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%I #14 Jan 02 2017 02:10:48
%S 0,1,1,2,1,2,1,3,2,3,1,3,1,3,2,4,1,4,1,4,2,4,1,4,2,4,3,4,1,4,1,5,3,5,
%T 2,5,1,5,3,5,1,5,1,5,3,5,1,5,2,5,3,5,1,5,2,5,3,5,1,5,1,5,3,6,2,6,1,6,
%U 3,6,1,6,1,6,3,6,2,6,1,6,4,6,1,6,2,6,4,6,1,6,2,6,4,6,2,6,1,6,4,6,1,6,1,6,4,6,1,6,1,6,4,6,1,6,2,6,4,6,2,6
%N a(n) = floor(log_p(n)) where p = A020639(n), i.e., the least prime factor of n.
%C a(1) = 0 since 1 is the empty product.
%C a(p) = 1 since the exponent e of the largest power p^e of the prime divisor p is p^1 (i.e., p itself).
%C a(p^m) = m since the largest power p^e of the prime divisor p is p^m, (p^m itself), i.e., e = m.
%C a(n) is the greatest value of the power e of p^e across the prime divisors p of n such that p^e <= n.
%C Consider integers 1<=r<=n with all prime divisors p of r also dividing n. Let m be the smallest power n^m | r, and let e be the largest value of m across 1<=r<=n. This is A280274(n). This sequence underlies A280274: A280274(1) = 0, A280274(n) = 1 with n having omega(n) = 1. A280274(n) = a(n) for squarefree n. A280274(n) for all other n is ceiling(a(n)/k), with k being the multiplicity of p = A020639(n) in the prime decomposition of n.
%H Michael De Vlieger, <a href="/A280363/b280363.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric W. Weisstein World of Mathematics, <a href="http://mathworld.wolfram.com/LeastPrimeFactor.html">Least Prime Factor</a>
%e a(10) = 3, because 2^3 = 8 and 5^1 = 5 are less than 10 = 2*5, and of the multiplicities of these numbers, 3 is the greatest.
%e a(12) = 3, because 2^3 = 8 and 3^2 = 9 are less than 12 = 2*2*3, and of the multiplicities of these numbers, 3 is the greatest.
%e a(16) = 4, because 2^4 = 16 = n, and is the largest power of the distinct prime divisor 2 of 16.
%t Table[If[n == 1, 0, Floor[Log[FactorInteger[n][[1, 1]], n]]], {n, 120}]
%o (PARI) a(n) = if (n==1, 0, logint(n, vecmin(factor(n)[,1]))); \\ _Michel Marcus_, Jan 01 2017
%Y Cf. A020639, A007947, A280274.
%K nonn,easy
%O 1,4
%A _Michael De Vlieger_, Jan 01 2017
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