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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 (list; graph; refs; listen; history; text; internal format)
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(k-1)*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 m-dimensional 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(x-1) = (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(x-1), 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/(1-t)^(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(k-1)*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) - (1-1/2+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, Springer-Verlag.

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), 1-19.

FORMULA

a(n) = n!*p(n)*sum {k = 1..n} (-1)^(k+1)/(k*p(k-1)*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+...+(n-1)^2/3)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(3+1^2/(3+2^2/(3+3^2/(3+...+(n-1)^2/(3+...))))) = sum {k = 1..inf} (-1)^(k+1)/(k*(4k^2-1)) = 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(k-1)*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 A298611

Adjacent sequences:  A142976 A142977 A142978 * A142980 A142981 A142982

KEYWORD

easy,nonn

AUTHOR

Peter Bala, Jul 17 2008

STATUS

approved

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Last modified January 21 11:21 EST 2019. Contains 319354 sequences. (Running on oeis4.)