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 A275949 Number of distinct nonzero digits that occur multiple times in factorial base representation of n: a(n) = A056170(A275735(n)). 5
 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, 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,42 LINKS Antti Karttunen, Table of n, a(n) for n = 0..40320 FORMULA a(n) = A056170(A275735(n)). Other identities and observations. For all n >= 0. a(n) = A275947(A225901(n)). A275806(n) = A275948(n) + a(n). a(n) <= A275964(n). EXAMPLE For n=0, with factorial base representation (A007623) also 0, there are no nonzero digits, thus a(0) = 0. For n=2, with factorial base representation "10", there are no nonzero digits that are present multiple times, thus a(2) = 0. For n=3 ("11") there is one distinct nonzero digit which occurs more than once, thus a(3) = 1. For n=41 ("1221") there are two distinct nonzero digits ("1" and "2"), and both occur more than once, thus a(41) = 2. 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. PROG (Scheme) (define (A275949 n) (A056170 (A275735 n))) (Python) from sympy import prime, factorint from operator import mul import collections def a056170(n):     f = factorint(n)     return sum([1 for i in f if f[i]!=1]) def a007623(n, p=2): return n if n

0). Cf. also A007623, A225901, A275806, A275947, A275948, A275964. Sequence in context: A116929 A059984 A046675 * A090418 A076754 A101659 Adjacent sequences:  A275946 A275947 A275948 * A275950 A275951 A275952 KEYWORD nonn,base AUTHOR Antti Karttunen, Aug 15 2016 STATUS approved

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Last modified November 19 02:01 EST 2018. Contains 317332 sequences. (Running on oeis4.)