A331259 Numerator of harmonic mean of 3 consecutive primes. Denominators are A331260.
90, 315, 1155, 3003, 7293, 12597, 22287, 38019, 62031, 99789, 141081, 195693, 248583, 321339, 146969, 572241, 723399, 870531, 1041783, 1228371, 1435983, 1750719, 2149617, 2615799, 3027273, 3339363, 3603867, 3953757, 4692777, 5639943, 6837807, 7483899, 8512221
Offset: 1
Examples
b(1) = a(1)/A331260(1) = 3*2*3*5 / (3*5 + 2*5 + 2*3) = 90/31, b(2) = a(2)/A331260(2) = 3*3*5*7 / (5*7 + 3*7 + 3*5) = 315/71, ... b(15) = a(15)/A331260(15) = 3*47*53*59 / (53*59 + 47*59 + 47*53) = 440907/8391 = 146969/2797. The common factor of 3 (see A292530) makes the denominator different from A127345(15).
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
-
Maple
q:= proc(a,b,c) if nops({a,b,c} mod 3) = 1 then a*b*c else 3*a*b*c fi end proc: P:= [seq(ithprime(i),i=1..102)]: seq(q(P[i],P[i+1],P[i+2]),i=1..100); # Robert Israel, Jul 29 2024 -
PARI
hm3(x,y,z)=3/(1/x+1/y+1/z); p1=2; p2=3; forprime(p3=5,150, print1(numerator(hm3(p1,p2,p3)),", ");p1=p2;p2=p3)
Formula
a(n) = numerator ((3*p1*p2*p3)/(p2*p3 + p1*p3 + p1*p2)) with p1 = prime(n), p2 = prime(n + 1), p3 = prime(n + 2).
a(n) = p1*p2*p3 if p1 == p2 == p3 (mod 3), otherwise 3*p1*p2*p3. - Robert Israel, Jul 29 2024