cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

Antti Karttunen, Dec 22 2015

Keywords

Examples

			   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.
		

Crossrefs

Cf. A265349 (positions of terms <= 1), A265350 (positions of term > 1).
Cf. also A266117, A266118.

Programs

  • 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 *)
  • Python
    from sympy import prime, factorint
    from operator import mul
    import collections
    def a007623(n, p=2): return n if n

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)

Extensions

Name changed by Antti Karttunen, Aug 15 2016

A266117 Lexicographically first injection of positive integers beginning with a(1) such that a(n)*a(n+1) is a term of A265349, i.e., has no multiple occurrences of any nonzero digit when viewed in factorial base (A007623).

Original entry on oeis.org

1, 2, 3, 4, 5, 10, 11, 6, 7, 12, 8, 9, 24, 13, 18, 19, 26, 14, 17, 22, 21, 16, 15, 32, 30, 20, 23, 29, 42, 28, 25, 48, 34, 53, 41, 54, 27, 40, 33, 36, 37, 61, 65, 44, 49, 72, 39, 38, 51, 52, 55, 59, 47, 46, 58, 50, 60, 62, 31, 66, 56, 57, 64, 45, 80, 63, 74, 69, 68, 35, 79, 100, 73, 43, 67, 70, 71, 78, 84, 87, 92, 90, 76, 96, 75
Offset: 1

Views

Author

Antti Karttunen, Dec 22 2015

Keywords

Comments

After a(1) = 1, always choose for a(n+1) the least unused k such that in factorial base representation (A007623), the product a(n)*a(n+1) will not show any of the nonzero digits present twice (or more times), regardless of the positions of the digits.

Examples

			For n = 6, we start searching from the least not yet used number in range a(1) .. a(5) [which is 6, because all the previous terms are fixed] for the first number whose product with a(5) = 5 results a number in A265349.
Multiplying 5 (in factorial base "21") with 6 (in factorial base "100") results 30, which in factorial base is "1100", containing digit "1" twice, thus 6 is disqualified.
Similarly, products 5*7, 5*8 and 5*9 result 35 = "1121", 40 = "1220" and 45 = "1311", where in all cases one of the nonzero digits occur more than once, so 7, 8 and 9 are also all disqualified.
But 5*10 = 50, which has a factorial base representation ("2010") that matches the criterion, thus a(6) = 10.
		

Crossrefs

Left inverse: A266118 (also the right inverse if this sequence is a permutation of the positive integers).
Cf. also A266121, A266191 and A266195 for similar permutations.
Showing 1-2 of 2 results.