A034887 -id:A034887 - cp's OEIS frontend

A158911 Numbers of the form 2^i*5^j - 1.

0, 1, 3, 4, 7, 9, 15, 19, 24, 31, 39, 49, 63, 79, 99, 124, 127, 159, 199, 249, 255, 319, 399, 499, 511, 624, 639, 799, 999, 1023, 1249, 1279, 1599, 1999, 2047, 2499, 2559, 3124, 3199, 3999, 4095, 4999, 5119, 6249, 6399, 7999, 8191, 9999, 10239

Offset: 1

Ctibor O. Zizka, Mar 30 2009

Numbers n such that 10^n is divisible by n+1.
Numbers n such that the prime divisors of n+1 are also divisors of the numbers m obtained by the concatenation of n and n+1. For example, for n=39, m = 3940, the divisors of 40 are {2, 5} and the divisors of 3940 are {2, 5, 197}. - Michel Lagneau, Dec 20 2011
The entries correspond to positional information of A156703, which stem from ratios of consecutive integers. For example, A156703(4)=875 yields a(5). This is because 875 was produced from n/(n+1) where n=7, i.e., 7/8 = 0.875.  Similarly, a(23)=399 stems from 399/400=0.9975 (A156703(22)). - Bill McEachen, Jan 05 2014

Robert Israel, Table of n, a(n) for n = 1..10000 (first 134 terms from Peter Pein)

Cf. A158520, A034887, A129344.

[n: n in [0..10^5] | Modexp(10, n, n+1) eq 0]; // Vincenzo Librandi, Mar 07 2018

N:= 20000: # to get all terms <= N
sort([seq(seq(2^i*5^j-1, j=0..floor(log[5]((N+1)/2^i))),i=0..ilog2(N+1))]); # Robert Israel, Mar 06 2018

fQ[n_] := PowerMod[10, n, n + 1] == 0; Select[ Range[0, 11000], fQ] (* Robert G. Wilson v, Sep 08 2010 *)

is(n)=n=(n+1)>>valuation(n+1,2);ispower(n,,&n);n==1||n==5 \\ Charles R Greathouse IV, Jan 12 2012

list(lim)=my(v=List(), N); lim++; for(n=0, log(lim)\log(5), N=5^n; while(N<=lim, listput(v, N-1); N<<=1)); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jan 12 2012

a(n) = A003592(n) - 1.

Corrected by Claudio Meller, Rick L. Shepherd, Charles R Greathouse IV and R. J. Mathar, Aug 23 2010

Edited by N. J. A. Sloane, Aug 25 2010, Oct 04 2010

A317873 Number of digits in 2^(n!).

Original entry on oeis.org

1, 1, 1, 2, 8, 37, 217, 1518, 12138, 109238, 1092378, 12016155, 144193850, 1874520045, 26243280622, 393649209329, 6298387349264, 107072584937472, 1927306528874488, 36618824048615255, 732376480972305082, 15379906100418406713, 338357934209204947674

Offset: 0

Wolfdieter Lang and Robert G. Wilson v, Aug 09 2018

The old definition (which did not match the data) was "Number of digits in the numerators of partial sums for Liouville's constant, read as base-2 (binary) numbers (A145572)."

Kevin Ryde, Table of n, a(n) for n = 0..300

Cf. A000142, A034887, A050923, A055642, A145572.

Digits := 900: # for n <= 300
a := n -> ceil(exp(lnGAMMA(n + 1))*log10(2)):
seq(a(n), n = 0..30);  # Peter Luschny, Apr 18 2024

Mathematica
```
Array[ Floor[#! Log10@2 + 1] &, 22]
```

a(n) = A034887(n!).

Better definition suggested by Martin Renner, Mar 24 2024

a(0)=1 prepended by Alois P. Heinz, Jul 27 2025

A348553 Number of digits in 11^n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75

Offset: 0

Seiichi Manyama, Oct 22 2021

			a(24) = 25 because 11^24 = 9849732675807611094711841, which has 25 digits.
a(25) = 27 because 11^25 = 108347059433883722041830251, which has 27 digits.

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Number of digits in b^n: A034887 (b=2), A034888 (b=3), A210434 (b=4), A210435 (b=5), A210436 (b=6), A210062 (b=7), this sequence (b=11).

Cf. A001020, A055642.

a[n_] := IntegerLength[11^n]; Array[a, 100, 0] (* Amiram Eldar, Oct 22 2021 *)

PARI
```
a(n) = #Str(11^n);
```

def a(n): return len(str(11**n))
print([a(n) for n in range(98)]) # Michael S. Branicky, Oct 22 2021

a(n) = A055642(A001020(n)) = A055642(11^n).

A371887 a(1) = 1; for n > 1, a(n) is the smallest positive integer k such that the digits of 2^k contain 2^a(n-1) as a proper substring.

Original entry on oeis.org

1, 5, 15, 507

Offset: 1

Adam Vulic, Apr 11 2024

From David A. Corneth, Apr 11 2024: (Start)
This sequence is well defined as A030000 is well defined; every finite string of digits is contained in some power of 2.
An upper bound for a(n), n > 1, can be found by solving 2^k == 2^a(n-1) (mod 10^m) where m is the number of digits of 2^a(n-1) (cf. A034887). This gives a(n) <= k = a(n-1) + 4*5^(m-1) (cf. A005054). So a(5) <= 507 + 4*5^152, which is about 7*10^106. (End)

			a(2) is the smallest k > 0 such that the digits of 2^k contain 2^a(1) = 2^1 = 2 as a proper substring, so a(2) = 5. (2^5 = 32.)
a(3) is the smallest k > 0 such that the digits of 2^k contain 2^a(2) = 32 as a proper substring, so a(3) = 15. (2^15 = 32768.)

Brady Haran, Apocalyptic Numbers, Numberphile video, 2024.

Cf. A018856, A005054, A030000, A034887.

k = 0; Rest@ NestList[(While[SequenceCount[IntegerDigits[2^k], IntegerDigits[2^#]] == 0, k++]; k++; k - 1) &, 1, 4] (* Michael De Vlieger, Apr 19 2024 *)

A106432 Levenshtein distance between successive powers of 2 in decimal representation.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 5, 5, 5, 6, 6, 5, 6, 6, 6, 6, 8, 8, 8, 9, 9, 8, 8, 8, 9, 8, 10, 10, 8, 10, 10, 11, 11, 11, 11, 10, 11, 13, 14, 13, 13, 14, 12, 11, 14, 10, 12, 14, 12, 16, 17, 16, 17, 17, 16, 15, 18, 17, 17, 18, 18, 17, 18, 20, 17, 16, 21, 19, 19, 20, 22, 20, 22, 21

Offset: 0

Reinhard Zumkeller, Jan 22 2006

a(n) = minimal number of editing steps (delete, insert or substitute) to transform 2^n into 2^(n+1) in decimal representation;

a(n) <= A034887(n).

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Michael Gilleland, Levenshtein Distance [It has been suggested that this algorithm gives incorrect results sometimes. - _N. J. A. Sloane_]
Haskell Wiki, Edit distance
WikiBooks: Algorithm Implementation, Levenshtein Distance
Wikipedia, Edit Distance
Wikipedia, Levenshtein Distance

Cf. A000079.

-- import Data.Function (on)
a106432 n = a106432_list !! n
a106432_list = zipWith (levenshtein `on` show)
                       a000079_list $ tail a000079_list where
   levenshtein us vs = last $ foldl transform [0..length us] vs where
      transform xs@(x:xs') c = scanl compute (x+1) (zip3 us xs xs') where
         compute z (c', x, y) = minimum [y+1, z+1, x + fromEnum (c' /= c)]
-- Reinhard Zumkeller, Nov 10 2013

levenshtein[s_List, t_List] := Module[{d, n = Length@s, m = Length@t}, Which[s === t, 0, n == 0, m, m == 0, n, s != t, d = Table[0, {m + 1}, {n + 1}]; d[[1, Range[n + 1]]] = Range[0, n]; d[[Range[m + 1], 1]] = Range[0, m]; Do[ d[[j + 1, i + 1]] = Min[d[[j, i + 1]] + 1, d[[j + 1, i]] + 1, d[[j, i]] + If[ s[[i]] === t[[j]], 0, 1]], {j, m}, {i, n}]; d[[ -1, -1]] ]]; Table[ levenshtein[IntegerDigits[2^n], IntegerDigits[2^(n + 1)]], {n, 0, 80}] (* Robert G. Wilson v *)

More terms from Robert G. Wilson v, Jan 25 2006

A115566 Numbers k such that 2^k, 2^(k+1) and 2^(k+2) have the same number of digits.

Original entry on oeis.org

1, 4, 7, 10, 11, 14, 17, 20, 21, 24, 27, 30, 31, 34, 37, 40, 41, 44, 47, 50, 51, 54, 57, 60, 61, 64, 67, 70, 71, 74, 77, 80, 81, 84, 87, 90, 91, 94, 97, 100, 103, 104, 107, 110, 113, 114, 117, 120, 123, 124, 127, 130, 133, 134, 137, 140, 143, 144, 147, 150, 153, 154

Offset: 1

Stefan Steinerberger, Mar 11 2006

The density of this sequence is 1 - 2*log_10(2) = 0.3979400086720376...

			2^4 = 16, 2^5 = 32, 2^6 = 64: all these numbers have two digits.
2^10 = 1024, 2^11 = 2048, 2^12 = 4096: all these numbers have three digits.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A001682 (same definition with 3 instead of 2).

Cf. A034887 (number of digits in 2^n).

[k:k in [1..160]|#Intseq(2^k) eq #Intseq(2^(k+2))]; // Marius A. Burtea, May 20 2019

select(n -> ilog10(2^n)=ilog10(2^(n+2)), [$1..1000]); # Robert Israel, May 19 2019

Select[Range[220], Floor[Log[10, 2]*# ] == Floor[Log[10, 2]*(# + 2)] &]

floor(log_10(2)*k) = floor(log_10(2)*(k+1)) = floor(log_10(2)*(k+2)).

A158520 Numbers n such that 2^n divided by the number of digits of 2^n is an integer.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 10, 11, 12, 13, 24, 25, 26, 50, 51, 52, 53, 103, 104, 105, 106, 210, 211, 212, 422, 423, 424, 425, 848, 849, 850, 1698, 1699, 1700, 3399, 3400, 3401, 6800, 6801, 6802, 6803, 13604, 13605, 13606, 27210, 27211, 27212, 27213

Offset: 1

Ctibor O. Zizka, Mar 20 2009

Also n such that A000079(n)/A034887(n) is an integer.

Cf. A000079, A034887

ndnQ[n_]:=Module[{n2=2^n},Divisible[n2,IntegerLength[n2]]]; Select[Range[ 0,28000],ndnQ] (* Harvey P. Dale, Apr 16 2013 *)

Extended by R. J. Mathar, Mar 26 2009

A192543 Let r be the largest real zero of x^n - x^(n-1) - x^(n-2) - ... - 1 = 0. Then a(n) is the value of k which satisfies the equation 0.5/10^k < 2 - r < 5/10^k.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22

Offset: 1

Ruskin Harding, Dec 31 2012

Same as A034887 except for the offset and a(1). - T. D. Noe, Feb 11 2013

			For n = 5, the root is approximately r = 1.96594823. The value of k that satisfies 0.5/10^k < 2-r < 5/10^k is 2 as 0.005 < 0.03405177 < 0.05. So a(5) = 2.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

a(n)=if(n>1, -log(4-2*solve(x=1.5,2,x^n-(1-x^n)/(1-x)))\log(10)+1, 0) \\ Charles R Greathouse IV, Jan 15 2013

A227689 a(n) is the least integer k such that 2^k - 1 has at least 10^n digits.

Original entry on oeis.org

1, 30, 329, 3319, 33216, 332190, 3321925, 33219278, 332192807, 3321928092, 33219280946, 332192809486, 3321928094885, 33219280948871, 332192809488733, 3321928094887360, 33219280948873621, 332192809488736232, 3321928094887362345, 33219280948873623476

Offset: 0

Olivier de Mouzon, Jul 19 2013

			For n = 2, A000225(328) has 99 digits and A000225(329) has 100 digits, so a(2) = 329.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Wikipedia, Great Internet Mersenne Prime Search

See A000225, A020862, A034887.

a(n) = ceil(log(10^(10^n-1)+1)/log(2)); \\ Michel Marcus, Jun 28 2021

a(n) = ceiling(log_2(10^(10^n-1)+1)).

Limit_{n -> oo} a(n)/10^n = log_2(10) = A020862. - Alois P. Heinz, Jun 28 2021

a(7)-a(19) from Alois P. Heinz, Jun 28 2021

A354782 Second digit from left in decimal expansion of 2^n (n >= 4).

Original entry on oeis.org

6, 2, 4, 2, 5, 1, 0, 0, 0, 1, 6, 2, 5, 3, 6, 2, 0, 0, 1, 3, 6, 3, 7, 3, 6, 3, 0, 1, 2, 5, 7, 4, 8, 3, 7, 4, 0, 1, 3, 7, 7, 5, 0, 4, 8, 6, 1, 2, 5, 0, 8, 6, 2, 4, 8, 7, 1, 3, 6, 2, 8, 6, 3, 4, 9, 9, 1, 3, 7, 4, 8, 7, 5, 5, 0, 0, 2, 4, 8, 6, 9, 8, 7, 5, 0, 1, 2, 4, 9, 9, 9, 9, 9, 5, 1, 3, 2, 5, 0, 0, 0, 0, 1, 6, 2, 4, 2, 5, 1

Offset: 4

N. J. A. Sloane, Jul 07 2022, following a suggestion from Alexander Wajnberg

			2^4 = 16, so a(4) = 6. 2^5 = 32, so a(5) = 2.

Alois P. Heinz, Table of n, a(n) for n = 4..10000

Cf. A000079, A000689, A008952, A034887, A160590.

a:= n-> parse(""||(2^n)[2]):
seq(a(n), n=4..112);  # Alois P. Heinz, Jul 07 2022

A354782[n_]:=IntegerDigits[2^n][[2]];Array[A354782,100,4] (* Paolo Xausa, Oct 22 2023 *)

def A354782(n): return int(str(1<Chai Wah Wu, Jul 07 2022

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A158911 Numbers of the form 2^i*5^j - 1.

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A317873 Number of digits in 2^(n!).

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A348553 Number of digits in 11^n.

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A371887 a(1) = 1; for n > 1, a(n) is the smallest positive integer k such that the digits of 2^k contain 2^a(n-1) as a proper substring.

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A106432 Levenshtein distance between successive powers of 2 in decimal representation.

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A115566 Numbers k such that 2^k, 2^(k+1) and 2^(k+2) have the same number of digits.

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A158520 Numbers n such that 2^n divided by the number of digits of 2^n is an integer.

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