cp's OEIS Frontend

A182471 If n = Product (p_j^k_j), j=1 to m, then a(n) = Floor( Sum (p_j + k_j) / 2m ) (a(1) = 0 by convention).

%I A182471 #10 May 05 2012 23:46:12
%S A182471 0,1,2,2,3,1,4,2,2,2,6,2,7,2,2,3,9,2,10,2,3,3,12,2,3,4,3,3,15,2,16,3,
%T A182471 4,5,3,2,19,5,4,2,21,2,22,4,2,6,24,2,4,2,5,4,27,2,4,3,6,8,30,2,31,8,3,
%U A182471 4,5,3,34,5,7,2,36,2,37,10,2,6,5,3,40,3,3
%N A182471 If n = Product (p_j^k_j), j=1 to m, then a(n) = Floor( Sum (p_j + k_j) / 2m )  (a(1) = 0 by convention).
%C A182471 a(n) = k (a selected constant) for only a finite number of values of n. The maximal value of n for a desired value of a(n) grows rapidly.
%C A182471 Thus for a(n)=2, n=2^1*3^1*5^5 is maximal.
%C A182471 For a(n)=3, n=2^1*3^1*5^1*7^11 is maximal.
%e A182471 21 = (3^1)(7^1), with 2 distinct prime factors. Floor(((3+1)+(7+1))/(2*2)) = 3, thus a(21) = 3.
%o A182471 (PARI) {for(k=2, 175,
%o A182471 M=factor(k); smt = 0; h = 0;
%o A182471 for(i=1, matsize(M)[1], for(j=1, matsize(M)[2], smt=smt+M[i,j]; h++));
%o A182471 print1( floor(smt/h), ", "))} \\\ _Douglas Latimer_, Apr 30 2012
%Y A182471 A008475
%K A182471 nonn
%O A182471 1,3
%A A182471 _Douglas Latimer_, May 01 2012