Olivier de Mouzon - cp's OEIS frontend

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

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

A227683 Number of digits in n-th Mersenne number.

Original entry on oeis.org

1, 1, 2, 3, 4, 4, 6, 6, 7, 9, 10, 12, 13, 13, 15, 16, 18, 19, 21, 22, 22, 24, 25, 27, 30, 31, 32, 33, 33, 35, 39, 40, 42, 42, 45, 46, 48, 50, 51, 53, 54, 55, 58, 59, 60, 60, 64, 68, 69, 69, 71, 72, 73, 76, 78, 80, 81, 82, 84, 85, 86, 89, 93, 94, 95, 96, 100

Offset: 1

Olivier de Mouzon, Jul 19 2013

			For n = 5, the fifth prime number is 11, so 2^11 - 1 = 2047 is the fifth Mersenne number, which has 4 digits, so a(5) = 4.

Cf. A000040, A001348.

a227683(n) = floor(log(2^prime(n)-1)/log(10)) + 1 \\ Michael B. Porter, Jul 20 2013

a(n) = floor(log_10(A001348(n))) + 1.

A227616 Number of bits set to 1 in the binary representation of the n-th term of the Lucas-Lehmer sequence (A003010).

Original entry on oeis.org

1, 3, 3, 5, 12, 30, 58, 128, 237, 476, 975, 1956, 3899, 7798, 15534, 31270, 62262, 124635, 248944, 497797, 995730, 1990576, 3983767, 7969049, 15935289, 31870309, 63739461, 127519282, 254994762, 510016513, 1020092276, 2040066241, 4080236749

Offset: 0

Olivier de Mouzon, Jul 17 2013

			For n = 2, A003010(2) = 11000010 (in binary), so a(2) = 3.

Cf. A003010, A227608, A227615.

read("transforms") :
A227616 := proc(n)
    wt(A003010(n)) ;
end proc: # R. J. Mathar, Jul 20 2013

First@ DigitCount[#, 2] & /@ NestList[#^2 - 2 &, 4, 28] (* Michael De Vlieger, Apr 04 2016 *)

lista(nn) = {a = 4; print1(hammingweight(a), ", "); for (n=1, nn, a = a^2-2; print1(hammingweight(a), ", "););} \\ Michel Marcus, Apr 04 2016

a(n) = A000120(A003010(n)).

Terms from a(19) on from Michel Marcus, Apr 04 2016

A227615 Number of bits necessary to represent u(n) in binary, where u is the Lucas-Lehmer sequence: u(0) = 100 (in binary); for n>0, u(n) = u(n-1)^2 - 2.

Original entry on oeis.org

3, 4, 8, 16, 31, 61, 122, 244, 487, 973, 1946, 3892, 7783, 15565, 31130, 62259, 124517, 249033, 498066, 996131, 1992262, 3984524, 7969047, 15938093, 31876185, 63752369, 127504737, 255009473, 510018945, 1020037890, 2040075780

Offset: 0

Olivier de Mouzon, Jul 17 2013

a(0)=3, a(1)=4 and for n>=1, a(n+1) is 2*a(n) or 2*a(n)-1.
It seems the rule to decide between the 2 is not straightforward. So you actually need to compute u(n) to have its required number of bits.
Yet, for n>=1, we have a lower bound: a(n) >= 2^n and an upper bound: a(n) <= 2^(n+1).

			For n=2, u(2) = 194, log_2(u(2)) is between 7.5 and 7.6, so E(log_2(u(2))) = 7, so a(2) = E(log_2(u(2))) + 1 = 8. And indeed, u(2) = 194 (in base 10) = 11000010 in base 2 requires 8 bits (all bits above are 0).

Cf. A003010, A070939, A227608.

lista(nn) = {a = 4; print1(#binary(a), ", "); for (n=1, nn, a = a^2-2; print1(#binary(a), ", "););} \\ Michel Marcus, Apr 04 2016

a(n) = E(log_2(u(n))) + 1, where E(x) is the integer part of x and u is defined by: u(0) = 4 (or 100 in binary) and for n>0, u(n) = u(n-1)^2 - 2.

a(n) = A070939(A003010(n)). - Michel Marcus, Apr 04 2016

Terms from a(19) on from Michel Marcus, Apr 04 2016

A227537 Number of Mersenne primes that have between 10^n and 10^(n+1) - 1 digits.

Original entry on oeis.org

7, 5, 6, 8, 5, 6, 7

Offset: 0

Olivier de Mouzon, Jul 18 2013

The nice property of this sequence is that (at least up to n = 6) there seems to be a rather stable number of Mersenne primes for each digit number group [10^n ... 10^(n+1) - 1].
At the moment (Jul 18 2013), there are already 4 Mersenne primes in the next group (n = 7), the last one was discovered on Jan 25 2013 and has 17425170 digits.
Note that for n = 6, a(n) = 7 still needs full confirmation, as tests for all factors between M42 = M_25964951 and M_44457869 (more than 10^7 digits) have only made once and a double check is needed to confirm a(6) = 7.
If this sequence were to actually be stable, this would mean that the number of Mersenne primes having between 10^n and 10^(n+1) - 1 digits is always around 6, when the number of prime numbers in the same digit number group constantly increases: around 2.3*10^(10^(n+1)-(n+1)). Also the number of Mersenne numbers in the same digit group constantly increases (though much less than the number of prime numbers): 9*10^n/[(n+1)*log(2) + log(log(10)/log(2))*log(2)/log(10)]. So, if a(n) is really rather stable (around 6), Mersenne primes frequency among Mersenne numbers lower than x is converging towards 0 in the magnitude of [log(log(x))]^2/log(x). Hence primes are still around 6*[log(log(x))]^2 more frequent among Mersenne numbers than among numbers.

			For n = 1, a(n) = 5 Mersenne primes with 10 to 99 digits, which are:
* M8 = M_31 = 2147483647,
* M9 = M_61 = 2305843009213693951,
* M10 = M_89 = 618970019642690137449562111,
* M11 = M_107 = 162259276829213363391578010288127,
* M12 = M_127 = 170141183460469231731687303715884105727.

GIMPS, Great Internet Mersenne Prime Search official Home Page and Great Internet Mersenne Prime Search milestones
Wikipedia, Great Internet Mersenne Prime Search or more up to date the French version: Great Internet Mersenne Prime Search (FR)

Cf. A000043, A000668, A028335.

A049064 Describe the previous term in binary (method A - initial term is 0).

Original entry on oeis.org

0, 10, 1110, 11110, 100110, 1110010110, 111100111010110, 100110011110111010110, 1110010110010011011110111010110, 1111001110101100111001011010011011110111010110, 1001100111101110101100111100111010110111001011010011011110111010110

Offset: 1

Olivier de Mouzon

Method A = 'frequency' (in binary mode) followed by 'digit'-indication.

The number of digits of a(n) is A001609(n) except for n = 2. See the link from T. Sillke below. - Jianing Song, Mar 16 2019

			E.g., the term after 11110 is obtained by saying "four (i.e., 100 in binary mode) 1, one 0", which gives 100110.

Kade Robertson, Table of n, a(n) for n = 1..18
Kade Robertson, Table of n, a(n) for n = 1..31
T. Sillke, The binary form of Conway's sequence

Cf. A001387 (initial term is 1), A001391, A001609 (number of digits), A259710 (written in decimal).

Decimal look-and-say sequences: A005150, A006751, A006715, A001140, A001141, A001143, A001145, A001151, A001154.

a(n) = A001391(n-1), n > 1. - R. J. Mathar, Oct 15 2008

Edited by Charles R Greathouse IV, Apr 06 2010
a(11) from Kade Robertson, Jun 24 2015
Offset corrected by Jianing Song, Mar 16 2019

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User: Olivier de Mouzon

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

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A227683 Number of digits in n-th Mersenne number.

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A227616 Number of bits set to 1 in the binary representation of the n-th term of the Lucas-Lehmer sequence (A003010).

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A227615 Number of bits necessary to represent u(n) in binary, where u is the Lucas-Lehmer sequence: u(0) = 100 (in binary); for n>0, u(n) = u(n-1)^2 - 2.

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A227537 Number of Mersenne primes that have between 10^n and 10^(n+1) - 1 digits.

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A049064 Describe the previous term in binary (method A - initial term is 0).

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