A051397 a(n) = (2*n-2)*(2*n-1)*a(n-1)+1.
0, 1, 7, 141, 5923, 426457, 46910271, 7318002277, 1536780478171, 418004290062513, 142957467201379447, 60042136224579367741, 30381320929637160076947, 18228792557782296046168201, 12796612375563171824410077103, 10390849248957295521420982607637
Offset: 0
Links
- Harvey P. Dale, Table of n, a(n) for n = 0..225
- Romeo Mestrovic, Variations of Kurepa's left factorial hypothesis, arXiv preprint arXiv:1312.7037 [math.NT], 2013.
- Romeo Mestrovic, The Kurepa-Vandermonde matrices arising from Kurepa's left factorial hypothesis, Filomat 29:10 (2015), 2207-2215; DOI 10.2298/FIL1510207M.
- Aleksandar Petojevic, On Kurepa's Hypothesis for the Left Factorial, FILOMAT (Nis), 12:1 (1998), p. 29-37.
Crossrefs
Programs
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Mathematica
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 *)
Formula
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
From Peter Bala, Sep 02 2016: (Start)
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)