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A051396
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a(n) = (2*n-2)*(2*n-3)*a(n-1)+1.
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12
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0, 1, 3, 37, 1111, 62217, 5599531, 739138093, 134523132927, 32285551902481, 9879378882159187, 3754163975220491061, 1734423756551866870183, 957401913616630512341017, 622311243850809833021661051, 470467300351212233764375754557, 409306551305554643375006906464591
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OFFSET
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0,3
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COMMENTS
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The sequence 1,0,3,0,37,... has e.g.f. cosh(x)/(1-x^2) with a(n) = Sum_{k=0..n} C(n,k)k!(1+(-1)^k)(1+(-1)^(n-k))/4. - Paul Barry, May 01 2005
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LINKS
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FORMULA
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a(n) = Sum_{k=0..n-1} (2*n-2)!/(2*k)! = floor((2*n-2)!*cosh(1)), n>=1. - Vladeta Jovovic, Aug 10 2002
a(n+1) = Sum_{k=0..2n}, C(2n, k)*k!*(1+(-1)^k)^2. - Paul Barry, May 01 2005
a(n) +(-4*n^2+10*n-7)*a(n-1) +2*(n-2)*(2*n-5)*a(n-2)=0. - R. J. Mathar, Nov 26 2012
The sequence b(n) := (2*n - 2)! also satisfies Mathar's recurrence with b(1) = 1, b(2) = 2. This leads to the continued fraction representation a(n) = (2*n - 2)!*(1 + 1/(2 - 2/(13 - 12/(31 - ... - (2*n - 4)*(2*n - 5)/(4*n^2 - 10*n + 7) )))) for n >= 3. Taking the limit gives the continued fraction representation cosh(1) = A073743 = 1 + 1/(2 - 2/(13 - 12/(31 - ... - (2*n - 4)*(2*n - 5)/((4*n^2 - 10*n + 7) - ... )))). (End)
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MAPLE
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A051396 := proc(n) option remember; if n <= 1 then n else (2*n-2)*(2*n-3)*A051396(n-1)+1; fi; end;
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MATHEMATICA
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a[0] = 0; a[n_] := a[n] = (2*n-2)*(2*n-3)*a[n-1] + 1;
nxt[{n_, a_}]:={n+1, a(4n^2-2n)+1}; NestList[nxt, {0, 0}, 20][[;; , 2]] (* Harvey P. Dale, Sep 26 2023 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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