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%I #17 Feb 05 2019 09:44:28
%S 1,2,3,4,5,2,7,8,9,4,11,3,13,6,5,16,17,8,19,4,6,10,23,8,25,12,27,7,29,
%T 3,31,32,11,16,14,8,37,18,12,15,41,6,43,11,9,22,47,15,49,24,17,12,53,
%U 26,10,7,18,28,59,3,61,30,27,64,25,9,67,16,23,5,71,8
%N a(n) is the smallest k such that there exists a permutation b_0, b_1, ... of the maximal prime powers dividing n satisfying b_i | k+i for all i.
%C Old name was: "If the prime factorization of n is n=product{p|n} p^b(n,p), then a(n) = the smallest positive integer such that each p^b(n,p) divides a different integer k where a(n) <= k <= a(n)+{number distinct prime divisors of n}-1. a(1)=1."
%C Clarification of definition: Each b(n,p) is a positive integer. Let the number of distinct primes dividing n be P. There is a permutation (c(1),c(2),c(3),...c(P)) of the p^b(n,p)'s such that c(1)|a(n), c(2)|(a(n)+1), c(3)|(a(n)+2),..., c(P)|(a(n)+P-1).
%H Charlie Neder, <a href="/A162961/b162961.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = n iff n is a prime power (A000961). - _Charlie Neder_, Feb 03 2019
%e 3150 is factored as 2^1 * 3^2 * 5^2 * 7^1. a(3150) = 25, which can be seen from 5^2|25, 2^1|26, 3^2|27, and 7^1|28.
%o (PARI) isokp(m, perm, f, nb) = {for (k=1, nb, if ((m+k-1) % f[perm[k],1]^f[perm[k],2], return (0)); ); return (1);}
%o isokm(m, f, nb) = {for (j=0, nb!-1, my(perm = numtoperm(nb, j)); if (isokp(m, perm, f, nb), return (1));); return (0);}
%o a(n) = {my(f = factor(n), nb = #f~, m = 1); while (! isokm(m, f, nb), m++); m;} \\ _Michel Marcus_, Feb 05 2019
%Y Cf. A000961.
%K nonn
%O 1,2
%A _Leroy Quet_, Jul 19 2009
%E Name changed and a(31)-a(72) by _Charlie Neder_, Feb 03 2019
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