OFFSET
1,4
COMMENTS
a(1) = 0 since 1 is the empty product.
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).
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.
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.
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.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..10000
Eric W. Weisstein World of Mathematics, Least Prime Factor
EXAMPLE
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.
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.
a(16) = 4, because 2^4 = 16 = n, and is the largest power of the distinct prime divisor 2 of 16.
MATHEMATICA
Table[If[n == 1, 0, Floor[Log[FactorInteger[n][[1, 1]], n]]], {n, 120}]
PROG
(PARI) a(n) = if (n==1, 0, logint(n, vecmin(factor(n)[, 1]))); \\ Michel Marcus, Jan 01 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Michael De Vlieger, Jan 01 2017
STATUS
approved