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A102188
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Sum_{0 <= m <= n} (-1)^m binom(n,m)(1.3.5...(4m-1)).
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0
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1, -2, 100, -10088, 1986064, -644696864, 312335967808, -211258086400640, 190199937621590272, -219923664429290840576, 317623165714668087632896, -560356047603329076188997632, 1185822908596734257450734981120
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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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)^2f(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).
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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;
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CROSSREFS
| Sequence in context: A174646 A092699 A121975 * A126135 A016034 A171988
Adjacent sequences: A102185 A102186 A102187 * A102189 A102190 A102191
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KEYWORD
| sign
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), Feb 17 2005
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