A144301 a(0) = a(1) = 1; thereafter a(n) = (2*n-3)*a(n-1) + a(n-2).
1, 1, 2, 7, 37, 266, 2431, 27007, 353522, 5329837, 90960751, 1733584106, 36496226977, 841146804577, 21065166341402, 569600638022431, 16539483668991901, 513293594376771362, 16955228098102446847, 593946277027962411007, 21992967478132711654106, 858319677924203716921141
Offset: 0
Examples
G.f. = 1 + x + 2*x^2 + 7*x^3 + 37*x^4 + 266*x^5 + 2431*x^6 + 27007*x^7 + ...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..400
- E. Grosswald, Bessel Polynomials, Lecture Notes Math., Vol. 698, 1978.
- Toufik Mansour, Matthias Schork and Mark Shattuck, On the Stirling numbers associated with the meromorphic Weyl algebra, Applied Mathematics Letters, Volume 25, Issue 11, November 2012, Pages 1767-1771. - From _N. J. A. Sloane_, Sep 15 2012
- W. Mlotkowski and A. Romanowicz, A family of sequences of binomial type, Probability and Mathematical Statistics, Vol. 33, Fasc. 2 (2013), pp. 401-408.
Programs
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Magma
[1] cat [n le 1 select n+1 else (2*n-1)*Self(n) + Self(n-1): n in [0..20]]; // Vincenzo Librandi, Jul 23 2015
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Mathematica
a[n_]:= HypergeometricPFQ[{n, 1 - n}, {}, -1/2]; (* Michael Somos, Nov 22 2013 *) a[n_]:= With[{m= If[n<1, -n, n-1]}, Sum[(m+k)!/((m-k)! k! 2^k), {k,0,m}]]; (* Michael Somos, Nov 22 2013 *) RecurrenceTable[{a[0]==a[1]==1, a[n]==(2*n-3)*a[n-1] +a[n-2]}, a, {n, 25}] (* Vincenzo Librandi, Jul 23 2015 *) nxt[{n_,a_,b_}]:={n+1,b,b(2n-1)+a}; NestList[nxt,{1,1,1},30][[All,2]] (* Harvey P. Dale, Nov 29 2022 *)
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PARI
{a(n) = my(m = if( n<1, -n, n-1)); sum( k=0, m, (m+k)! / (k! * (m-k)! * 2^k))}; /* Michael Somos, Nov 22 2013 */
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SageMath
def A144301(n): return int(n==0) + sum(binomial(n-1,k)*factorial(n+k-1)/(2^k*factorial(n-1)) for k in range(n)) [A144301(n) for n in range(31)] # G. C. Greubel, Sep 29 2023
Formula
a(n) = A001515(n-1) for n>= 1.
E.g.f.: A(x) = exp(1-sqrt(1-2*x)) satisfies A'(x) = A(x)/(1-sqrt(1-2*x)).
Hence a(n+1) = Sum_{k=0..n} ( a(n-k)*binomial(n,k)*(2*k)!/(k!*2^k) ).
A''(x) = (A'(x)/(1-2*x))*(1 + 1/sqrt(1-2*x)).
A''(x) = 2*x*A''(x) + A'(x) + A(x), which is equivalent to the recurrence in the definition.
a(n) = Sum_{k=0..n-1} binomial(n+k-1,2*k)*(2*k)!/(k!*2^k). [See Grosswald, p. 6, Eq. (8).]
a(n) ~ exp(1)*(2n-1)!/(n!*2^n) as n -> oo. [See Grosswald, p. 124]
From Sergei N. Gladkovskii, Oct 06 2012: (Start)
G.f.: 1+x/U(0) where U(k) = 1 - x - x*(2*k+1)/(1 - x - 2*x*(k+1)/U(k+1)); (continued fraction).
G.f.: 1+x*(1-x)/U(0) where U(k) = 1 - 3*x + x^2 - x*4*k - x^2*(2*k+1)*(2*k+2)/U(k+1) ; (continued fraction). (End)
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)*(1+sqrt(1-2*x))/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 23 2013
G.f.: conjecture: 1 + x*(1-x)/(1-3*x+x^2)*Q(0), where Q(k) = 1 - 2*(k+1)*(2*k+1)*x^2/(2*(k+1)*(2*k+1)*x^2 - (1 - 3*x + x^2 - 4*x*k)*(1 - 7*x + x^2 - 4*x*k)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 19 2013
a(1 - n) = a(n) for all n in Z. (a(n+1) + a(n+2))^2 = a(n)*a(n+2) + a(n+1)*a(n+3) for all integer n. - Michael Somos, Nov 22 2013
G.f.: 1 + x/(1-x)*T(0), where T(k) = 1 - x*(k+1)/( x*(k+1) - (1-x)^2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 26 2013
Extensions
More terms from Vincenzo Librandi, Jul 23 2015
Comments