A189998 Numerator of h(n+5) - h(n) where h(n) = Sum_{k=1..n} (1/k) are the Harmonic numbers.

137, 29, 153, 743, 1879, 1627, 15797, 2021, 11899, 25381, 7793, 2627, 124877, 26987, 68879, 65003, 107699, 66167, 482897, 16167, 77293, 412561, 323959, 94781, 1323137, 255127, 587299, 504563, 255733, 145209, 2956637, 277681, 1247459, 2094661, 1558379, 433501

Offset: 0

Author

Gary Detlefs, May 03 2011

Keywords

• nonn

Comments

a(n) = Numerator of (5*n^4 + 60*n^3 + 255*n^2 + 450*n + 274)/((n+1)*(n+2)*(n+3)*(n+4)*(n+5)).

(5*n^4 + 60*n^3 + 255*n^2 + 450*n + 274)/a(n) can be factored into 2^p(n)* 3^q(n) where p(n) is a sequence of period 4 repeating [1,2,1,3] and q(n) is of period 9,repeating [0,2,2,0,1,1,0,1,1].

p(n) = A131743(n) + 1.

q(n) = A011655(n) + [0,2,2,0,0,0,0,0,0]

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A011655, A131743.

Programs

Magma

[137] cat [Numerator(HarmonicNumber(n+5)-HarmonicNumber(n)): n in [0..30]]; // G. C. Greubel, Jan 11 2018

Maple

h:= n->sum(1/k,k=1..n):seq(numer(h(n+5)-h(n)), n=0..28);
q:=n-> (1-(-1)^n)*(3+I^(n+1))/4+1:
P:=(k,n)-> floor(1/2*cos(2*n*Pi/k)+1/2):
seq( (5*n^4+60*n^3+255*n^2+450*n+274)/(2^q(n)*3^(P(9,n-1)+P(9,n-2)+1-P(3,n))),n=0..28)

Mathematica

Numerator[Table[HarmonicNumber[n+5]-HarmonicNumber[n],{n,0,30}]] (* Harvey P. Dale, Sep 15 2016 *)

Python

from sympy import harmonic,numer
print([numer(harmonic(n+5) - harmonic(n)) for n in range(0, 30)])
# Javier Rivera Romeu, May 22 2023

Formula

a(n) = (5*n^4 + 60*n^3 + 255*n^2 + 450*n + 274)/(2^q(n)*3^(P(9,n-1) + P(9,n-2) + 1 - P(3,n))), where q(n) = (1-(-1)^n)*(3+i^(n+1))/4 + 1 and P(k,n) = floor(1/2*cos(2*n*Pi/k)+1/2).