A244849 Least number k > 0 such that 2^k begins with exactly n consecutive increasing digits.

1, 7, 90, 155, 8290, 63293, 4338436, 5194868, 62759188

Offset

1,2

Comments

The leading digit of the resulting powers of 2 are: 2, 1, 1, 4, 3, 1, 2, 1, 1. - Michel Marcus, Jul 11 2014

Links

Table of n, a(n) for n=1..9.

Example

2^7 = 128 begins with 2 consecutive increasing digits ('12'). Thus a(2) = 7.

Prog

(Python)
def a(n):
  for k in range(1, 10**5):
    st = str(2**k)
    count = 0
    if len(st) > n:
      for i in range(len(st)):
        if int(st[i]) == int(st[i+1])-1:
          count += 1
        else:
          break
      if count == n:
        return k
n = 0
while n < 10:
  print(a(n), end=', ')
  n += 1

Crossrefs

Cf. A244848, A244851, A244852.

Sequence in context: A350937 A200832 A103064 * A355591 A360914 A321164

Adjacent sequences: A244846 A244847 A244848 * A244850 A244851 A244852

Keyword

nonn,base,fini,full

Author

Derek Orr, Jul 07 2014

Extensions

a(7)-a(9) from Hiroaki Yamanouchi, Jul 10 2014

Status

approved