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%I #26 Feb 14 2024 01:05:06
%S 0,0,0,1,0,0,0,1,1,1,0,1,0,0,0,1,1,1,0,0,0,1,0,0,0,1,1,1,0,1,1,1,1,1,
%T 1,1,0,1,1,1,1,2,0,1,1,1,0,1,0,0,0,1,1,1,0,1,1,1,1,2,1,1,1,2,1,1,0,0,
%U 0,1,1,1,0,0,0,1,0,0,0,1,1,1,0,1,0,0,0,1,1,1,1,1,1,2,1,1,0,0,0,1,0,0,0,1,1,1,0,1,0,0,0,1,1,1,0,0,0,1,0,0,0
%N Number of distinct nonzero digits that occur multiple times in factorial base representation of n: a(n) = A056170(A275735(n)).
%H Antti Karttunen, <a href="/A275949/b275949.txt">Table of n, a(n) for n = 0..40320</a>
%H <a href="/index/Fa#facbase">Index entries for sequences related to factorial base representation</a>.
%F a(n) = A056170(A275735(n)).
%F Other identities and observations. For all n >= 0.
%F a(n) = A275947(A225901(n)).
%F A275806(n) = A275948(n) + a(n).
%F a(n) <= A275964(n).
%e For n=0, with factorial base representation (A007623) also 0, there are no nonzero digits, thus a(0) = 0.
%e For n=2, with factorial base representation "10", there are no nonzero digits that are present multiple times, thus a(2) = 0.
%e For n=3 ("11") there is one distinct nonzero digit which occurs more than once, thus a(3) = 1.
%e For n=41 ("1221") there are two distinct nonzero digits ("1" and "2"), and both occur more than once, thus a(41) = 2.
%e For n=44 ( "1310") there are two distinct nonzero digits ("1" and "3"), but only the other (1) occurs more than once, thus a(44) = 1.
%t a[n_] := Module[{k = n, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; Count[Tally[Select[s, # > 0 &]][[;;, 2]], _?(# > 1 &)]]; Array[a, 100, 0] (* _Amiram Eldar_, Feb 14 2024 *)
%o (Scheme) (define (A275949 n) (A056170 (A275735 n)))
%o (Python)
%o from sympy import prime, factorint
%o from operator import mul
%o from functools import reduce
%o import collections
%o def a056170(n):
%o f = factorint(n)
%o return sum([1 for i in f if f[i]!=1])
%o def a007623(n, p=2): return n if n<p else a007623(n//p, p+1)*10 + n%p
%o def a275735(n):
%o y=collections.Counter(map(int, list(str(a007623(n)).replace("0", "")))).most_common()
%o return 1 if n==0 else reduce(mul, [prime(y[i][0])**y[i][1] for i in range(len(y))])
%o def a(n): return a056170(a275735(n))
%o print([a(n) for n in range(201)]) # _Indranil Ghosh_, Jun 20 2017
%Y Cf. A056170, A275735.
%Y Cf. A265349 (indices of zeros), A265350 (of terms > 0).
%Y Cf. also A007623, A225901, A275806, A275947, A275948, A275964.
%K nonn,base
%O 0,42
%A _Antti Karttunen_, Aug 15 2016
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