

A140997


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, m1) + G(n+1, m) + G(n+2, m) + G(n+3, m) for n >= 0 and m = 1..n+1.


24



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, 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, 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
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OFFSET

0,5


COMMENTS

From Petros Hadjicostas, Jun 12 2019: (Start)
This is a mirror of image of 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 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.)
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 that has only 2 infinite sets of initial conditions, namely, G(n, 0) = G(n+1, n+1) = 1 for n >= 0).
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).
(End)


LINKS

Robert Price, Table of n, a(n) for n = 0..1325
JuriStepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...


FORMULA

From Petros Hadjicostas, Jun 12 2019: (Start)
G(n, k) = A140994(n, nk) for 0 <= k <= n.
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)).
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.
(End)


EXAMPLE

Triangle begins:
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 19 9 4 2 1
1 96 83 41 19 9 4 2 1
1 177 170 88 41 19 9 4 2 1
1 326 345 188 88 41 19 9 4 2 1
1 600 694 400 189 88 41 19 9 4 2 1
...
E.g., G(14, 2) = G(11, 1) + G(11, 2) + G(12, 2) + G(13, 2) = 600 + 694 + 1386 + 2751 = 5431.


MATHEMATICA

nlim = 50;
Do[G[n, 0] = 1, {n, 0, nlim}];
Do[G[n + 1, n + 1] = 1, {n, 0, nlim}];
Do[G[n + 2, n + 1] = 2, {n, 0, nlim}];
Do[G[n + 3, n + 1] = 4, {n, 0, nlim}];
Do[G[n + 4, m] =
G[n + 1, m  1] + G[n + 1, m] + G[n + 2, m] + G[n + 3, m], {n, 0,
nlim}, {m, 1, n + 1}];
A140997 = {}; For[n = 0, n <= nlim, n++,
For[k = 0, k <= n, k++, AppendTo[A140997, G[n, k]]]];
A140997 (* Robert Price, Aug 25 2019 *)


CROSSREFS

Cf. A007318, A008937, A140993, A140994, A140995, A140996, A140998, A141015, A141018, A141020, A141021, A141065, A141066, A141067.
Sequence in context: A048004 A114394 A059623 * A140996 A141020 A152568
Adjacent sequences: A140994 A140995 A140996 * A140998 A140999 A141000


KEYWORD

nonn,tabl


AUTHOR

JuriStepan Gerasimov, Jul 08 2008


EXTENSIONS

Typo in definition corrected by R. J. Mathar, Sep 19 2008
Name edited by and more terms from Petros Hadjicostas, Jun 12 2019
Deleted extraneous term at a(29) by Robert Price, Aug 25 2019
Added 13 missing terms at a(79) by Robert Price, Aug 25 2019


STATUS

approved



