A238448 -id:A238448 - cp's OEIS frontend

A238449 Smallest numbers m such that 2^m contains a string of n consecutive decreasing integers in its decimal representation.

0, 5, 25, 78, 161, 341, 1315, 28238, 56047, 283789

Offset: 1

Derek Orr, Feb 26 2014

This is an increasing sequence (not necessarily strictly increasing).

			5 is the smallest exponent such that 2^5 contains two consecutive decreasing integers (2^5 = 32).
25 is the smallest exponent such that 2^25 contains three consecutive decreasing integers (2^25 = 33554432).

Cf. A045875, A238448.

a[1] = 0; a[n_] := Block[{k = 4, p = 16}, While[Max[ Length /@  Select[ Split@ Differences@ IntegerDigits@p, First@# == -1 &]] < n-1, k++; p *= 2]; k]; a/@ Range[7] (* Giovanni Resta, Feb 26 2014 *)

def StrDec(x):
  for n in range(10**5):
    count = 0
    i = 0
    if len(str(2**n)) == x and x == 1:
      return n
    while i < len(str(2**n))-1:
      if int(str(2**n)[i]) == int(str(2**n)[i+1])-1:
        count += 1
        i += 1
      else:
        if count == x-1:
          return n
        else:
          count = 0
          i += 1
    if count == x-1:
      return n
x = 1
while x < 50:
  print(StrDec(x))
  x += 1

a(8)-a(10) from Giovanni Resta, Feb 26 2014

A238507 Smallest number m such that 3^m contains a string of n consecutive increasing integers in its decimal representation.

Original entry on oeis.org

0, 8, 20, 57, 332, 332, 6814, 7926, 16724, 200633

Offset: 1

Derek Orr, Feb 27 2014

			8 is the smallest exponent such that 3^8 contains two consecutive increasing integers (3^8 = 6561).
20 is the smallest exponent such that 3^20 contains three consecutive increasing integers (3^20 = 3486784401).

Cf. A238448, A215727.

def StrInc(x):
  for n in range(10**5):
    count = 0
    i = 0
    string = str(3**n)
    if len(string) == x and x == 1:
      return n
    while i < len(string)-1:
      if int(string[i]) == int(string[i+1])-1:
        count += 1
        i += 1
      else:
        if count >= x-1:
          return n
        else:
          count = 0
          i += 1
    if count >= x-1:
      return n
x = 1
while x < 15:
  print(StrInc(x))
  x += 1

a(8)-a(10) from Giovanni Resta, Mar 02 2014

A243150 Least number k > 0 such that 2^k contains an n-digit long substring of the infinite string "0123456789012345678901234567890123456....".

Original entry on oeis.org

1, 7, 28, 106, 391, 992, 1178, 7255, 15975, 67143, 333212, 333212, 1641257

Offset: 1

Derek Orr, May 31 2014

By A238448, a(10) <= 244178.

			2^7 = 128 contains the 2-digit substring "12". Thus a(2) = 7.

Cf. A238448.

import sys
sys.set_int_max_str_digits(5000)
def a(n):
  for k in range(1, 10**5):
    for i in range(10):
      s = ''
      for j in range(i, i+n):
        dig=j%10
        s+=str(dig)
      if str(2**k).find(s) > -1:
        return k
n=1
while n < 10:
  print(a(n))
  n+=1

a(10)-a(13) from Hiroaki Yamanouchi, Sep 26 2014

cp's OEIS Frontend

A238449 Smallest numbers m such that 2^m contains a string of n consecutive decreasing integers in its decimal representation.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

Extensions

A238507 Smallest number m such that 3^m contains a string of n consecutive increasing integers in its decimal representation.

Views

Author

Keywords

Examples

Crossrefs

Programs

Python

Extensions

A243150 Least number k > 0 such that 2^k contains an n-digit long substring of the infinite string "0123456789012345678901234567890123456....".

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Python

Extensions