A142982 a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1) + (n + 1)^2*a(n).
1, 9, 85, 846, 8974, 101916, 1240308, 16156656, 224789616, 3331795680, 52465122720, 875333381760, 15432978107520, 286828144485120, 5606317009440000, 114993185594112000, 2470155824763648000, 55464433059571200000, 1299510384759562752000, 31718253797341267968000
Offset: 1
References
- Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..446
Programs
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Maple
p := n -> (2*n^4+4*n^3+10*n^2+8*n+3)/3: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20);
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Mathematica
Module[{a, n}, RecurrenceTable[{a[n+2] == 9*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 9}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)
Formula
a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = (2*n^4 + 4*n^3 + 10*n^2 + 8*n + 3)/3 = A001846(n) is the Ehrhart polynomial for the 4-dimensional cross polytope (the 16-cell).
Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1) + (n + 1)^2*a(n).
The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 9, b(2) = 82.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + (n-1)^2/9)))), for n >= 2.
The behavior of a(n) for large n is given by limit_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p(k-1)*p(k)) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + n^2/(9 + ...))))) = log(2) - (1 - 1/2 + 1/3 - 1/4); the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
Thus a(n) ~ c*n^4*n! as n -> oo, where c = (12*log(2) - 7)/18.
E.g.f.: A(x) = (3*(x + 1)^4*log(1 + x) - 4*x^2*(2*x^2 + 2*x + 3))/(3*(1 - x)^5) satisfies the differential equation 1 + (x + 9)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0. - Peter Bala, Dec 09 2024
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