A250328 - OEIS

%I #14 Nov 06 2017 04:58:19

%S 1,3,77,877,6271,36049,36423,422137,49691099,1448086909,11631128477,

%T 2334008785,44471893747,1827784004699,832564679309,39202882860913,

%U 196334425398149,3473612060358899,3478128507653999,205449856947685261,303604578504856471

%N Denominator of the harmonic mean of the first n pentagonal numbers.

%C a(n+1) is, for n >= 0, also the numerator of the partial sums of the reciprocals of twice the pentagonal numbers {A049450(k+1)}_{k>=0} with the denominators given in A294513(n) (assuming that A250327(n+1)/(n+1) = A294513(n)/2). - _Wolfdieter Lang_, Nov 02 2017

%H Colin Barker, <a href="/A250328/b250328.txt">Table of n, a(n) for n = 1..1000</a>

%e a(3) = 77 because the pentagonal numbers A000326(n), for n = 1,2,3 are 1, 5, 12 and 3/(1/1+1/5+1/12) = 180/77.

%t With[{s = Array[PolygonalNumber[5, #] &, 21]}, Denominator@ Array[HarmonicMean@ Take[s, #] &, Length@ s]] (* _Michael De Vlieger_, Nov 02 2017 *)

%o (PARI)

%o harmonicmean(v) = #v / sum(k=1, #v, 1/v[k])

%o s=vector(30); for(k=1, #s, s[k]=denominator(harmonicmean(vector(k, i, (3*i^2-i)/2)))); s

%Y Cf. A000326, A250327 (numerators).

%K nonn,frac,easy

%O 1,2

%A _Colin Barker_, Nov 18 2014