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).
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
KEYWORD
sign
AUTHOR
N. J. A. Sloane, Feb 17 2005
STATUS
approved