A188386 a(n) = numerator(H(n+2)-H(n-1)), where H(n) = Sum_{k=1..n} 1/k is the n-th harmonic number.

11, 13, 47, 37, 107, 73, 191, 121, 299, 181, 431, 253, 587, 337, 767, 433, 971, 541, 1199, 661, 1451, 793, 1727, 937, 2027, 1093, 2351, 1261, 2699, 1441, 3071, 1633, 3467, 1837, 3887, 2053, 4331, 2281, 4799, 2521, 5291, 2773, 5807, 3037, 6347, 3313, 6911, 3601

Offset: 1

Author

Gary Detlefs, Mar 29 2011

Keywords

• nonn
• easy
• look

Comments

Denominators are listed in A033931.

A027446 appears to be divisible by a(n).

The sequence lists also the largest odd divisors of 3*m^2-1 (A080663) for m>1. In fact, for m even, the largest odd divisor is 3*m^2-1 itself; for m odd, the largest odd divisor is (3*m^2-1)/2. From this follows the second formula given in Formula field. - Bruno Berselli, Aug 27 2013

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Crossrefs

Cf. A033931 (denominators), A001008, A001711, A027446, A002805, A003154, A080663, A158463, A213998.

Programs

Haskell

import Data.Ratio ((%), numerator)
a188386 n = a188386_list !! (n-1)
a188386_list = map numerator $ zipWith (-) (drop 3 hs) hs
   where hs = 0 : scanl1 (+) (map (1 %) [1..])
-- Reinhard Zumkeller, Jul 03 2012

Magma

[Numerator((3*n^2+6*n+2)/((n*(n+1)*(n+2)))): n in [1..50]]; // Vincenzo Librandi, Mar 30 2011

Maple

seq((3-(-1)^n)*(3*n^2+6*n+2)/4, n=1..100);

Mathematica

Table[(3 - (-1)^n)*(3*n^2 + 6*n + 2)/4, {n, 40}] (* Wesley Ivan Hurt, Jan 29 2017 *)
Numerator[#[[4]]-#[[1]]]&/@Partition[HarmonicNumber[Range[0,50]],4,1] (* or *) LinearRecurrence[{0,3,0,-3,0,1},{11,13,47,37,107,73},50] (* Harvey P. Dale, Dec 31 2017 *)

Formula

a(n) = numerator((3*n^2+6*n+2)/(n*(n+1)*(n+2))).

a(n) = (3-(-1)^n)*(3*n^2+6*n+2)/4.

a(2n+1) = A158463(n+1), a(2n) = A003154(n+1).

G.f.: -x*(11+13*x+14*x^2-2*x^3-x^4+x^5) / ( (x-1)^3*(1+x)^3 ). - R. J. Mathar, Apr 09 2011

a(n) = numerator of coefficient of x^3 in the Maclaurin expansion of sin(x)*exp((n+1)*x). - Francesco Daddi, Aug 04 2011

H(n+3) = 3/2 + 2*f(n)/((n+2)*(n+3)), where f(n) = Sum_{k=0..n}((-1)^k*binomial(-3,k)/(n+1-k)). - Gary Detlefs, Jul 17 2011

a(n) = A213998(n+2,2). - Reinhard Zumkeller, Jul 03 2012

Sum_{n>=1} 1/a(n) = c*(tan(c) - cot(c)/2) - 1/2, where c = Pi/(2*sqrt(3)). - Amiram Eldar, Sep 27 2022