A264990 a(n) = number of occurrences of a most frequent nonzero digit in factorial base representation (A007623) of n.

0, 1, 1, 2, 1, 1, 1, 2, 2, 3, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 3, 1, 2, 2, 3, 3, 4, 2, 3, 1, 2, 2, 3, 2, 2, 1, 2, 2, 3, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 3, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 3, 1, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 3, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2

Offset

0,4

Links

Antti Karttunen, Table of n, a(n) for n = 0..10080

Index entries for sequences related to factorial base representation.

Formula

a(0) = 0; for n >= 1, a(n) = max(A257511(n), a(A257684(n)).

Other identities. For all n >= 0:

From Antti Karttunen, Aug 15 2016: (Start)

a(n) = A275811(A225901(n)).

a(n) = A051903(A275735(n)).

(End)

Example

   n  A007623(n)   a(n) [highest number of times any nonzero digit occurs].
   0 =   0           0 (because no nonzero digits present)
   1 =   1           1
   2 =  10           1
   3 =  11           2
   4 =  20           1
   5 =  21           1
   6 = 100           1
   7 = 101           2
   8 = 110           2
   9 = 111           3
  10 = 120           1
  11 = 121           2
  12 = 200           1
  13 = 201           1
  14 = 210           1
  15 = 211           2
  16 = 220           2
  17 = 221           2
  18 = 300           1
and for n=63 we have:
  63 = 2211          2.

Mathematica

a[n_] := Module[{k = n, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; Max[Tally[Select[s, # > 0 &]][[;; , 2]]]]; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Jan 24 2024 *)

Prog

(Scheme with memoization-macro definec)
(definec (A264990 n) (if (zero? n) n (max (A257511 n) (A264990 (A257684 n)))))
(Python)
from sympy import prime, factorint
from operator import mul
import collections
def a007623(n, p=2): return n if n<p else a007623(n//p, p+1)*10 + n%p
def a051903(n): return 0 if n==1 else max(factorint(n).values())
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 0 if n==0 else a051903(a275735(n))
print([a(n) for n in range(201)]) # Indranil Ghosh, Jun 20 2017

Crossrefs

Cf. A007623, A051903, A225901, A257511, A257684, A275735, A275811.

Cf. A265349 (positions of terms <= 1), A265350 (positions of term > 1).

Cf. also A266117, A266118.

Sequence in context: A110730 A198339 A262561 * A277315 A277326 A050431

Adjacent sequences: A264987 A264988 A264989 * A264991 A264992 A264993

Keyword

nonn,base

Author

Antti Karttunen, Dec 22 2015

Extensions

Name changed by Antti Karttunen, Aug 15 2016

Status

approved