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A051397
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a(n) = (2*n-2)*(2*n-1)*a(n-1)+1.
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11
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0, 1, 7, 141, 5923, 426457, 46910271, 7318002277, 1536780478171, 418004290062513, 142957467201379447, 60042136224579367741, 30381320929637160076947, 18228792557782296046168201, 12796612375563171824410077103, 10390849248957295521420982607637
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OFFSET
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0,3
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LINKS
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FORMULA
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a(n) = Sum_{k=0..n-1} (2*n-1)!/(2*k+1)!. a(n) = floor((2*n-1)!*sinh(1)). - Vladeta Jovovic, Aug 10 2002
Conjecture: a(n) +(-4*n^2+6*n-3)*a(n-1) +2*(2*n-3)*(n-2)*a(n-2)=0. - R. J. Mathar, Jan 31 2014
G.f. sinh(x)/(1 - x^2) = x + 7*x^3/3! + 141*x^5/5! + 5923*x^7/7! + ....
Mathar's conjectured recurrence a(n) = (4*n^2 - 6*n + 3)*a(n-1) - (2*n - 3)*(2*n - 4)*a(n-2) follows easily from the defining recurrence. The sequence b(n) := (2*n - 1)! also satisfies Mathar's recurrence but with b(1) = 1, b(2) = 6. This leads to the continued fraction representation a(n) = (2*n - 1)!*(1 + 1/(6 - 6/(21 - 20/(43 - ... - (2*n - 3)*(2*n - 4)/(4*n^2 - 6*n + 3) )))) for n >= 3. Taking the limit gives the continued fraction representation sinh(1) = A073742 = 1 + 1/(6 - 6/(21 - 20/(43 - ... - (2*n - 3)*(2*n - 4)/((4*n^2 - 6*n + 3) - ... )))). (End)
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MATHEMATICA
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nxt[{n_, a_}]:={n+1, (2(n+1)-2)(2(n+1)-1)a+1}; Transpose[NestList[nxt, {0, 0}, 20]][[2]] (* Harvey P. Dale, Jun 13 2016 *)
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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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