Search: id:A154030
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%I A154030
%S A154030 0,2,6,12,16,120,30,1680,48,30240,70,665280,96,17297280,126,518918400,
%T A154030 160,17643225600,198,670442572800,240,28158588057600,286,
%U A154030 1295295050649600,336,64764752532480000,390,3497296636753920000,448
%N A154030 List of pairs of numbers:{2*(n^2 - 1), (2*n)!/n!} such that F((2*n)!/
               n!)=2*(n^2 - 1).
%C A154030 If you have two recursions ( addition and multiplication):
%C A154030 a(0)=-1;a(n)=2*(2*n-1)+a(n-1);a(n)=2*(n^2-1);
%C A154030 and
%C A154030 b(0)=1;b(n)=2*(2*n-1)*a(n-1):a(n)=(2*n)!/n!;
%C A154030 then you can form a function F such that:
%C A154030 F((2*n)!/n!)=2*(n^2 - 1).
%C A154030 This line of thought came from the row sum sequence in Pascal-like
%C A154030 triangular sequence:
%C A154030 {2^n,n!,2^n*n!,(2*n-1)!!,(2*n)!/n!,...}.
%C A154030 These field transforms are a way to get a functionally reduced
%C A154030 form of a very large number one for one.
%F A154030 {2*(n^2 - 1), (2*n)!/n!}
%t A154030 Clear[a,b,n];
%t A154030 Flatten[Table[{2*(n^2 - 1), (2*n)!/n!}, {n, 1, 20}]]
%t A154030 (*addition*)
%t A154030 a[0] = -2; a[n_] := a[n] = 2*(2*n - 1) + a[n - 1];
%t A154030 Table[a[n] - 2*(n^2 - 1), {n, 0, 20}]
%t A154030 (*multiplication*)
%t A154030 b[0] = 1; b[n_] := b[n] = 2*(2*n - 1)*b[n - 1];
%t A154030 Table[b[n] - (2*n)!/n!, {n, 0, 20}]
%Y A154030 Sequence in context: A084790 A130237 A053457 this_sequence A055560 A108727 
               A138630
%Y A154030 Adjacent sequences: A154027 A154028 A154029 this_sequence A154031 A154032 
               A154033
%K A154030 nonn,tabf
%O A154030 0,2
%A A154030 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 04 2009

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