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%I #12 Feb 13 2018 15:16:44
%S 1,2,3,4,5,2,1,7,8,9,2,3,11,3,1,13,2,5,3,2,16,17,2,7,19,4,1,3,4,2,9,
%T 23,3,5,25,2,11,27,4,3,29,2,1,2,31,32,3,8,2,15,5,2,4,5,37,2,17,3,10,5,
%U 3,41,2,1,4,43,4,7,5,4,2,21,47,3,13,49,2,23,3,14,4,9,53,2,25,5,6,7,1,3,16,2
%N Irregular array: For any prime p that divides n, if the highest power of the prime p that divides n is p^b(n,p), then p^b(n,p) = Sum_{k=1..m} a(n,k), where m is the order of the prime-power p^b(n,p) among the prime-powers (each being the highest power of each prime q that divides n, where q divides n) when they are ordered by size. Row 1 = (1).
%C Row n contains A001221(n) terms.
%e The prime factorization of 300 is 2^2 *3^1 *5^2. So the prime powers ordered by size are 3, 4, 25. Therefore row 300 is (3,1,21), because 3=3, 3+1 = 4, 3+1+21 = 25.
%o (PARI) { A141404row(n) = my(f); if(n==1,return([1])); f=factorint(n); f=vecsort(vector(matsize(f)[1],i,f[i,1]^f[i,2])); vector(#f,i,f[i]-if(i>1,f[i-1])); } \\ _Max Alekseyev_, May 07 2009
%Y Cf. A141810, A001221.
%K nonn,tabf
%O 1,2
%A _Leroy Quet_, Aug 03 2008
%E Extended by _Max Alekseyev_, May 07 2009
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