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A275964
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Total number of nonzero digits with multiple occurrences in factorial base representation of n (counted with multiplicity): a(n) = A275812(A275735(n)).
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8
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0, 0, 0, 2, 0, 0, 0, 2, 2, 3, 0, 2, 0, 0, 0, 2, 2, 2, 0, 0, 0, 2, 0, 0, 0, 2, 2, 3, 0, 2, 2, 3, 3, 4, 2, 3, 0, 2, 2, 3, 2, 4, 0, 2, 2, 3, 0, 2, 0, 0, 0, 2, 2, 2, 0, 2, 2, 3, 2, 4, 2, 2, 2, 4, 3, 3, 0, 0, 0, 2, 2, 2, 0, 0, 0, 2, 0, 0, 0, 2, 2, 3, 0, 2, 0, 0, 0, 2, 2, 2, 2, 2, 2, 4, 2, 2, 0, 0, 0, 2, 0, 0, 0, 2, 2, 3, 0, 2, 0, 0, 0, 2, 2, 2, 0, 0, 0, 2, 0, 0, 0
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OFFSET
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0,4
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LINKS
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FORMULA
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Other identities and observations. For all n >= 0.
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EXAMPLE
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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 nonzero digit which occurs more than once, and it occurs two times in total, thus a(3) = 2.
For n=41 ("1221") there are two distinct nonzero digits ("1" and "2"), and both occur more than once, namely twice each, thus a(41) = 2+2 = 4.
For n=44 ("1310") there are two distinct nonzero digits ("1" and "3"), but only the other (1) occurs more than once (two times), thus a(44) = 2.
For n=279 ("21211") there are two distinct nonzero digits present that occur more than once, digit 2 twice, and digit 1 for three times, thus a(279) = 2+3 = 5.
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MATHEMATICA
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a[n_] := Module[{k = n, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; Total[Select[Tally[Select[s, # > 0 &]][[;; , 2]], # > 1 &]]]; Array[a, 100, 0] (* Amiram Eldar, Feb 07 2024 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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