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%I #17 Aug 17 2024 18:17:49
%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) { for (n=0, 100, p=1; x=n; until (x==0, p*=(x - 10*(x\10))!; x\=10); write("b061603.txt", n, " ", n!/p) ) } \\ _Harry J. Smith_, Jul 25 2009
%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,changed
%O 0,11
%A _Amarnath Murthy_, May 19 2001
%E Corrected and extended by _Vladeta Jovovic_, May 19 2001
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