A256450 Numbers that have at least one 1-digit in their factorial base representation (A007623).

1, 2, 3, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 17, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 65, 67, 68, 69, 71, 73, 74, 75, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87, 89, 91, 92, 93, 95, 97, 98, 99, 101

Offset: 0

Author

Antti Karttunen, Apr 27 2015

Keywords

• nonn
• base

Comments

Numbers n for which A257679(n) = 1, i.e., numbers n such that the least nonzero digit in their factorial base representation (A007623) is 1.

Involution A225901 maps each term of this sequence to a unique term of A273670, and vice versa.

Starting offset is zero (with a(0) = 1) because it is the most natural offset for the given fast recurrence.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..10000
• Index entries for sequences related to factorial base representation

Crossrefs

Cf. A007623, A257679.

Complement of A255411.

Cf. A257680 (characteristic function), A273662 (left inverse).

First row of A257503, first column of A257505.

Subsequences: A059590 (apart from its zero-term), A255341, A255342, A255343, A257262, A257263, A258198, A258199.

Cf. also A227187 (numbers with at least one nonleading zero) and A273670, A225901.

Programs

Mathematica

Select[Range@ 101, MemberQ[IntegerDigits[#, MixedRadix[Reverse@ Range@ 12]], 1] &] (* Michael De Vlieger, May 30 2016, Version 10.2 *)
r = MixedRadix[Reverse@ Range[2, 12]]; Select[Range@ 101, Min[IntegerDigits[#, r] /. 0 -> Nothing] == 1 &]  (* Michael De Vlieger, Aug 14 2016, Version 10.2 *)

Python

def A(n, p=2): return n if n
=1]) # Indranil Ghosh, Jun 19 2017

Formula

a(0) = 1, and for n >= 1, if A257511(1+a(n-1)) > 0, then a(n) = a(n-1) + 1, otherwise a(n-1) + 2. [In particular, if the previous term is 2k, then the next term is 2k+1, because all odd numbers are members.]

Other identities:

For all n >= 0, A273662(a(n)) = n. [A273662 works as the left inverse for this sequence.]

From Antti Karttunen, May 26 2015: (Start)

Alternative recurrence for the same sequence:

Set k = A258198(n), d = n - A258199(n) and f = A000142(k+1) = (k+1)! If d < f then b(n) = f+d, otherwise b(n) = ((2+floor((d-f)/A258199(n))) * f) + b((d-f) mod A258199(n)). For offset=1 sequence, define a(n) = b(n-1).

(End)

Extensions

Starting offset changed from 1 to 0 by Antti Karttunen, May 30 2016