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On the Gap-Sum and Gap-Product Sequences of Integer Sequences

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On the Gap-Sum and Gap-Product Sequences of Integer Sequences On the Gap-sum and Gap-product Sequences of Integer Sequences Paul Barry School of Science Waterford Institute of Technology Ireland [email protected] Abstract In this note, we explore two families of sequences associated to a suitable integer se- quence: the gap-sum sequence and the gap-product sequence. These are the sums and the products of consecutive numbers not in the original sequence. We give closed forms for both, in terms of the original sequence, and in the case of Horadam sequences, we find the generating function of the gap-sum sequence. For some elementary sequences, we indicate that the gap-product sequences are given by the Fuss-Catalan-Raney num- bers. 1 Introduction Let A be the integer sequence (an)n≥0. The consecutive numbers between, but not including, an to an+1 will be called the nth A-gap, or just the nth gap, when the sequence in question is clear. Note that for sequences with an+1 > an + 1, each gap is an arithmetic sequence with initial term an + 1, with an+1 an 1 terms, and increment +1. If an+1 = an + 1, then we say that the gap is 0. − − Example 1. We consider the sequence an = Jn+2 which begins 1, 3, 5, 11, 21, 43, 85, 171,.... arXiv:2104.05593v1 [math.CO] 7 Apr 2021 Here, Jn is the n-th Jacobsthal number, and hence 2n ( 1)n an =4 − . 3 − 3 The first gaps of this sequence are then 1, 2 , 3, 4 , 5, 6, 7, 8, 9, 10, 11, 12, 13, , 20, 21,.... ··· We define the gap-sum sequence of a to be the sequence whose n-th term is the sum of |{z} |{z} n the elements of the nth-gap. Thus for| the{z sequence} | above,{z the gap-sum} sequence begins 2, 4, 40, 144,.... 1 This sequence is often called the sequence of sums of consecutive non-A numbers. Similarly, we define the gap-product sequence of an to be the sequence whose n-th term is the product of the elements of the n-th gap. For the sequence of the above example, we have that the gap-product sequence begins 2, 4, 30240, 60949324800,.... A first result is the following. Proposition 2. Let Sn be the gap-sum sequence of an. Then we have an+1−an−1 Sn = (an + j). j=1 X Let Pn be the gap-product sequence of an. Then we have an+1−an−1 Pn = (an + j). j=1 Y Proof. This is a reformulation of the definitions of the gap-sum and the gap-product se- quences. Example 3. We consider the Fibonacci numbers A000045 n n 1 1 √5 1 1 √5 Fn = + . √ 2 2 − √ 2 − 2 5 ! 5 ! Fn+1−Fn−1 We find that the formula for the gap-sum sequence j=1 Fn + j above in this case gives us 0, 0, 0, 0, 4, 13, 42, 119, 330, 890P, 2376,.... This is A109454 in the On-Line Encyclopedia of Integer Sequences (OEIS) [4, 5]. Example 4. We let A denote the sequence A000040 of prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,.... The prime gap-sum sequence then begins 0, 4, 6, 27, 12, 45, 18, 63, 130, 30, 170,.... This is the sequence A054265 in the OEIS, where it is described as the sequence of sums of composite numbers between successive primes. Using the formula for the sum of an arithmetic series, we obtain the following result for the gap-sum sequence. 2 Proposition 5. Let an be a sequence such that an+1 > an. Then an+1 an 1 Sn = − − (an + an ) . 2 +1 Proof. The n-th gap is an arithmetic sequence beginning with an + 1 and having step d = 1, with an an 1 terms. Thus we have +1 − − an+1 an 1 an+1 an 1 Sn = − − (2(an +1)+(an an 1) 1) = − − (an + an ) . 2 +1 − − − 2 +1 For instance, if pn denotes the n-th prime, then the prime gap-sum sequence has general term pn pn 1 S = +1 − − (p + p ) . n 2 n n+1 We note that the gap-sum sequence can be the zero sequence. For example, if an = n, then the gap-sum sequence is the zero sequence 0, 0, 0, 0,.... Corollary 6. Let an = rn, r N. Then the gap-sum sequence of an is given by ∈ 1 r Sn = (2n + 1)r(r 1) = (2n + 1) . 2 − 2 n Corollary 7. Let an = k , then the gap-sum sequence for an is given by 1 2 2n n Sn = (k 1)k (k + 1)k , 2 − − with generating function 1 (k + 1)(kx + k 2) S(x)= − . 2 (1 kx)(1 k2x) − − For k = 2, this is the gap-sum sequence for 2n. This is A103897, which begins 0, 3, 18, 84, 360, 1488, 6048, 24384,.... This sequence has generating function 3x . (1 2x)(1 4x) − − n We note that if an =2 1, then Sn is the sequence that begins − 0, 2, 15, 77, 345, 1457, 5985,.... This sequence has the generating function x(2 + x) . (1 x)(1 2x)(1 4x) − − − 3 2 Example 8. We consider the case of an =2n . We find that 3 2 Sn =8n + 10n +6n +1, with generating function 1+21x + 33x2 +3x3 . (1 x)4 − This sequence begins 1, 25, 117, 325, 697, 1281, 2125, 3277, 4785,.... 2 The gap-sum sequence for an = n is A048395. For this sequence we have 3 2 Sn =2n +2n + n, with generating function x(1 + x)(5 + x) . (1 x)4 − We give below a short table of figurate numbers and their gap-sum sequences. G.f. of an an Sn G.f. of Sn x(1+x) 2 3 2 x(1+x)(5+x) (1−x)3 n 2n +2n + n (1−x)4 x n(n+1) n(n+1)2 x(2+x) (1−x)3 2 2 (1−x)4 2x 2 1+8x+3x2 (1−x)3 n(n + 1) (2n + 1)(n + 1) (1−x)4 x(2+x) n(3n+1) (3n+1)(3n2+4n+2) 1+14x+11x2+x3 (1−x)3 2 2 (1−x)4 x n n(n+1)(n+2)(n+3)(2n+3) 5x(1+x) (1−x)4 6 (n + 1)(n + 2) 4! (1−x)6 x n+3 n(n+1)(n+2)2(n+3)(n2+6n+11) x(9+18x+7x2+x3) (1−x)5 4 144 (1−x)8 x(1+2x) n(3n−1) 3n(3n2+2n+1) 3x(x2+5x+3) (1−x)3 2 3 (1−x)4 n(n+1) n+1 The gap-sum sequence of the triangular numbers 2 = 2 is A006002. The gap-sum sequence of the tetrahedral numbers n (n+1)(n+2) = n+2 can be expressed 6 3 as n +3 n +3 5 + 10 . 4 5 n(n+1)(n+2)(n+3)(2n+3) This is 5 times A005585, the 5-dimensional pyramidal numbers 5! . n+3 The gap-sum sequence of 4 can be expressed as n +4 n +4 n +4 n +3 9 + 36 + 34 + . 5 6 7 7 4 2 The gap-product sequence Proposition 9. Let an be a sequence with an+1 an. Then the gap-product sequence Pn is given by ≥ (an+1 1)! Pn = − . an! Proof. The product of the terms of an arithmetic sequence with initial term t1, with r terms and increment d, is given by Γ t1 + r P = d . Γ(t1) For Pn, we have t = an + 1, r = an an 1, and d = 1. The result follows from this. 1 +1 − − Example 10. We consider the Fibonacci numbers (whose terms satisfy an+1 > an after n = 2). We obtain that the gap-product sequence (Fn+1 1)! Pn = − Fn! begins 1, 1, 1, 1, 4, 42, 11880, 390700800, 169958063987712000,.... Since F n +1= Fn + Fn−1, we have that Fn−1−1 Fn−1−1 j=1 (Fn+1 j)Fn! Pn = − = (Fn+1 j). Fn! − Q j=1 Y n 2n (−1) Example 11. For the sequence an =4 , we have 3 − 3 n 2n (−1) (8 3 + 3 1)! (Jn+3 1)! Pn = n −n = − , 2 (−1) (4 )! Jn+2! 3 − 3 n 2n (−1) where Jn = is the n-th Jacobsthal number (A001045). 3 − 3 3 The gap-sum sequences of Horadam sequences A Horadam sequence is a second-order recurrence sequence an = an(α,β,r,s) that is defined by an = ran−1 + san−2, with a0 = α, a1 = β. For instance, Fn = Fn(0, 1, 1, 1) and Jn = Jn(0, 1, 1, 2). The generating function of anan(α,β,r,s) is given by α (αr β)x − − . 1 rx sx2 − − 5 In this section we wish to find the generating function of the gap-sum sequence for a general Horadam sequence. We know that the gap-sum sequence for a general sequence an is given by 1 1 2 2 Sn = (an + an )(an an 1) = a a an an . 2 +1 +1 − − 2 n+1 − n − − +1 2 2 Thus the generating function of Sn is determined by those of an, an+1, an, and an+1. In the case of Horadam sequences, the sequence an+1 is again a Horadam sequence, with generating function β + αsx , 1 rx sx2 − − with initial terms β,rα + sβ.
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