

A142979


a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1)+(n+1)^2*a(n).


15



1, 3, 13, 66, 406, 2868, 23220, 210192, 2116656, 23375520, 281792160, 3673814400, 51599514240, 775673176320, 12440524320000, 211848037632000, 3820318338816000, 72685037892096000, 1455838255452672000
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OFFSET

1,2


COMMENTS

This is the case m = 1 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) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of Mercator's series for the constant log(2). For other cases see A024167 (m=0), A142980 (m=2), A142981 (m=3) and A142982 (m=4). The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n)*sum {k = 1..n} (1)^(k+1)/(k*p_m(k1)*p_m(k)), where p_m(x) := sum {k = 0..m} 2^k*C(m,k)*C(x,k) = sum {k = 0..m} C(m,k)*C(x+k,m), is the Ehrhart polynomial of the mdimensional cross polytope (the hyperoctahedron). The first few values are p_0(x) = 1, p_1(x) = 2*x+1, p_2(x) = 2*x^2+2*x+1 and p_3(x) = (4*x^3+6*x^2+8*x+3)/3.
The sequence {p_m(k)},k=0,1,2,3,... is the crystal ball sequence for the product lattice A_1 x... x A_1 (m copies). The table of values [p_m(k)]m,k>=0 is the array of Delannoy numbers A008288.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)*f(x+1)  x*f(x1) = (2*m+1)*f(x), normalized so that f(0) = 1. These polynomials have their zeros on the vertical line Re x = 1/2 in the complex plane, that is, the polynomials p_m(x1), m = 1,2,3,..., satisfy a Riemann hypothesis [BUMP et al., Theorems 4 and 6]. The o.g.f. for the p_m(x) is (1+t)^x/(1t)^(x+1) = 1 + (2*x+1)*t + (2*x^2+2*x+1)*t^2 + ... .
The general recurrence in the first paragraph above also has a second solution b(n) = n!*p_m(n) with initial conditions b(1) = 2*m+1, b(2) = (2*m+1)^2+1. Hence the behavior 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_m(k1)*p_m(k)) = 1/((2*m+1)+ 1^2/((2*m+1)+ 2^2/((2m+1)+ 3^2/((2*m+1)+...+ n^2/((2*m+1)+...))))) = (1)^m * (log(2)  (11/2+1/3 ...+ (1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
For other sequences defined by similar recurrences and related to log(2) see A142983 and A142988. See also A142992 for the connection between log(2) and the C_n lattices. For corresponding results for the constants e, zeta(2) and zeta(3) see A000522, A142995 and A143003 respectively.


REFERENCES

Bruce C. Berndt, Ramanujan's Notebooks Part II, SpringerVerlag.


LINKS

Table of n, a(n) for n=1..19.
D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 119.


FORMULA

a(n) = n!*p(n)*sum {k = 1..n} (1)^(k+1)/(k*p(k1)*p(k)), where p(n) = 2*n+1. Recurrence: a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1)+(n+1)^2*a(n). The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 3, b(2) = 10. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3+1^2/(3+2^2/(3+3^2/(3+...+(n1)^2/3)))), for n >=2. Lim n > infinity a(n)/b(n) = 1/(3+1^2/(3+2^2/(3+3^2/(3+...+(n1)^2/(3+...))))) = sum {k = 1..inf} (1)^(k+1)/(k*(4k^21)) = 1  log(2). Thus a(n) ~ c*n*n! as n > infinity, where c = 2*(1  log(2)).


MAPLE

p := n > 2*n+1: a := n > n!*p(n)*sum ((1)^(k+1)/(k*p(k1)*p(k)), k = 1..n): seq(a(n), n = 1..20)


MATHEMATICA

RecurrenceTable[{a[1]==1, a[2]==3, a[n+2]==3a[n+1]+(n+1)^2 a[n]}, a, {n, 20}] Harvey P. Dale, May 20 2012


CROSSREFS

Cf. A008288, A024167, A142980, A142981, A142982, A142983, A142988, A142992.
Sequence in context: A219537 A045743 A110530 * A302303 A201713 A333083
Adjacent sequences: A142976 A142977 A142978 * A142980 A142981 A142982


KEYWORD

easy,nonn


AUTHOR

Peter Bala, Jul 17 2008


STATUS

approved



