A093826 In binary representation: least number, k, which occurs n times in its factorial.

5, 1, 16, 12, 49, 58, 60, 110, 209, 117, 240, 430, 255, 1423, 921, 980, 511, 1847, 3737, 3692, 3998, 7265, 15267, 15651, 15722, 31457, 32659, 64248, 57927, 64448, 64171, 250068, 129013, 501578, 256159, 510732, 980930, 979883

Offset: 0

Author

Robert G. Wilson v and Reinhard Zumkeller, Apr 16 2004

Keywords

• nonn
• base

Comments

Overlapping occurrences are counted. - Michael S. Branicky, May 01 2021

a(47) = 262143. - Michael S. Branicky, May 02 2021

Examples

			12!_b = 11100100011001111110000000000 and 12_b = 1100 and the later string appears thrice in the former string.
		

Crossrefs

Cf. A093685.

Programs

Mathematica

f[n_] := ToString[ FromDigits[ IntegerDigits[n, 2]]]; g[n_] := Length[ StringPosition[ f[n! ], f[n]]]; a = Table[0, {30}]; Do[ b = g[n]; If[a[[b + 1]] == 0, a[[b + 1]] = n], {n, 29000}]; a

Python

from itertools import count, takewhile
def count_overlaps(subs, s):
  c = i = 0
  while i != -1:
    i = s.find(subs, i)
    if i != -1: c += 1; i += 1
  return c
def afind(limit):
  kfact, adict = 1, dict()
  for k in range(1, limit+1):
    kb, kfact = bin(k)[2:], kfact * k
    kfactb = bin(kfact)[2:]
    n = count_overlaps(kb, kfactb)
    if n not in adict: adict[n] = k
  return [adict[n] for n in takewhile(lambda i: i in adict, count(0))]
print(afind(16000))  # Michael S. Branicky, May 01 2021

Extensions

a(25)-a(37) from Michael S. Branicky, May 03 2021