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Numbers n such that n! has a record number of zeros.
4

%I #34 Feb 17 2018 10:29:21

%S 0,5,7,12,18,19,20,22,25,28,34,37,38,50,57,61,73,85,94,105,114,115,

%T 122,124,127,133,153,154,162,172,176,182,185,186,194,203,213,216,241,

%U 249,254,257,264,273,285,304,327,337,345,353,357,394,395,402,420,425,426

%N Numbers n such that n! has a record number of zeros.

%C All zeros are counted, not just the trailing zeros. So a particular n! might have more zeros than (n - 1)! (e.g., n = 10), but that's not enough for it to be in the sequence. - _Alonso del Arte_, Apr 30 2017

%H David A. Corneth, <a href="/A031144/b031144.txt">Table of n, a(n) for n = 1..476</a>

%e Since 0! = 1, 0! has no significant zeros, and so 0 is the first term of the sequence.

%e It isn't until 5! = 120 that n! gets its first significant zero, so 5 is the second term of the sequence.

%t Function[s, -1 + Map[First@ Position[s, #] &, Union@ FoldList[Max, s]]]@ Array[DigitCount[#!, 10, 0] &, 430, 0] // Flatten (* _Michael De Vlieger_, May 12 2017 *)

%o (PARI) lista(n) = my(l = List([0]), m=0, p=1, d); for(i=2,n, p*=i; d = digits(p); s = sum(i=1, #d, d[i]==0); if(s > m, listput(l, i); m=s));l \\ _David A. Corneth_, May 19 2017

%Y Cf. A031145.

%K nonn,base,easy

%O 1,2

%A _N. J. A. Sloane_

%E Corrected and extended by _Erich Friedman_.

%E Name clarified by _Alonso del Arte_, Apr 30 2017

%E Offset changed by _N. J. A. Sloane_, May 20 2017