cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A141072 Sum of diagonal numbers in a Pascal-like triangle with index of asymmetry y = 3 and index of obliquity z = 0 (going upwards, left to right).

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%I A141072 #45 Jul 26 2019 18:48:05
%S A141072 1,1,2,3,6,11,22,42,83,162,319,626,1231,2419,4756,9349,18380,36133,
%T A141072 71036,139652,274549,539748,1061117,2086100,4101165,8062677,15850806,
%U A141072 31161863,61262610,120439119,236777074,465491470,915132135,1799102406,3536942203,6953445286
%N A141072 Sum of diagonal numbers in a Pascal-like triangle with index of asymmetry y = 3 and index of obliquity z = 0 (going upwards, left to right).
%C A141072 For the Pascal-like triangle G(n, k) with index of asymmetry y = 3 and index of obliqueness z = 0, which is read by rows, we have G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) + G(n+4, k) for n >= 0 and k = 1..(n+1). (This is array A140996.)
%C A141072 For the Pascal-like triangle G(n, k) with index of asymmetry y = 3 and index of obliqueness z = 1, which is read by rows, we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, k) = G(n+1, k-3) + G(n+1, k-4)  + G(n+2, k-3) + G(n+3, k-2) + G(n+4, k-1) for n >= 0 and for k = 4..(n+4). (This is array A140995.)
%C A141072 From _Petros Hadjicostas_, Jun 13 2019: (Start)
%C A141072 In the example below the author uses array A140996 to create this sequence. If we use array A140995, which is the mirror image of A140996, and follow the same process, we get a different sequence: 1, 1, 2, 3, 4, 7, 8, 15, 16, 31, 33, 63, 68, 127, 140, 255, 288, 512, 592, ...
%C A141072 Even though array A140996 starts at row n = 0, the offset of the current sequence was set at n = 1. In other words, a(n) = Sum_{ceiling((n-1)/2) <= i <= n-1} G(i, n-1-i) = G(n-1, 0) + G(n-2, 1) + ... + G(ceiling((n-1)/2), floor((n-1)/2)) for n >= 1, where G(n, k) = A140996(n, k).
%C A141072 To get the g.f. of this sequence, we take the bivariate g.f. of sequence A140996, and set x = y. We multiply the result by x because the offset here was set at n = 1.
%C A141072 Finally, we mention that in the attached photograph about Stepan's triangle, the index of asymmetry is denoted by s (rather than y) and the index of obliqueness is denoted by e (rather than z). For the Pascal triangle, s = y = 0.
%C A141072 (End)
%H A141072 Juri-Stepan Gerasimov, <a href="/A140998/a140998.jpg">Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...</a>
%F A141072 From _Petros Hadjicostas_, Jun 13 2019: (Start)
%F A141072 a(n) = Sum_{ceiling((n-1)/2) <= i <= n-1} G(i, n-1-i) for n >= 1, where G(n, k) = A140996(n, k) for 0 <= k <= n.
%F A141072 G.f.: x*(1 - x^2 - x^3 - x^4 - x^5)/((1 - x)*(1 + x)*(1 - x - x^2 - x^3 - x^4 - x^5)) = x*(1 - x^2 - x^3 - x^4 - x^5)/(1 - x - 2*x^2 + x^6 + x^7).
%F A141072 Recurrence: a(n) = -(3 + (-1)^n)/2 + a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) for n >= 6 with a(1) = a(2) =  1, a(3) = 2, a(4) = 3, and a(5) = 6.
%F A141072 (End)
%e A141072 Pascal-like triangle with y = 3 and z = 0 (i.e, A140996) begins as follows:
%e A141072   1, so a(1) = 1.
%e A141072   1   1, so a(2) = 1.
%e A141072   1   2   1, so a(3) = 1 + 1 = 2.
%e A141072   1   4   2   1, so a(4) = 1 + 2 = 3.
%e A141072   1   8   4   2  1, so a(5) = 1 + 4 + 1 = 6.
%e A141072   1  16   8   4  2  1, so a(6) = 1 + 8 + 2 = 11.
%e A141072   1  31  17   8  4  2  1, so a(7) = 1 + 16 + 4 + 1 = 22.
%e A141072   1  60  35  17  8  4  2 1, so a(8) = 1 + 31 + 8 + 2 = 42.
%e A141072   1 116  72  35 17  8  4 2 1, so a(9) = 1 + 60 + 17 + 4 + 1 = 83.
%e A141072   1 224 148  72 35 17  8 4 2 1, so a(10) = 1 +  116 + 35 + 8 + 2 = 162.
%e A141072   1 432 303 149 72 35 17 8 4 2 1, so a(11) = 1 + 224 + 72 + 17 + 4 + 1 = 319.
%e A141072 ... [edited by _Petros Hadjicostas_, Jun 13 2019]
%Y A141072 Cf. A140993, A140994, A140995, A140996, A140997, A140998, A141065, A141066, A141067, A141068, A141069, A141073.
%K A141072 nonn
%O A141072 1,3
%A A141072 _Juri-Stepan Gerasimov_, Jul 16 2008
%E A141072 Partially edited by _N. J. A. Sloane_, Jul 18 2008
%E A141072 Name edited by and more terms from _Petros Hadjicostas_, Jun 13 2019