

A102188


a(n) = Sum_{m=0..n} (1)^m * binomial(n,m)*(1*3*5*...*(4m1)).


0



1, 2, 100, 10088, 1986064, 644696864, 312335967808, 211258086400640, 190199937621590272, 219923664429290840576, 317623165714668087632896, 560356047603329076188997632, 1185822908596734257450734981120
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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 2adic 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 oneline 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*i1, 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



