

A253259


Number of factorizations of m^n into 4 factors, where m is a product of exactly 4 distinct primes and each factor is a product of n primes (counted with multiplicity).


2



1, 1, 17, 93, 465, 1746, 5741, 16238, 41650, 97407, 212412, 434767, 845366, 1569344, 2801696, 4828140, 8069053, 13114785, 20796651, 32242621, 48986553, 73052382, 107114645, 154621230, 220021932, 308940815, 428492880, 587520315, 797019526, 1070458096
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OFFSET

0,3


LINKS



FORMULA

[A^n B^n C^n D^n] Z(S_4)(Z(S_n)(A+B+C+D)) with Z(S_q) the cycle index of the symmetric group; parenthesis denote the canonical substitution of the argument into the cycle index.  Marko Riedel, Feb 06 2016
G.f.: (x^18 +6*x^17 +58*x^16 +213*x^15 +646*x^14 +1415*x^13 +2515*x^12 +3554*x^11 +4296*x^10 +4248*x^9 +3578*x^8 +2452*x^7 +1421*x^6 +628*x^5 +240*x^4 +61*x^3 +12*x^2x+1) /((1x)^10 *(1+x)^5 *(1+x+x^2)^3 *(1+x^2)). [This was found by Will Orrick and confirmed by Marko Riedel, see the StackExchange link above.]  Alois P. Heinz, Feb 09 2016


EXAMPLE

a(2) = 17: (2*3*5*7)^2 = 44100 = 15*15*14*14 = 21*15*14*10 = 21*21*10*10 = 25*14*14*9 = 25*21*14*6 = 25*21*21*4 = 35*14*10*9 = 35*15*14*6 = 35*21*10*6 = 35*21*15*4 = 35*35*6*6 = 35*35*9*4 = 49*10*10*9 = 49*15*10*6 = 49*15*15*4 = 49*25*6*6 = 49*25*9*4.


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



