A024200 a(0) = 1, a(1) = 0, a(n+1) = 2*a(n) + (2*n-1)^2*a(n-1).

1, 0, 1, 2, 29, 156, 2661, 24198, 498105, 6440760, 156833865, 2638782090, 74441298645, 1544798322900, 49615408298925, 1225388793991950, 44177335967379825, 1265953302961023600, 50641025474398676625, 1652074847076051263250, 72631713568603890826125, 2658069269539881753055500

Offset

0,4

Comments

a(n) = s(1)s(2)...s(n)(1/s(2) - 1/s(3) + ... + c/s(n)) where c=(-1)^n and s(k) = 2k-1 for k = 1,2,3,...

References

A. E. Jolliffe, Continued Fractions, in Encyclopaedia Britannica, 11th ed., pp. 30-33; see p. 31.

Links

Table of n, a(n) for n=0..21.

Formula

A024199(n) + a(n) = A001147(n) = (2n-1)!! - Max Alekseyev, Sep 23 2007

A024199(n)/a(n) -> Pi/(4-Pi) as n -> oo. - Max Alekseyev, Sep 23 2007

E.g.f.: (1-Pi/4)/sqrt(1-2*x) + 1/2*log(2*x+sqrt(4*x^2-1))/sqrt(2*x-1). - Vaclav Kotesovec, Mar 18 2014

a(n) ~ (4-Pi) * 2^(n-3/2) * n^n / exp(n). - Vaclav Kotesovec, Mar 18 2014

Mathematica

CoefficientList[Series[(1-Pi/4)/Sqrt[1-2*x] + 1/2*Log[2*x+Sqrt[4*x^2-1]]/Sqrt[2*x-1], {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Mar 18 2014 *)
RecurrenceTable[{a[n+1] == 2*a[n] + (2*n-1)^2*a[n-1], a[0] == 1, a[1] == 0}, a, {n, 0, 20}] (* Vaclav Kotesovec, Mar 18 2014 *)

Crossrefs

Sequence in context: A128842 A308899 A028883 * A137625 A132412 A369384

Adjacent sequences: A024197 A024198 A024199 * A024201 A024202 A024203

Keyword

nonn

Author

Clark Kimberling

Extensions

Revised by N. J. A. Sloane, Jul 19 2002

Initial terms changed by Max Alekseyev, Sep 23 2007

Status

approved