A319204 Sequence used for the Boas-Buck type recurrence for Riordan triangle A319203.

0, -2, -3, 6, 20, -5, -105, -98, 420, 1008, -990, -6501, -2574, 31603, 52052, -107250, -411944, 81328, 2343042, 2413456, -9883800, -25327722, 23371634, 168185131, 77113020, -835281800, -1452148815, 2847865635, 11561517870, -1613666430, -66318892875, -72637680690, 280330495200, 750725215020

Offset

0,2

Comments

See A319203 for the Boas-Buck type recurrence.

Links

Table of n, a(n) for n=0..33.

Formula

O.g.f.: (log(f(x))' = (1/(1/f(x) + x^2*f(x) + 2*x^3*f(x)^2) - 1)/x, with the expansion of f given in A319201. f(x) = F^{[-1]}(x)/x, where F(t) = t/(1 - t^2 - t^3).

a(n) = (1/(n+1)!)*[d^(n+1)/dx^(n+1) (1 - x^2 - x^3)^(n+1)] evaluated at x = 0, for n >= 0. (Cf. Joerg Arndt's conjecture for A176806, which is proved there.)

a(n-1) = Sum_{2*e + 3*e3 = n} (-1)^(e2+e3)*n!/((n - (e2+e3))!*e2!*e3!), n >= 2, with a(0) = 0. The pairs (e2, e3) are given in A321201; see also the multinomial coefficient table A321203 and add the sign factors.

Example

a(5) =  (1/6!)*[d^6/dx^6 (1 - x^2 - x^3)^6] for x  = 0, which is -5.
a(5) =  +15 - 20  = -5; from the sum of the signed row n=6 in A321203, with parity of e2 + e3 from A321201 even and odd.

Crossrefs

Cf. A319203, A321201, A321203.

Sequence in context: A141048 A326030 A124066 * A093447 A321203 A259456

Adjacent sequences: A319201 A319202 A319203 * A319205 A319206 A319207

Keyword

sign

Author

Wolfdieter Lang, Oct 29 2018

Status

approved