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A142982
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a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)^2*a(n).
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3
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1, 9, 85, 846, 8974, 101916, 1240308, 16156656, 224789616, 3331795680, 52465122720, 875333381760, 15432978107520, 286828144485120, 5606317009440000, 114993185594112000, 2470155824763648000
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| This is the case m = 4 of the more general recurrence a(1) = 1, a(2) = 2*m+1, a(n+2) = (2*m+1)*a(n+1)+(n+1)^2*a(n), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.
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REFERENCES
| Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
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FORMULA
| a(n) = n!*p(n)*sum {k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = (2n^4+4n^3+10n^2+8n+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 behaviour of a(n) for large n is given by lim n -> infinity a(n)/b(n) = sum {k = 1..inf} (-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 -> infinity, where c = (12*log(2) - 7)/18.
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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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CROSSREFS
| Cf. A024167, A142979, A142980, A142981.
Sequence in context: A160112 A108427 A152106 * A196955 A029711 A204465
Adjacent sequences: A142979 A142980 A142981 * A142983 A142984 A142985
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KEYWORD
| easy,nonn
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AUTHOR
| Peter Bala (pbala(AT)toucansurf.com), Jul 17 2008
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