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%I #6 Jun 24 2014 01:08:21
%S 2,3,4,5,3,7,8,9,4,11,9,13,5,4,16,17,9,19,16,5,7,23,9,25,8,27,25,29,3,
%T 31,32,7,10,6,9,37,11,8,16,41,4,43,49,16,13,47,27,49,16,10,64,53,9,8,
%U 25,11,16,59,3,61,17,25,64,9,5,67,100,13,5,71,27,73,20,16,121,9,6,79
%N a(n) = p^e, where p and e are the rounded means of the prime factors p_i and the exponents e_i, respectively, in the factorization n = p_1^e_1 * ... * p_r^e_r of n into distinct primes p_i. Each mean is rounded to the nearest integer, rounding up if there's a choice.
%e 24 = 2^3 * 3^1. The prime factors have mean (2+3)/2 = 2 1/2, which rounds up to 3. The exponents have mean (3+1)/2 = 2. So a(24) = 3^2 = 9.
%p with(numtheory): for n from 2 to 100 do pmean := round(sum(ifactors(n)[2][i][1], i=1..nops(ifactors(n)[2]))/nops(ifactors(n)[2])): emean := round(sum(ifactors(n)[2][i][2], i=1..nops(ifactors(n)[2]))/nops(ifactors(n)[2])): printf(`%d,`,pmean^emean) od:
%t a[n_] := Floor[1/2+(Plus@@First/@(fn=FactorInteger[n]))/(lth=Length[fn])]^Floor[1/2+(Plus@@Last/@fn)/lth]
%K easy,nonn
%O 2,1
%A _Joseph L. Pe_, Feb 02 2002
%E Edited by _Dean Hickerson_ and _James A. Sellers_, Feb 12 2002
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