

A140994


Triangle G(n, k), for 0 <= k <= n, read by rows, where G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, m) = G(n+1, m2) + G(n+1, m3) + G(n+2, m2) + G(n+3, m1) for n >= 0 and m = 3..(n+3).


22



1, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 2, 4, 8, 1, 1, 2, 4, 9, 15, 1, 1, 2, 4, 9, 19, 28, 1, 1, 2, 4, 9, 19, 40, 52, 1, 1, 2, 4, 9, 19, 41, 83, 96, 1, 1, 2, 4, 9, 19, 41, 88, 170, 177, 1, 1, 2, 4, 9, 19, 41, 88, 188, 345, 326, 1, 1, 2, 4, 9, 19, 41, 88, 189, 400, 694, 600, 1, 1, 2, 4, 9, 19, 41, 88, 189, 406, 846, 1386, 1104, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,5


COMMENTS

From Petros Hadjicostas, Jun 12 2019: (Start)
This is a mirror of image of triangular array A140997. The current array has index of asymmetry s = 2 and index of obliqueness (obliquity) e = 1. Array A140997 has the same index of asymmetry, but has index of obliqueness e = 0. (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).
If A(x,y) = Sum_{n,k >= 0} G(n, k)*x^n*y^k is the bivariate g.f. for this array (with G(n, k) = 0 for 0 <= n < k) and B(x, y) = Sum_{n, k} A140997(n, k)*x^n*y^k, then A(x, y) = B(x*y, y^(1)). This can be proved using formal manipulation of double series expansions and the fact G(n, k) = A140997(n, nk) for 0 <= k <= n.
If we let b(k) = lim_{n > infinity} G(n, k) for k >= 0, then b(0) = 1, b(1) = 2, b(2) = 4, and b(k) = b(k1) + 2*b(k2) + b(k3) for k >= 3. (The existence of the limit can be proved by induction on k.) It follows that b(k) = A141015(k) for k >= 0.
(End)


LINKS

Table of n, a(n) for n=0..90.
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) = A140997(n, nk) for 0 <= k <= n.
Bivariate g.f.: Sum_{n,k >= 0} G(n, k)*x^n*y^k = (x^4*y^3  x^3*y^3  x^2*y^2 + x^2*y  x*y + 1)/((1 x*y)*(1  x)*(1 x*y  x^2*y^2  x^3*y^3  x^3*y^2)).
(End)


EXAMPLE

Triangle begins:
1
1 1
1 2 1
1 2 4 1
1 2 4 8 1
1 2 4 9 15 1
1 2 4 9 19 28 1
1 2 4 9 19 40 52 1
1 2 4 9 19 41 83 96 1
1 2 4 9 19 41 88 170 177 1
1 2 4 9 19 41 88 188 345 326 1
1 2 4 9 19 41 88 189 400 694 600 1
1 2 4 9 19 41 88 189 406 846 1386 1104 1
... [corrected by Petros Hadjicostas, Jun 12 2019]
E.g., G(12, 9) = G(9, 7) + G(9, 6) + G(10, 7) + G(11, 8) = 170 + 88 + 188 + 400 = 846.


MAPLE

G := proc(n, k) if k=0 or n =k then 1; elif k= 1 then 2 ; elif k =2 then 4; elif k > n or k < 0 then 0 ; else procname(n3, k2)+procname(n3, k3)+procname(n2, k2)+procname(n1, k1) ; end if; end proc: seq(seq(G(n, k), k=0..n), n=0..15) ; # R. J. Mathar, Apr 14 2010


CROSSREFS

Cf. A007318, A008937, A140993, A140995, A140996, A140997, A140998, A141015, A141018, A141020, A141021, A141031, A141065, A141066, A141067.
Sequence in context: A232089 A140995 A141021 * A245163 A140993 A027935
Adjacent sequences: A140991 A140992 A140993 * A140995 A140996 A140997


KEYWORD

nonn,tabl


AUTHOR

JuriStepan Gerasimov, Jul 08 2008


EXTENSIONS

Entries checked by R. J. Mathar, Apr 14 2010


STATUS

approved



