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a(n) = gcd(A001008(m(n)), m(n)), with m(n) = A256102(n), n >= 1.
4

%I #12 Dec 24 2018 08:49:19

%S 5,7,11,11,13,17,19,23,29,31,43,37,41,43,47,53,59,61,67,71,73,79,83,

%T 89,97,11,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,

%U 179,181,191,193,197,199,211,223

%N a(n) = gcd(A001008(m(n)), m(n)), with m(n) = A256102(n), n >= 1.

%C See A256102. The entry a(n) gives the quotient of the numerator of the harmonic sum of the first A256102(n) positive integers and the denominator of the harmonic mean of the same numbers. For each positive integer values m not from A256102 this quotient is 1.

%F a(n) = gcd(A001008(m(n)), m(n)), with m(n) = A256102(n), n >= 1.

%F a(n) = A001008(m(n))/A175441(m(n)), with m(n) = A256103(n), n >= 1.

%e n = 1: gcd(A001008(20), 20) = gcd(55835135, 20) = 5. A001008(20)/A175441(20) = 55835135/11167027 = 5.

%e Because 19 is not from A256102 one has A001008(19) = A175441(19) = 275295799.

%Y Cf. A256102, A002805, A001008, A175441.

%K nonn,easy

%O 1,1

%A _Wolfdieter Lang_, Apr 16 2015