A020862 -id:A020862 - cp's OEIS frontend

A129344 a(n) is the number of powers of 2 that have n decimal digits.

4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3

Offset: 1

Tanya Khovanova, May 28 2007

Ignoring the first term, first differences of A066343. - Andrew Woods, Jun 10 2013

			a(1) is 4 because there are 4 one-digit powers of 2: 1, 2, 4, 8.

Amiram Eldar, Table of n, a(n) for n = 1..10000

First differences of A067497.

Cf. A000079, A020862, A066343.

Table[Transpose[ Select[Table[{n, 2^n}, {n, 0, 310}], IntegerDigits[ #[[2]]][[1]] == 1 &]][[1]][[k]] - Transpose[ Select[Table[{n, 2^n}, {n, 0, 310}], IntegerDigits[ #[[2]]][[1]] == 1 &]][[1]][[k - 1]], {k, 2, 94}]
Join[{4}, Differences @ Table[Floor[n*Log2[10]], {n, 100}]] (* Amiram Eldar, Apr 09 2021 *)

a(n) = my(k=0, i=0); while(#Str(2^k)!=n, k++); while(#Str(2^k)==n, i++; k++); i \\ Felix Fröhlich, Jan 19 2016

def A129344(n): return -(m:=5**(n-1)).bit_length()+(5*m).bit_length()+1 if n>1 else 4 # Chai Wah Wu, Sep 08 2024

For n>1, a(n) = floor(n*L)-floor((n-1)*L) where L = log(10)/log(2). - Andrew Woods, Jun 10 2013

Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log_2(10) (A020862). - Amiram Eldar, Apr 09 2021

A137284 a(0)=1 and a(n) for n > 0 equals the minimal positive integer such that addition of 2^(-a(n)) to Sum_{k = 0,1,...,n-1} 2^(-a(k)) changes only trailing zeros in its decimal representation.

Original entry on oeis.org

1, 4, 14, 47, 157, 522, 1735, 5764, 19148, 63609, 211305, 701941, 2331798, 7746066, 25731875, 85479439, 283956550, 943283242, 3133519104, 10409325148, 34579029658, 114869050115, 381586724811, 1267603661786, 4210888217270, 13988267873380, 46468020047392

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Mar 14 2008

First and last nonzero decimal digits of 2^(-m) appear respectively at the ceiling(m/log_2(10))-th and m-th positions after the point. Hence a(n+1) equals the minimum solution to ceiling(x/log_2(10)) = a(n) + 1, which is x = ceiling(a(n)*log_2(10)).

			Start from 0;
0 + 2^(-1) = 0.5;
0.5 + 2^(-4) = 0.5625 (first digit "5" is equal to the decimal of previous number);
0.5625 + 2^(-14) = 0.56256103515625 (first digits "5625" are equal to the decimals of previous number);
etc.

a(n+1) = ceiling(a(n)*log_2(10)) = ceiling(a(n)*A020862). - Conjectured by R. J. Mathar, proved by Max Alekseyev

Edited and extended by Max Alekseyev, May 13 2009

A242347 Number of decimal digits of A008559.

Original entry on oeis.org

1, 2, 4, 10, 31, 100, 330, 1093, 3628, 12049, 40023, 132951, 441651, 1467130, 4873698, 16190071, 53782249, 178660761, 593498199, 1971558339

Offset: 1

J. Lowell, May 11 2014

a(n+1)/a(n) is approximately log_2(10) = A020862. - André Engels, Apr 01 2021

			a(3) = 4 because 1010 has 4 decimal digits.

Code Golf StackExchange, Length of Binary as Base 10 [OEIS A242347], coding challenge started Oct 28 2022.

Cf. A008559, A055642, A020862.

a242347(n) = {my (k=2, d=digits); while(n--, k=fromdigits(d(k,2))); #d(k)} \\ Hugo Pfoertner, Nov 04 2022

A242347_list, l = [1], 2
for _ in range(10):
    l = int(bin(l)[2:])
    A242347_list.append(len(str(l))) # Chai Wah Wu, Dec 26 2014

a(n) = A055642(A008559(n)). - Michel Marcus, May 11 2014

a(1), a(18)-a(20) from Chai Wah Wu, Dec 26 2014

A352040 a(n) is the least number k such that 2^k contains each of the 10 digits at least n times.

Original entry on oeis.org

68, 88, 119, 200, 209, 246, 291, 318, 396, 398, 443, 443, 495, 586, 592, 622, 646, 707, 758, 758, 813, 866, 875, 903, 923, 1001, 1022, 1022, 1105, 1111, 1111, 1231, 1243, 1245, 1327, 1342, 1419, 1453, 1453, 1454, 1534, 1536, 1537, 1626, 1676, 1699, 1699, 1763

Offset: 1

Amiram Eldar, Apr 16 2022

a(1)-a(8) were given in the solution to Problem 410 (Crux Mathematicorum, 1979), but a(2) = 170 is wrong.

a(1) was calculated by Rudolph Ondrejka in 1976.

			a(1) = 68 since 2^68 = 295147905179352825856 is the least power of 2 that contains all the 10 digits at least once.

Amiram Eldar, Table of n, a(n) for n = 1..10000
James Gary Propp, Problem 410, Crux Mathematicorum, Vol. 5, No. 1 (1979), p. 17; Solution to Problem 410, ibid., Vol. 5, No. 10 (1979), pp. 296-299.

Cf. A000079, A007377, A020862, A130694, A130696.

a:= proc(n) option remember; local k; for k from a(n-1)
      while min((p-> seq(coeff(p, x, j), j=0..9))(add(
            x^i, i=convert(2^k, base, 10))))Alois P. Heinz, Apr 22 2022

s = Table[Min[DigitCount[2^n, 10, Range[0, 9]]], {n, 1, 2500}]; Table[FirstPosition[s, _?(# >= n &)], {n, 1, Max[s]}] // Flatten

from sympy import ceiling, log
def A352040(n):
    k = 10*n-1+int(ceiling((10*n-1)*log(5,2)))
    s = str(c := 2**k)
    while any(s.count(d) < n for d in '0123456789'):
        c *= 2
        k += 1
        s = str(c)
    return k   # Chai Wah Wu, Apr 16 2022

Conjecture: a(n) ~ c*n, where c = 10*log_2(10) = 33.21928... .

a(n) >= (10n-1)*log_2(10), i.e., c = 10*log_2(10) is a lower bound on the asymptotic growth rate. - Chai Wah Wu, Apr 16 2022

A120357 a(n) is the smallest prime p such that 2^p-1 (a Mersenne number) contains 10^n or more decimal digits.

Original entry on oeis.org

2, 31, 331, 3319, 33223, 332191, 3321937, 33219281, 332192831, 3321928097, 33219280951, 332192809589, 3321928094941, 33219280948907, 332192809488739, 3321928094887411, 33219280948873687, 332192809488736253

Offset: 0

G. L. Honaker, Jr., Jun 25 2006

For n>0 almost all digits of a(n) from the left are equal to the first terms of the expansion Log[10]/Log[2] = {3, 3, 2, 1, 9, 2, 8, 0, 9, 4, 8, 8, 7, 3, 6, 2, 3, 4, 7, 8, 7, 0, 3, 1, 9, 4, 2, 9, 4, 8, 9, 3, 9, ...} = A020862(n). - Alexander Adamchuk, Jan 16 2007

			E.g. a(7)=33219281 because 2^33219281-1 is the smallest Mersenne number that contains 10^7 (ten million) or more decimal digits.

Farideh Firoozbakht, Table of n, a(n) for n = 0..29
G. L. Honaker, Jr. and Chris Caldwell, Prime Curios! 33219281
Author?, GIMPS [Broken link?]

Cf. A001348.

Cf. A020862 = decimal expansion of log(10)/log(2).

More terms from Farideh Firoozbakht, Jul 22 2006

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

cp's OEIS Frontend

A129344 a(n) is the number of powers of 2 that have n decimal digits.

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A137284 a(0)=1 and a(n) for n > 0 equals the minimal positive integer such that addition of 2^(-a(n)) to Sum_{k = 0,1,...,n-1} 2^(-a(k)) changes only trailing zeros in its decimal representation.

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A242347 Number of decimal digits of A008559.

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A352040 a(n) is the least number k such that 2^k contains each of the 10 digits at least n times.

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Maple

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A120357 a(n) is the smallest prime p such that 2^p-1 (a Mersenne number) contains 10^n or more decimal digits.

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Extensions

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

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PARI

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Extensions