but also the integer . Theorem 2.18 ( Fundamental Theorem of Arithmetic).
Every positive integer n > iseither a prime or a product of primes; this representation is unique, apart from the order inwhich the factors occur. J.R.M.Antalan
Theorem 2.18 is a well known result in elementary Number Theory, its proof is includedin most of elementary Number Theory textbooks. Consider Burton in [12] for instance.
Illustration 2.19.
Let us consider the positive integer . From Theorem 2.18, either is a prime or a product of primes. The latter holds true since = × × . After a brief recall in some essential results in Elementary Number Theory, we are nowready to present the solution of the “Repetitious Number Puzzle”.
The solution discussed in this subsection is due to the solution presented by Gardner in [10].Any three digit number takes the form d d d where d , d and d are non-negativeintegers with bounds 0 < d ≤ ≤ d ≤ ≤ d ≤ d d d d d d . Thisinteger can be factored into 1 001 × d d d as shown in the following computation1 0 0 1 × d d d d d d d + d d d d d d d d .Thus, d d d and 1 001 divides d d d d d d and that d d d d d d = × d d d .From Illustration 2.19, 1 001 can be expressed as a product of primes 7 ,
11 and 13. Hence7 ,
11 and 13 divides d d d d d d and that d d d d d d = × × × d d d . In lieu of Theorem 2.14, we have d d d d d d = × ( × × d d d ) + . So, dividing the integer d d d d d d by 7 gives the integer 11 × × d d d with remain-der 0.Next, we consider the integer 11 × × d d d . In lieu of Theorem 2.14, we have11 × × d d d = × ( × d d d ) + . ecreationalApplicationofTwoIntegerSequences... × × d d d by 11 gives the integer 13 × d d d with remainder0. Finally, we consider the integer 13 × d d d . Note that this integer can be written as13 × d d d = × ( d d d ) + . So, dividing the integer 13 × d d d by 13 gives the integer d d d with remainder 0.Hence, dividing the six-digit repetitive number d d d d d d in succession by the in-tegers 7 ,
11 and 13 returns the repetitive number into its generator d d d . This solves thepuzzle. Remark 2.20.
The order of dividing the integer d d d d d d by the integers , and do not matter in the puzzle. For d d d d d d can be written as × ( × × d d d ) , × ( × × d d d ) × ( × × d d d ) , × ( × × d d d ) , × ( × × d d d ) and × ( × × d d d ) . We now present our results in the next section.
The goal of this subsection is to show that the k th term a ( k ) of the sequence A000533 dividesthe 2 k − digit repetitive number n generated by g with l ( g ) = k . Hence, when a k − digit number g is duplicated resulting to n , dividing n with the prime factors of a ( k ) gives theoriginal number g . This result is due to Theorem 3.1.
Let n = ( d d . . . d k ) be a repetitive number generated by g = d d . . . d k oflength k. Then there is a finite sequence of divisors D i such that n upon division by all of D i becomes g.Proof. Given a repetitive number n = ( d d . . . d k ) , we express it as a sum of two positiveintegers both divisible by g = d d . . . d k . In particular n can be expressed as the sum d d . . . d k . . . + d d . . . d k d d . . . d k d d . . . d k .Note that since g | g and g | d d . . . d k . . . | {z } k-zeros , by Corollary 2.12 we have g | ( g + d d . . . d k . . . | {z } k-zeros ) . J.R.M.Antalan
But g + d d . . . d k . . . | {z } k-zeros = n . So, g | n .After factoring out the common factor g in both summands we have n = g × ( + . . . | {z } k-zeros )= g × ( . . . | {z } k − )= g × a ( k ) . By the Fundamental Theorem of Arithmetic (Theorem 2.18), a ( k ) is either a prime ora product of primes. If a ( k ) is prime, then the finite sequence of divisors to be divided to n to become g is a ( k ) itself. If a ( k ) is non-prime then the finite sequence of divisors tobe divided to n to become g is the finite sequence whose terms are the prime divisors of a ( k ) .The proof of Theorem 3.1 gives us a method on solving a particular extension of the repetitious number puzzle . Problem.
Suppose that in the repetitious number puzzle spectator A was asked to writedown any k − digit positive integer. To what sequence of prime numbers does the resulting2 k − digit number be divided in order to return to the original k − digit number? Solution.
Let g be the k − digit positive integer and let n be the resulting 2 k − digit num-ber. Note that n is a repetitive number generated by g of length k with replication number2. That is, n = g with l ( g ) = k . Using the result contained in the proof of Theorem 3.1, wemust divide n by the prime divisors of 1 00 . . . | {z } k-1-zeros
1, or the k th term of the sequence A000533in order to return to the original number g . Illustration 3.2.
Suppose that spectator A wrote the number g =
451 220 125 . Duplicatingg gives the number n =
451 220 125 451 220 125 . Dividing n by the numbers , , , and
52 579 , the prime divisors of which is the th term of the sequenceA000533 (See Table 1), gives the original number g =
451 220 125 . The goal of this subsection is to show that the ( r − ) st term b ( r − ) of the sequenceA261544 divides the 3 r − digit repetitive number n generated by g with l ( g ) =
3. Hence,when a 3 − digit number g is replicated r -times resulting to n , dividing n with the primefactors of b ( r − ) gives the original number g . This result is due to Theorem 3.3.
Let n = ( d d d ) r be a repetitive number generated by g = d d d of length . Then there is a finite sequence of divisors D i such that n upon division by all of D i becomes g. ec r ea ti on a l A pp li ca ti ono f T w o I n t e g e r S e qu e n ce s ... Table 1:
Prime factorization of the first 25 terms of the sequence A000533 (Generated using Wolfram Alpha[13])
No. of Digits ( k ) Rep. No. ( r ) Terms of Sequence A000533 Prime Factorization0 2 1 -1 2 11 112 2 101 1013 2 1001 7 · ·
134 2 10001 73 · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · J.R.M.Antalan
Proof.
Given a repetitive number n = ( d d d ) r , we express it as a sum of r positive integersboth divisible by g = d d d . In particular n can be expressed as the sum n = d d d ( ) ( r − ) + d d d ( ) ( r − ) + . . . + d d d ( ) ( r − r ) . Note that since g | g and g | d d d ( ) j , for j = , , . . . r −
1, by Corollary 2.12 we have g | d d d ( ) ( r − ) + d d d ( ) ( r − ) + . . . + d d d ( ) ( r − r ) .So, g | n .Factoring out the common factor g in all of the summands we have n = g × (cid:16) ( ) ( r − ) + ( ) ( r − ) + . . . + ( ) ( r − r ) (cid:17) = g × b ( r − ) . By the Fundamental Theorem of Arithmetic (Theorem 2.18), b ( r − ) is either a primeor a product of primes. However, we know that (except for the zeroth term) the terms of thesequence A261544 are all composite. So the finite sequence of divisors to be divided to n inorder to become g is the finite sequence whose terms are the prime divisors of b ( r − ) .The proof of Theorem 3.2 gives us a method on solving another particular extension ofthe repetitious number puzzle . Problem.
Suppose that in the repetitious number puzzle spectator A was asked to writedown any 3 − digit positive integer and replicate it r − times. To what sequence of numbersdoes the resulting 3 r − digit number be divided in order to return to the original 3 − digitnumber? Solution.
Let g be the 3 − digit positive integer and let n be the resulting 3 r − digitnumber. Note that n is a repetitive number generated by g of length 3 with replicationnumber r . That is, n = g r with l ( g ) =
3. Using the result contained in the proof of Theorem3.2, we must divide n by the prime divisors of b ( r − ) , the ( r − ) st term of the sequenceA261544 in order to return to the original number g . Illustration 3.4.
Suppose that spectator A wrote the number g = . Replicating g 4-timesgives the number n =
721 721 721 721 . Dividing n by the numbers , , , , theprime divisors of the third term of the sequence A261544 which is (See Table2), gives the original number g = . In this subsection, we generalize the repetitious number puzzle by allowing spectator A towrite down any k − digit number and replicate it r − times to generate the integer n = g r with l ( g ) = k . The generalization is given in the next theorem. ec r ea ti on a l A pp li ca ti ono f T w o I n t e g e r S e qu e n ce s ... Table 2:
Prime factorization of the first nine terms of the sequence A261544 (Generated using Wolfram Alpha [13])
Number of Digits ( k ) Number of Repetitions ( r ) Terms of the Sequence A261544 Prime Factorization3 1 1 − · ·
133 3 1001001 3 · · · · · · · · · · · · · · · · · · · · · · · · · · · · J.R.M.Antalan
Theorem 3.5.
Let n = ( d d . . . d k ) r be a repetitive number generated by g = d d . . . d k oflength k. Then the sequence of prime factors of the integer (cid:16) ( ) k − (cid:17) r − is a finite sequence such that n upon division by all the sequence terms becomes g.Proof. Given a repetitive number n = ( d d . . . d k ) r , we express it as a sum of r positiveintegers both divisible by g = d d . . . d k . In particular n can be expressed as the sum n = d d . . . d k ( ) k ( r − ) + d d . . . d k ( ) k ( r − ) + . . . + d d . . . d k ( ) k ( r − r ) . Note that since g | g and g | d d . . . d k ( ) k j , for j = , , . . . r −
1, by Corollary 2.12 wehave g | d d . . . d k ( ) k ( r − ) + d d . . . d k ( ) k ( r − ) + . . . + d d . . . d k ( ) k ( r − r ) .So, g | n .Factoring out the common factor g in all of the summands we have n = g × (cid:16) ( ) k ( r − ) + ( ) k ( r − ) + . . . + ( ) k ( r − r ) (cid:17) = g × (cid:16) ( ) k − (cid:17) r − . By the Fundamental Theorem of Arithmetic (Theorem 2.18), (cid:16) ( ) k − (cid:17) r − (cid:16) ( ) k − (cid:17) r − n to become g is (cid:16) ( ) k − (cid:17) r − (cid:16) ( ) k − (cid:17) r − n to become g is the finite sequence whoseterms are the prime divisors of (cid:16) ( ) k − (cid:17) r − Puzzle 3.6. A Relay Involving Division of Large Numbers.
Sir DELTA is grade 7 math-ematics teacher in the Philippines. To test the proficiency of his students on performingdivision of large numbers, he grouped his students such that each group is consists of 10members.He then instruct the first student which we name S1 to write down in a 1/4 sheet ofpaper any − digit positive integer (say 2 019) and replicate it − times to get a − digitnumber (
20 192 019 201 920 192 019 201 920 192 019 ). Then he asked S1 to give the papercontaining the − digit number to S2. S2 then was asked to divide the − digit number by and write down the answer (1 187 765 835 407 070 118 776 583 540 707) in another1/4 sheet of paper. After S2 was done writing the answer in a 1/4 sheet of paper, Sir Deltaasked S2 to give the paper to S3. ecreationalApplicationofTwoIntegerSequences... Denote by A n the answer of student n. Suppose that the process continues with thefollowing given (See Table 3): • S3 performs A ÷ • S4 performs A ÷ • S5 performs A ÷ • S6 performs A ÷ • S7 performs A ÷ • S8 performs A ÷ • S9 performs A ÷
69 857 • S10 performs A ÷ .Sir DELTA then asked S10 to give his/her answer to him.If Sir DELTA wants to determine whether his students performed their assigned divisionproblem correctly or not, prove that it is enough for him to ask S1: “Is this your − digitnumber?” Illustration 3.7.
Given below are the correct answers for the assigned sequence of divisionin Puzzle 3.6 generated using Wolfram Alpha [13].
20 192 019 201 920 192 019 201 920 192 019 ÷ = ÷ =
16 270 764 868 590 001 627 076 486 85916 270 764 868 590 001 627 076 486 859 ÷ =
118 764 707 070 000 011 876 470 707118 764 707 070 000 011 876 470 707 ÷ =
336 443 929 376 770 571 888 019336 443 929 376 770 571 888 019 ÷ =
749 318 328 233 342 030 931749 318 328 233 342 030 931 ÷ = ÷ =
829 654 614 178 899829 654 614 178 899 ÷
69 857 =
11 876 470 70711 876 470 707 ÷ = . J . R . M . A n t a l a n Table 3:
Some integers of the form (cid:0) ( ) k − (cid:1) r − with their corresponding prime divisors (Generated using Wolfram Alpha [13]). ( j ) ( r ) (cid:0) ( ) j − (cid:1) r − ecreationalApplicationofTwoIntegerSequences... ( l , r ) co-divisor Number and the Family of ( l , r ) co-divisor Sequences We learned from the previous subsection that given a repetitive number n = ( d d . . . d k ) r that is generated by g = d d . . . d k of length k , we have n = g × (cid:16) ( ) k − (cid:17) r − . (2)The number (cid:16) ( ) k − (cid:17) r − Definition 3.8.
Let k , r ∈ Z + . The number (cid:16) ( ) k − (cid:17) r − in equation (2) is called the ( l , r ) co-divisor of g relative to n. Remark 3.9.
The name ( l , r ) co-divisor number defined on Definition 3.8 is base from theidea that the number (cid:16) ( ) k − (cid:17) r − is dependent to the length ( l ) of the generator and itsreplication number ( r ) . Example 3.10.
Recall that in the Repetitious Number Puzzle we have
394 394 = × . Hence, the ( , ) co-divisor of relative to
394 394 is . In general, given anypositive integer g of length when duplicated has the ( l , r ) co-divisor of . Remark 3.11.
To avoid redundancy, we drop the word “relative to n” in determining the ( l , r ) co-divisor of g. This is because the ( l , r ) co-divisor of an integer g is completelydetermined by the length of g which is k and the number of replications performed in gwhich is r. Example 3.12.
In Illustration 3.2, the ( , ) co-divisor of
451 220 125 is .In general, any positive integer g of length when duplicated has the ( l , r ) codivisor of . Example 3.13.
Let g be a -digit positive integer. The ( , ) codivisor of g is .(See Illustration 3.4) Example 3.14.
Let g be a -digit positive integer. The ( , ) codivisor of g is the number
10 001 000 100 010 001 000 100 010 001 . (See Puzzle 3.6)
The concept of ( l , r ) co-divisor allows us to view the sequence A000533 and the se-quence A261544 in the OEIS as a particular member of a family of sequence which we call ( l , r ) co-divisor sequences .In particular, if we let s ( k , r ) = (cid:0) ( ) k − (cid:1) r − s ( k , ) = a ( k ) , j = , , , . . . where a ( k ) is the k th term of the sequence A000533. We also have s ( , r ) = b ( r − ) , r = , , , . . . . where b ( r − ) is the ( r − ) st term of the sequence A261544.We end this paper by recommending further studies on the ( l , r ) co-divisor number andthe ( l , r ) co-divisor sequences and their applications.6 J.R.M.Antalan
In this article, we discussed the
Repetitious Number Puzzle and its solution. We establishedthat the
Repetitious Number Puzzle is equivalent to the problem:
Given a positive integer generator g of length k that is to be replicated r − times resulting tothe integer n of length kr, by what prime numbers must n be divided such that upondividing n by all of the prime numbers gives back g? where the length of g is 3 and the number of replication r is 2.We then provide a generalization to the puzzle by first taking k ≥
3. We showed that thesolution to the puzzle when k ≥ a ( k ) where a ( k ) is the k th term of the sequence A000533. Then fixing k =
3, we consider r ≥
2. In this case, weshowed that the solution to the puzzle when k = r ≥ b ( r − ) where b ( r − ) is the ( r − ) st term of the sequence A261544.For the general case where k ≥ r ≥
2, we showed that the solution to the puzzleis given by the prime divisors of the ( l , r ) co-divisor number (cid:16) ( ) k − (cid:17) r −
1. The conceptof ( l , r ) co-divisor number allowed the possibility to view the sequence A000533 and thesequence A261544 in the OEIS as a particular member of a family of sequences which wecall ( l , r ) co-divisor sequences. Further studies on on the ( l , r ) co-divisor number and the ( l , r ) co-divisor sequences and their applications are then recommended. The author is thankful to every individual and organizations who are responsible for the cre-ation of this research article. The creation of this article would not be possible without thesuggestion of Mr. Melchor A. Cupatan of the Department of Mathematics and Physics Cen-tral Luzon State University. The author would also like to express his sincere gratitude toMs. Josephine Joy V. Tolentino of Philippine Science High School Central Luzon Campusfor her motivation and valuable comments that lead to the improvement of the manuscript.Also, the author thanks the Central Luzon State University for their untarnished effort inencouraging its faculty members to do research. Finally, the author would like to thank thevarious referees for their valuable comments and suggestions that helps improve the contentof the paper. ecreationalApplicationofTwoIntegerSequences... References [1] N.J.A. S
LOANE , editor,
The On-Line Encyclopedia of Integer Sequences , publishedelectronically at https://oeis.org , https://oeis.org/wiki/Welcome , May 04, 2019.[2] N.J.A. S LOANE , editor,
The On-Line Encyclopedia of Integer Sequences , publishedelectronically at https://oeis.org , https://oeis.org/wiki/OEIS FAQ , May 04, 2019.[3] J.F.T. R ABAGO AND
R.P. T
AGLE , On the Area and Volume of a certain RectangularSolid and the Diophantine Equation / = / x + / y + / z , Notes on Number Theoryand Discrete Mathematics, Vol.19 No.3 pp.28-32, (2013).[4] J EAN -F RANCOIS A LCOVER , Sequence A229941 in The On-Line Encyclopedia of Inte-ger Sequences (Oct. 4, 2013) , published electronically at https://oeis.org .[5] N.J.A. S
LOANE , Sequence A000668 in The On-Line Encyclopedia of Integer Se-quences (n.d.) , published electronically at https://oeis.org .[6]
Great Internet Mersenne Prime Search GIMPS , , May 04,2019.[7] N.J.A. S LOANE , Sequence A000533 in The On-Line Encyclopedia of Integer Se-quences (n.d.) , published electronically at https://oeis.org .[8] I. G
UTKOVSKIY , Sequence A261544 in The On-Line Encyclopedia of Integer Se-quences (Aug 24, 2015) , published electronically at https://oeis.org .[9] A. G
ARDINER , The Mathematical Olympiad Handbook: An Introduction to ProblemSolving Based on the First 32 British Mathematical Olympiads 1965-1996 , OxfordUniv. Pr., Oxford, England, UK, (1997).[10] M. G
ARDNER , The Second Scientific American Book of Mathematical Puzzles andDiversions , Simon and Schuster, New York, USA, (1961).[11] Y. P
ERELMAN , Figures for Fun: Stories and Conundrums , Foreign Language Pub-lishing House, Moscow, Russia, (1957).[12] D.M. B
URTON , Elementary Number Theory, Seventh Edition , McGraw-Hill Educa-tion, (2010).[13] W
OLFRAM A LPHA ,,