|
%I #62 Jan 28 2024 14:31:20
%S 1,1,1,1,2,1,1,4,2,1,1,8,4,2,1,1,15,9,4,2,1,1,28,19,9,4,2,1,1,52,40,
%T 19,9,4,2,1,1,96,83,41,19,9,4,2,1,1,177,170,88,41,19,9,4,2,1,1,326,
%U 345,188,88,41,19,9,4,2,1,1,600,694,400,189,88,41,19,9,4,2,1,1,1104,1386,846,406,189,88,41,19,9,4,2,1,1,2031,2751,1779,871,406,189,88,41,19,9,4,2,1,1,3736,5431,3719,1866,872,406,189,88,41,19,9,4,2,1
%N Triangle G(n,k) read by rows, for 0 <= k <= n, where G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, and G(n+4, m) = G(n+1, m-1) + G(n+1, m) + G(n+2, m) + G(n+3, m) for n >= 0 and m = 1..n+1.
%C From _Petros Hadjicostas_, Jun 12 2019: (Start)
%C This is a mirror image of the triangular array A140994. The current array has index of asymmetry s = 2 and index of obliqueness (obliquity) e = 0. Array A140994 has the same index of asymmetry, but has index of obliqueness e = 1. (In other related sequences, the author uses the letter y for the index of asymmetry and the letter z for the index of obliqueness, but on the stone slab that appears over a tomb in a picture that he posted in those sequences, the letters s and e are used instead. See, for example, the documentation for sequences A140998, A141065, A141066, and A141067.)
%C In general, if the index of asymmetry (from the Pascal triangle A007318) is s, then the order of the recurrence is s + 2 (because the recurrence of the Pascal triangle has order 2). There are also s + 2 infinite sets of initial conditions (as opposed to the Pascal triangle, which has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0).
%C Pascal's triangle A007318 has s = 0 and is symmetric, arrays A140998 and A140993 have s = 1 (with e = 0 and e = 1, respectively), and arrays A140996 and A140995 have s = 3 (with e = 0 and e = 1, respectively).
%C (End)
%H Robert Price, <a href="/A140997/b140997.txt">Table of n, a(n) for n = 0..1325</a>
%H Juri-Stepan Gerasimov, <a href="/A140998/a140998.jpg">Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...</a>
%F From _Petros Hadjicostas_, Jun 12 2019: (Start)
%F G(n, k) = A140994(n, n-k) for 0 <= k <= n.
%F Bivariate g.f.: Sum_{n,k >= 0} G(n,k)*x^n*y^k = (1 - x - x^2 - x^3 + x^2*y + x^4*y)/((1 - x) * (1 - x*y) * (1 - x - x^2 - x^3 - x^3*y)).
%F Differentiating once w.r.t. y and setting y = 0, we get the g.f. of column k = 1: x/((1 - x) * (1 - x - x^2 - x^3)). This is the g.f. of sequence A008937.
%F (End)
%e Triangle begins:
%e 1
%e 1 1
%e 1 2 1
%e 1 4 2 1
%e 1 8 4 2 1
%e 1 15 9 4 2 1
%e 1 28 19 9 4 2 1
%e 1 52 40 19 9 4 2 1
%e 1 96 83 41 19 9 4 2 1
%e 1 177 170 88 41 19 9 4 2 1
%e 1 326 345 188 88 41 19 9 4 2 1
%e 1 600 694 400 189 88 41 19 9 4 2 1
%e ...
%e E.g., G(14, 2) = G(11, 1) + G(11, 2) + G(12, 2) + G(13, 2) = 600 + 694 + 1386 + 2751 = 5431.
%t nlim = 50;
%t Do[G[n, 0] = 1, {n, 0, nlim}];
%t Do[G[n + 1, n + 1] = 1, {n, 0, nlim}];
%t Do[G[n + 2, n + 1] = 2, {n, 0, nlim}];
%t Do[G[n + 3, n + 1] = 4, {n, 0, nlim}];
%t Do[G[n + 4, m] =
%t G[n + 1, m - 1] + G[n + 1, m] + G[n + 2, m] + G[n + 3, m], {n, 0,
%t nlim}, {m, 1, n + 1}];
%t A140997 = {}; For[n = 0, n <= nlim, n++,
%t For[k = 0, k <= n, k++, AppendTo[A140997, G[n, k]]]];
%t A140997 (* _Robert Price_, Aug 25 2019 *)
%Y Cf. A007318, A008937, A140993, A140994, A140995, A140996, A140998, A141015, A141018, A141020, A141021, A141065, A141066, A141067.
%K nonn,tabl
%O 0,5
%A _Juri-Stepan Gerasimov_, Jul 08 2008
%E Typo in definition corrected by _R. J. Mathar_, Sep 19 2008
%E Name edited by and more terms from _Petros Hadjicostas_, Jun 12 2019
%E Deleted extraneous term at a(29) by _Robert Price_, Aug 25 2019
%E Added 13 missing terms at a(79) by _Robert Price_, Aug 25 2019
|
