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Numerator of the harmonic mean of the first n primes.
4

%I #35 Feb 12 2021 16:45:27

%S 2,12,90,840,11550,180180,3573570,77597520,2007835830,64696932300,

%T 2206165391430,89048857617720,3955253425853730,183158658643380420,

%U 9223346738827371150,521426535635040715680,32686925952621614864190,2111190864469325477698860

%N Numerator of the harmonic mean of the first n primes.

%H Colin Barker, <a href="/A250130/b250130.txt">Table of n, a(n) for n = 1..300</a>

%F a(n) = n*A002110(n). - _L. Edson Jeffery_, Jan 04 2015

%e a(3) = 90 because the first 3 primes are [2,3,5] and 3 / (1/2+1/3+1/5) = 90/31.

%e The first fractions are 2/1, 12/5, 90/31, 840/247, 11550/2927, 180180/40361, 3573570/716167, 77597520/14117683, ...

%p N:= 100: # to get a(1) to a(N)

%p B:= ListTools:-PartialSums([seq](1/ithprime(i),i=1..N)):

%p seq(numer(n/B[n]), n=1..N); # _Robert Israel_, Nov 13 2014

%t Table[n/Sum[1/Prime[k],{k,1,n}],{n,1,20}]//Numerator (* _Vaclav Kotesovec_, Nov 13 2014 *)

%t Table[n*Product[Prime[j], {j, n}], {n, 17}] (* _L. Edson Jeffery_, Jan 04 2015 *)

%o (PARI)

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

%o s=vector(30); p=primes(#s); for(k=1, #p, s[k]=numerator( harmonicmean( vector(k, i, p[i])))); s

%o (PARI) n=0; P=1; forprime(p=2, 100, n++; P *= p; print1(n*P, ", ")) \\ _Jeppe Stig Nielsen_, Aug 11 2019

%o (Python)

%o from sympy import prime

%o from fractions import Fraction

%o def a(n):

%o return (n/sum(Fraction(1, prime(k)) for k in range(1, n+1))).numerator

%o print([a(n) for n in range(1, 19)]) # _Michael S. Branicky_, Feb 12 2021

%Y Cf. A024451 (denominators), A002110 (primorial numbers).

%K nonn,frac

%O 1,1

%A _Colin Barker_, Nov 13 2014