A215732
a(n) is the first digit to appear n times in succession in a power of 2.
Original entry on oeis.org
1, 5, 7, 5, 6, 8, 7, 1, 9, 9, 6, 3, 2, 9, 1, 4, 1
Offset: 1
2^16 = 65536 is the first power of 2 with a repeated digit (cf. A045875), with 5 repeated, so a(2) = 5. - _N. J. A. Sloane_, Aug 23 2012
-
n = 1; x = 1; lst = {};
For[i = 1, i <= 10000, i++,
z = Split[IntegerDigits[x]]; a = Length /@ z; b = Max[a];
For[j = n, j <= b, j++,
AppendTo[lst, First[First[Part[z, First[Position[a, b]]]]]]; n++
]; x = 2 x ]; lst (* Robert Price, Mar 16 2019 *)
-
def A215732(n):
l, x = [str(d)*n for d in range(10)], 1
for m in range(10**9):
s = str(x)
for k in range(10):
if l[k] in s:
return k
x *= 2
return 'search limit reached'
# Chai Wah Wu, Dec 17 2014
A215727
a(n) is the smallest m for which 3^m contains n consecutive identical digits.
Original entry on oeis.org
0, 11, 32, 33, 274, 538, 2124, 7720, 22791, 107187, 107187, 639226, 5756979, 8885853, 68353787, 78927180, 78927180
Offset: 1
3^11 = 177147, which has two digits in a row.
-
A215727[n_] := Module[{m = 0 , t}, t = Table[i, {i, 0, 9}, {n}];
While[True, If[ContainsAny[Subsequences[IntegerDigits[3^m], {n}], t], Return[m], m++]]; m]; Table[A215727[n], {n, 1, 14}] (* Robert Price, Oct 16 2018 *)
-
def A215727(n):
l, x = [str(d)*n for d in range(10)], 1
for m in range(10**9):
s = str(x)
for k in l:
if k in s:
return m
x *= 3
return 'search limit reached'
# Chai Wah Wu, Dec 17 2014
A215728
a(n) is the smallest m for which 5^m contains n consecutive identical digits.
Original entry on oeis.org
0, 11, 50, 95, 125, 1087, 2786, 2790, 2796, 2797, 2802, 2803, 1040703, 5547565, 7761841, 17876345
Offset: 1
-
def A215728(n):
l, x = [str(d)*n for d in range(10)], 1
for m in range(10**9):
s = str(x)
for k in l:
if k in s:
return m
x *= 5
return 'search limit reached'
# Chai Wah Wu, Dec 17 2014
A215729
a(n) is the smallest m for which 6^m contains (at least) n consecutive identical digits.
Original entry on oeis.org
0, 5, 5, 115, 226, 371, 1503, 8533, 27717, 27717, 69936, 848255, 1308931, 8255246, 32564822, 39063403
Offset: 1
-
def A215729(n):
l, x = [str(d)*n for d in range(10)], 1
for m in range(10**9):
s = str(x)
for k in l:
if k in s:
return m
x *= 6
return 'search limit reached'
# Chai Wah Wu, Dec 17 2014
A215730
a(n) is the smallest m for which 7^m contains n consecutive identical digits.
Original entry on oeis.org
0, 6, 31, 71, 172, 175, 1961, 6176, 33836, 61282, 305871, 856635, 2135396, 7291510, 11032874, 30775389
Offset: 1
7^31 = 157775382034845806615042743 contains 3 consecutive identical digits.
-
import sys
sys.set_int_max_str_digits(200000)
def a(n):
st = "0123456789"
for k in range(10**6):
s = str(7**k)
tot = 0
for i in st:
if s.count(i*n) > 0:
tot += 1
break
if tot > 0:
return k
n = 1
while n < 10:
print(a(n), end=', ')
n += 1
# Derek Orr, Jul 28 2014
-
def A215730(n):
l, x = [str(d)*n for d in range(10)], 1
for m in range(10**9):
s = str(x)
for k in l:
if k in s:
return m
x *= 7
return 'search limit reached'
# Chai Wah Wu, Dec 17 2014
A215733
a(n) is the first digit to appear n times in succession in a power of 3.
Original entry on oeis.org
1, 7, 8, 5, 5, 8, 2, 1, 2, 2, 2, 4, 5, 8, 2, 4, 4
Offset: 1
-
def A215733(n):
l, x = [str(d)*n for d in range(10)], 1
for m in range(10**9):
s = str(x)
for k in range(10):
if l[k] in s:
return k
x *= 3
return 'search limit reached'
# Chai Wah Wu, Dec 17 2014
A215731
a(n) is the smallest m for which the decimal representation of 11^m contains n consecutive identical digits.
Original entry on oeis.org
0, 1, 8, 39, 156, 482, 1323, 2983, 9443, 39879, 214747, 296095, 296095, 5541239, 8621384, 30789328
Offset: 1
The decimal representation of 11^39879 contains ten consecutive 6s, and is the least such power with such a string of digits.
-
mostDigits[t_] := Module[{lastDigit = t[[1]], record = 1, cnt = 1}, Do[If[t[[n]] == lastDigit, cnt++, If[cnt > record, record = cnt]; cnt = 1; lastDigit = t[[n]]], {n, 2, Length[t]}]; If[cnt > record, record = cnt] ; record]; nn = 10; t = Table[-1, {nn}]; n = -1; While[Min[t] == -1, n++; c = mostDigits[IntegerDigits[11^n]]; If[c > nn, c = nn]; While[c > 0 && t[[c]] == -1, t[[c]] = n; c--]]; t (* T. D. Noe, Apr 29 2013 *)
-
def A215731(n):
l, x = [str(d)*n for d in range(10)], 1
for m in range(10**9):
s = str(x)
for k in l:
if k in s:
return m
x *= 11
return 'search limit reached'
# Chai Wah Wu, Dec 17 2014
a(10) discovered by "Wick" (See http://www.mersenneforum.org/showpost.php?p=334789&postcount=89). Definition clarified and all terms to a(10) verified by
Daran Gill, Mar 24 2013
a(11) discovered by Tom Womack (See http://www.mersenneforum.org/showpost.php?p=337916&postcount=105),
Rick van der Hoorn, Apr 24 2013
A215734
a(n) is the first digit to appear n times in succession in a power of 5.
Original entry on oeis.org
1, 8, 8, 7, 4, 2, 8, 5, 5, 7, 5, 7, 4, 1, 5, 7
Offset: 1
A215735
a(n) is the first digit to appear n times in succession in a power of 6.
Original entry on oeis.org
1, 7, 7, 0, 2, 1, 2, 7, 1, 1, 0, 5, 7, 6, 3, 7
Offset: 1
A215737
a(n) is the first digit to appear n times in succession in a power of 11.
Original entry on oeis.org
1, 1, 8, 7, 6, 0, 6, 0, 9, 6, 6, 2, 2, 7, 6, 9
Offset: 1
-
n = 1; x = 1; lst = {};
For[i = 1, i <= 10000, i++,
z = Split[IntegerDigits[x]]; a = Length /@ z; b = Max[a];
For[j = n, j <= b, j++,
AppendTo[lst, First[First[Part[z, First[Position[a, b]]]]]]; n++
]; x = 11 x ]; lst (* Robert Price, Mar 16 2019 *)
-
def A215737(n):
a, s = 1, tuple(str(i)*n for i in range(10))
while True:
a *= 11
t = str(a)
for i, x in enumerate(s):
if x in t:
return i # Chai Wah Wu, Mar 30 2021
a(10)-a(13) added by
V. Raman, Apr 30 2012, in correspondence with
A215731.
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