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a(0) = 1, a(1) = 0, a(n+1) = 2*a(n) + (2*n-1)^2*a(n-1).
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%I #19 Dec 21 2024 15:31:33

%S 1,0,1,2,29,156,2661,24198,498105,6440760,156833865,2638782090,

%T 74441298645,1544798322900,49615408298925,1225388793991950,

%U 44177335967379825,1265953302961023600,50641025474398676625,1652074847076051263250,72631713568603890826125,2658069269539881753055500

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

%C 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,...

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

%H Harvey P. Dale, <a href="/A024200/b024200.txt">Table of n, a(n) for n = 0..404</a>

%F A024199(n) + a(n) = A001147(n) = (2n-1)!! - _Max Alekseyev_, Sep 23 2007

%F A024199(n)/a(n) -> Pi/(4-Pi) as n -> oo. - _Max Alekseyev_, Sep 23 2007

%F 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

%F a(n) ~ (4-Pi) * 2^(n-3/2) * n^n / exp(n). - _Vaclav Kotesovec_, Mar 18 2014

%t 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 *)

%t 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 *)

%t nxt[{n_,a_,b_}]:={n+1,b,2b+a (2n-1)^2}; NestList[nxt,{1,1,0},30][[;;,2]] (* _Harvey P. Dale_, Dec 21 2024 *)

%K nonn

%O 0,4

%A _Clark Kimberling_

%E Revised by _N. J. A. Sloane_, Jul 19 2002

%E Initial terms changed by _Max Alekseyev_, Sep 23 2007