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a(n) = n! / {product of factorials of the digits of n}.
2

%I #20 Dec 06 2024 21:43:48

%S 1,1,1,1,1,1,1,1,1,1,3628800,39916800,239500800,1037836800,3632428800,

%T 10897286400,29059430400,70572902400,158789030400,335221286400,

%U 1216451004088320000,25545471085854720000,281000181944401920000,2154334728240414720000

%N a(n) = n! / {product of factorials of the digits of n}.

%C It can be shown that the terms obtained by the above formula are positive integers using the fact that k! divides a product of k consecutive numbers.

%H Harry J. Smith, <a href="/A061603/b061603.txt">Table of n, a(n) for n = 0..100</a>

%F a(n) = A000142(n)/A066459(n). - _Michel Marcus_, Jul 02 2018

%e a(12) = (12!) / (1!*2!) = 239500800.

%t Table[n!/Times@@(IntegerDigits[n]!),{n,0,30}] (* _Harvey P. Dale_, Jan 19 2017 *)

%o (PARI) a(n) = my(d = digits(n)); n!/prod(k=1, #d, d[k]!); \\ _Michel Marcus_, Jul 02 2018

%Y Cf. A000142, A061602, A066459.

%K nonn,base,easy

%O 0,11

%A _Amarnath Murthy_, May 19 2001

%E Corrected and extended by _Vladeta Jovovic_, May 19 2001