A102188 a(n) = Sum_{m=0..n} (-1)^m * binomial(n,m)*(1*3*5*...*(4m-1)).

1, -2, 100, -10088, 1986064, -644696864, 312335967808, -211258086400640, 190199937621590272, -219923664429290840576, 317623165714668087632896, -560356047603329076188997632, 1185822908596734257450734981120

Offset

0,2

Comments

From a posting by Henri Cohen to the Number Theory List, Feb 17 2005. He says: Show that 2^n divides f(n) (in fact the 2-adic valuation is exactly n). I do not know a proof, but it must be true.

Comment from Kevin Buzzard (k.buzzard(AT)imperial.ac.uk), Feb 17 2005: 2^k exactly divides f(k). Applying the theory of Wilf and Zeilberger to this problem gives a one-line proof that 16*(k+1)*(k+2)*f(k) - 32*(k+2)^2*f(k+1) + (16k^2 + 80k + 98)*f(k+2) + f(k+3) = 0 for all k >= 0, from which the conjecture follows easily (check for the first few terms and then easy induction on k).

Links

Table of n, a(n) for n=0..12.

Maple

g:=proc(m) local i; mul(2*i-1, i=1..2*m); end; f:=proc(k) local m; add( (-1)^m* binomial(k, m) * g(m), m=0..k); end;

Crossrefs

Sequence in context: A174646 A092699 A121975 * A225601 A332567 A126135

Adjacent sequences: A102185 A102186 A102187 * A102189 A102190 A102191

Keyword

sign

Author

N. J. A. Sloane, Feb 17 2005

Status

approved