A103670 Smallest m such that A102730(m) = n.

0, 1, 2, 3, 4, 5, 8, 23, 117, 155, 1410, 3702, 29406, 104312, 680407

Offset

1,3

Comments

Reinhard Zumkeller conjectures (at A102730) that this sequence is finite. I conjecture the contrary, that a(n) exists for every n. Further, I expect a(n) << n^n. - Charles R Greathouse IV, Aug 21 2011

a(14) > 41000, if it exists. - Amiram Eldar, Apr 03 2025

The number of times that 1! through 10! appear in a(11)! is: 17417, 8777, 4361, 1118, 265, 31, 5, 1, 1, 1. It suggests factorials are normal in base 2 except for their trailing zeros. On this assumption, a(16) is expected around 3-4 million, and a(17) around 20-30 million. - Martin Fuller, Apr 14 2025

Links

Table of n, a(n) for n=1..15.

Example

a(6) = 5: A102730(5) = #{0,1,2,3,4,5} = 6.
a(7) = 8: A102730(8) = #{0,1,2,3,4,7,8} = 7.
a(8) = 23: A102730(23) = #{0,1,2,3,4,5,6,23} = 8.
a(9) = 117: A102730(117) = #{0,1,2,3,4,5,6,7,117} = 9.
a(10) = 155: A102730(155) = #{0,1,2,3,4,5,6,7,8,155} = 10.

Mathematica

A102730[n_] := Sum[Boole[StringContainsQ[IntegerString[n!, 2], IntegerString[k!, 2]]], {k, 0, n}];
seq[len_] := Module[{s = Table[0, {len}], c = 0, m = 0, i}, While[c < len, i = A102730[m]; If[i <= len && s[[i]] == 0, c++; s[[i]] = m]; m++]; s]; seq[10] (* Amiram Eldar, Apr 03 2025 *)

Crossrefs

Cf. A102730.

Sequence in context: A116657 A268274 A079383 * A289081 A105950 A227778

Adjacent sequences: A103667 A103668 A103669 * A103671 A103672 A103673

Keyword

nonn,base,more,hard

Author

Reinhard Zumkeller, Feb 12 2005

Extensions

a(11)-a(12) from D. S. McNeil, Aug 21 2011

a(13) from Amiram Eldar, Apr 03 2025

a(14)-a(15) from Martin Fuller, Apr 14 2025

Status

approved