A051396 a(n) = (2*n-2)*(2*n-3)*a(n-1)+1.

0, 1, 3, 37, 1111, 62217, 5599531, 739138093, 134523132927, 32285551902481, 9879378882159187, 3754163975220491061, 1734423756551866870183, 957401913616630512341017, 622311243850809833021661051, 470467300351212233764375754557, 409306551305554643375006906464591

Offset: 0

Author

Aleksandar Petojevic

Keywords

• nonn
• easy

Comments

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

Links

• Seiichi Manyama, 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

Bisection of abs(A009179(n)). Cf. A049470 (cos(1)), A073743 (cosh(1)), A275651.

Programs

Maple

A051396 := proc(n) option remember; if n <= 1 then n else (2*n-2)*(2*n-3)*A051396(n-1)+1; fi; end;

Mathematica

a[0] = 0; a[n_] := a[n] = (2*n-2)*(2*n-3)*a[n-1] + 1;
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Dec 11 2017 *)
nxt[{n_,a_}]:={n+1,a(4n^2-2n)+1}; NestList[nxt,{0,0},20][[;;,2]] (* Harvey P. Dale, Sep 26 2023 *)

Formula

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

From Peter Bala, Sep 05 2016: (Start)

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)