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%I #11 Nov 11 2019 21:36:15
%S 1,-2,100,-10088,1986064,-644696864,312335967808,-211258086400640,
%T 190199937621590272,-219923664429290840576,317623165714668087632896,
%U -560356047603329076188997632,1185822908596734257450734981120
%N a(n) = Sum_{m=0..n} (-1)^m * binomial(n,m)*(1*3*5*...*(4m-1)).
%C 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.
%C 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).
%p 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;
%K sign
%O 0,2
%A _N. J. A. Sloane_, Feb 17 2005
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