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