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Number of factorizations of m^n into 5 factors, where m is a product of exactly 5 distinct primes and each factor is a product of n primes (counted with multiplicity).
2

%I #10 Apr 20 2020 17:54:52

%S 1,1,73,1417,19834,190131,1398547,8246011,40837569,174901563,

%T 664006236,2274999093,7139338769,20758868781,56466073587,144806582536,

%U 352420554194,818441723112,1822255658908,3905392549491,8084697432576,16214886608495,31590986308810,59926495673780

%N Number of factorizations of m^n into 5 factors, where m is a product of exactly 5 distinct primes and each factor is a product of n primes (counted with multiplicity).

%H Andrew Howroyd, <a href="/A253263/b253263.txt">Table of n, a(n) for n = 0..50</a>

%e a(2) = 73: (2*3*5*7*11)^2 = 5336100 = 25*22*22*21*21 = 33*25*22*21*14 = 33*33*25*14*14 = 35*22*22*21*15 = 35*33*22*15*14 = 35*33*22*21*10 = 35*33*33*14*10 = 35*35*22*22*9 = 35*35*33*22*6 = 35*35*33*33*4 = 49*22*22*15*15 = 49*25*22*22*9 = 49*33*22*15*10 = 49*33*25*22*6 = 49*33*33*10*10 = 49*33*33*25*4 = 55*22*21*15*14 = 55*22*21*21*10 = 55*33*15*14*14 = 55*33*21*14*10 = 55*35*22*14*9 = 55*35*22*21*6 = 55*35*33*14*6 = 55*35*33*21*4 = 55*49*22*10*9 = 55*49*22*15*6 = 55*49*33*10*6 = 55*49*33*15*4 = 55*55*14*14*9 = 55*55*21*14*6 = 55*55*21*21*4 = 55*55*49*6*6 = 55*55*49*9*4 = 77*22*15*15*14 = 77*22*21*15*10 = 77*25*22*14*9 = 77*25*22*21*6 = 77*33*15*14*10 = 77*33*21*10*10 = 77*33*25*14*6 = 77*33*25*21*4 = 77*35*22*10*9 = 77*35*22*15*6 = 77*35*33*10*6 = 77*35*33*15*4 = 77*55*14*10*9 = 77*55*15*14*6 = 77*55*21*10*6 = 77*55*21*15*4 = 77*55*35*6*6 = 77*55*35*9*4 = 77*77*10*10*9 = 77*77*15*10*6 = 77*77*15*15*4 = 77*77*25*6*6 = 77*77*25*9*4 = 121*15*15*14*14 = 121*21*15*14*10 = 121*21*21*10*10 = 121*25*14*14*9 = 121*25*21*14*6 = 121*25*21*21*4 = 121*35*14*10*9 = 121*35*15*14*6 = 121*35*21*10*6 = 121*35*21*15*4 = 121*35*35*6*6 = 121*35*35*9*4 = 121*49*10*10*9 = 121*49*15*10*6 = 121*49*15*15*4 = 121*49*25*6*6 = 121*49*25*9*4.

%Y Row n=5 of A257463.

%K nonn

%O 0,3

%A _Alois P. Heinz_, Apr 30 2015