A319204 - OEIS

%I #13 May 31 2022 12:52:21

%S 0,-2,-3,6,20,-5,-105,-98,420,1008,-990,-6501,-2574,31603,52052,

%T -107250,-411944,81328,2343042,2413456,-9883800,-25327722,23371634,

%U 168185131,77113020,-835281800,-1452148815,2847865635,11561517870,-1613666430,-66318892875,-72637680690,280330495200,750725215020

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

%C See A319203 for the Boas-Buck type recurrence.

%F 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).

%F 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.)

%F 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.

%e a(5) =  (1/6!)*[d^6/dx^6 (1 - x^2 - x^3)^6] for x  = 0, which is -5.

%e 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.

%Y Cf. A319203, A321201, A321203.

%K sign

%O 0,2

%A _Wolfdieter Lang_, Oct 29 2018