A275949 Number of distinct nonzero digits that occur multiple times in factorial base representation of n: a(n) = A056170(A275735(n)).

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

Offset

0,42

Links

Antti Karttunen, Table of n, a(n) for n = 0..40320

Index entries for sequences related to factorial base representation.

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.

Mathematica

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 *)

Prog

(Scheme) (define (A275949 n) (A056170 (A275735 n)))
(Python)
from sympy import prime, factorint
from operator import mul
from functools import reduce
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<p else a007623(n//p, p+1)*10 + n%p
def a275735(n):
    y=collections.Counter(map(int, list(str(a007623(n)).replace("0", "")))).most_common()
    return 1 if n==0 else reduce(mul, [prime(y[i][0])**y[i][1] for i in range(len(y))])
def a(n): return a056170(a275735(n))
print([a(n) for n in range(201)]) # Indranil Ghosh, Jun 20 2017

Crossrefs

Cf. A056170, A275735.

Cf. A265349 (indices of zeros), A265350 (of terms > 0).

Cf. also A007623, A225901, A275806, A275947, A275948, A275964.

Sequence in context: A116929 A059984 A046675 * A357924 A351563 A362983

Adjacent sequences: A275946 A275947 A275948 * A275950 A275951 A275952

Keyword

nonn,base

Author

Antti Karttunen, Aug 15 2016

Status

approved