A007733 -id:A007733 - cp's OEIS frontend

A119513 a(n) = A119957(n) / n.

0, 0, 1, 0, 2, 1, 1, 0, 3, 2, 5, 1, 6, 1, 1, 0, 4, 3, 9, 2, 2, 5, 4, 1, 10, 6, 9, 1, 14, 1, 1, 0, 5, 4, 5, 3, 18, 9, 4, 2, 10, 2, 7, 5, 5, 4, 9, 1, 10, 10, 2, 6, 26, 9, 8, 1, 9, 14, 29, 1, 30, 1, 1, 0, 6, 5, 33, 4, 11, 5, 14, 3, 3, 18, 9, 9, 15, 4, 17, 2, 27, 10, 41, 2, 2, 7, 11, 5, 4, 5, 4, 4, 3, 9, 14

Offset: 1

Néstor Romeral Andrés, Aug 04 2006

			a(35) = A119957(35) / 35 = 175 / 35 = 5.

Andrei Zabolotskii, Table of n, a(n) for n = 1..8192

Cf. A119957, A007733, A038553. The positions of 1's are A175332.

a[n_] := Total[PowerMod[2, Range[n-MultiplicativeOrder[2, n/2^IntegerExponent[n, 2]], n-1], n]] / n;
Table[a[n], {n,95}] (* Andrei Zabolotskii, Jul 28 2025 *)

Offset corrected by Andrei Zabolotskii, Jul 28 2025

A163956 Multiplicative order of 2 in Z/mZ with m = A002997(n).

Original entry on oeis.org

40, 24, 36, 56, 60, 660, 198, 252, 45, 180, 60, 144, 153, 1012, 36, 120, 300, 72, 36, 1160, 60, 36, 300, 56, 36, 660, 4284, 264, 420, 3060, 2268, 180, 540, 1680, 120, 4900, 1080, 396, 72, 72, 60, 60, 144, 2970, 612, 396, 324, 210, 180, 540, 504, 792, 198, 180

Offset: 1

A.K. Devaraj, Aug 28 2009

nonn

Related sequence: A162990. - A.K. Devaraj, Aug 31 2009

A. K. Devaraj, "Minimum universal exponent generalisation of Fermat's theorem" (ISSN 1550-3747)

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

Cf. A002326, A002997, A007733, A104016, A162990, A165139.

MultiplicativeOrder[2, #] & /@ Select[Range[1, 10^6, 2], CompositeQ[#] && Divisible[# - 1, CarmichaelLambda[#]] &] (* Amiram Eldar, Jul 30 2020 *)

a(n) = A002326((A002997(n)-1)/2) = A007733(A002997(n)). - Amiram Eldar, Jul 30 2020

Corrected and extended by M. F. Hasler, Sep 23 2009
Edited by N. J. A. Sloane, Sep 23 2009, following suggestions from M. F. Hasler
More terms from Amiram Eldar, Jul 30 2020

A204983 a(n) = 2^(k-1)-2^(j-1), where (2^(k-1),2^(j-1)) is the least pair of distinct positive powers of 2 for which n divides 2^(k-1)-2^(j-1).

Original entry on oeis.org

1, 2, 3, 4, 15, 6, 7, 8, 63, 30, 1023, 12, 4095, 14, 15, 16, 255, 126, 262143, 60, 63, 2046, 2047, 24, 1048575, 8190, 262143, 28, 268435455, 30, 31, 32, 1023, 510, 4095, 252, 68719476735, 524286, 4095, 120, 1048575, 126, 16383, 4092, 4095

Offset: 1

Clark Kimberling, Jan 21 2012

nonn

For a guide to related sequences, see A204892.

(Conjecture) Equivalently, the solution set of 2^p * (2^q - 1) = x * y, OR 2^q - 1 = 2^p * x * y, for at most one of the naturals x and y being given; unknown p and q in the integers; then a(n) = 2^p * (2^q - 1) where p and q are directly related to n (see formula). - Andrew T. Porter, Dec 20 2022

Cf. A204979, A204892, A204984.
Cf. A000079, A173787.
Cf. A007733, A007814.

Mathematica
```
(See the program at A204979.)
```

a(n) = for (k=1, oo, for (j=1, k-1, my(d=2^(k-1)-2^(j-1)); if (!(d % n), return(d)););); \\ Michel Marcus, Sep 16 2023

Conjecture: a(n) = 2^A007814(n) * (2^A007733(n) - 1). - Andrew T. Porter, Dec 20 2022

A226181 Primes p such that p-1 divided by the period of the binary expansion of 1/p equals 2^x for some nonnegative integer x.

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 19, 23, 29, 37, 41, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 113, 131, 137, 139, 149, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 227, 233, 239, 257, 263, 269, 271, 281, 293, 311, 313, 317, 337, 347, 349

Offset: 1

Lear Young, May 30 2013

Equivalently, p-1 divided by the period of the decimal expansion of 1/p equals 2^x for some nonnegative integer x. Composite numbers satisfying this condition are given in A243050. - Lear Young, May 30 2013

Let pi_1(x) and pi(x) be the numbers of primes of this sequence and all primes not exceeding x, respectively. Then, for x>=3, p_1(x)/pi(x) >= C_Artin = 0.37395581... Numerical results suggest that it is likely lim pi_1(x)/pi(x) = 2*C_Artin. - Peter J. C. Moses and Vladimir Shevelev, May 29 2014

			(41-1)/20 = 2. 20 is the period of the binary representation of 1/n, the odd part of 2 is 1.

Lear Young, Table of n, a(n) for n = 1..1000

Cf. A007733, A136042.

Select[Prime[Range[2, 100]], # == 2^IntegerExponent[#, 2] &[(# - 1)/MultiplicativeOrder[2, #]] &] (* Peter J. C. Moses, May 28 2014 *)

is(n) = {
  m = valuation(n+1,2);
      k=(n+1)>>m;
      if(k!=1, for(i=0,(n-1)>>1,
        l=valuation(k+n,2);
        k=(k+n)>>l;
        m+=l;if(k==1,break)));
       ((n-1)/m)>>valuation((n-1)/m, 2)==1
       \\ m  equals znorder(Mod(2,n))
    }
forstep(i=3,1e3,2,if(is(i),print1(i, ", ")))
\\ Lear Young May 30 2013

forstep(i=1,1e3,2,j = (i-1)/znorder(Mod(2,i));if(j>>valuation(j, 2)==1,print1(i, ", "))) \\ Lear Young May 31 2013

A246503 Numbers m such that m^2 divides 2^k - 1 for some k, 0 < k <= m.

Original entry on oeis.org

1, 1093, 3279, 3511, 5465, 7651, 9837, 10533, 14209, 16395, 17555, 18581, 22953, 24577, 31599, 31697, 38255, 38621, 42627, 45643, 46999, 49185, 52665, 53557, 55743, 57929, 60115, 62301, 66709, 68859, 71045, 73731, 84161, 86347, 92905, 94797, 95091, 99463

Offset: 1

Max Alekseyev, Nov 29 2014

nonn

All terms are odd. m=1 is the only term with k=m.
Odd numbers m such that A007733(m^2) = A002326((m^2-1)/2) <= m.
Prime terms are Wieferich primes (A001220).

Chai Wah Wu, Table of n, a(n) for n = 1..127

Cf. A001220, A002326.

A246503_list = [1]
for i in range(2, 10**4):
    d, n = i*i, 1
    for _ in range(i):
        n = (2*n) % d
        if n == 1:
            A246503_list.append(i)
            break # Chai Wah Wu, Dec 04 2014

A250211 Square array read by antidiagonals: A(m,n) = multiplicative order of m mod n, or 0 if m and n are not coprime.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 1, 2, 1, 1, 0, 0, 0, 1, 1, 1, 1, 2, 4, 1, 1, 0, 2, 0, 4, 0, 1, 1, 1, 0, 1, 2, 0, 3, 1, 1, 0, 1, 0, 0, 0, 6, 0, 1, 1, 1, 2, 2, 1, 2, 3, 2, 6, 1, 1, 0, 0, 0, 4, 0, 6, 0, 0, 0, 1, 1, 1, 1, 1, 4, 1, 2, 2, 3, 4, 10, 1, 1, 0, 2, 0, 2, 0, 0, 0, 6, 0, 5, 0, 1, 1, 1, 0, 2, 0, 0, 1, 2, 0, 0, 5, 0, 12, 1

Offset: 1

Eric Chen, Dec 29 2014

Read by antidiagonals:
m\n   1    2    3    4    5    6    7    8    9    10    11    12    13
1     1    1    1    1    1    1    1    1    1     1     1     1     1
2     1    0    2    0    4    0    3    0    6     0    10     0    12
3     1    1    0    2    4    0    6    2    0     4     5     0     3
4     1    0    1    0    2    0    3    0    3     0     5     0     6
5     1    1    2    1    0    2    6    2    6     0     5     2     4
6     1    0    0    0    1    0    2    0    0     0    10     0    12
7     1    1    1    2    4    1    0    2    3     4    10     2    12
8     1    0    2    0    4    0    1    0    2     0    10     0     4
9     1    1    0    1    2    0    3    1    0     2     5     0     3
10    1    0    1    0    0    0    6    0    1     0     2     0     6
11    1    1    2    2    1    2    3    2    6     1     0     2    12
12    1    0    0    0    4    0    6    0    0     0     1     0     2
13    1    1    1    1    4    1    2    2    3     4    10     1     0
etc.
A(m,n) = Least k>0 such that m^k=1 (mod n), or 0 if no such k exists.
It is easy to prove that column n has period n.
A(1,n) = 1, A(m,1) =1.
If A(m,n) differs from 0, it is period length of 1/n in base m.
The maximum number in column n is psi(n) (A002322(n)), and all numbers in column n (except 0) divide psi(n), and all factors of psi(n) are in column n.
Except the first row, every row contains all natural numbers.

			A(3,7) = 6 because:
3^0 = 1 (mod 7)
3^1 = 3 (mod 7)
3^2 = 2 (mod 7)
3^3 = 6 (mod 7)
3^4 = 4 (mod 7)
3^5 = 5 (mod 7)
3^6 = 1 (mod 7)
...
And the period is 6, so A(3,7) = 6.

Cf. A002322, A111076, A111725, A001918, A008330, A007733, A002326, A007732, A051626, A066799.

See A139366 for another version.

f:= proc(m,n)
  if igcd(m,n) <> 1 then 0
  elif n=1 then 1
  else numtheory:-order(m,n)
  fi
end proc:
seq(seq(f(t-j,j),j=1..t-1),t=2..65); # Robert Israel, Dec 30 2014

a250211[m_, n_] = If[GCD[m, n] == 1, MultiplicativeOrder[m, n], 0]
Table[a250211[t-j, j], {t, 2, 65}, {j, 1, t-1}]

A279189 Primes p such that L(p^2) = (p-1)*L(p), where L(i) = A279186(i).

Original entry on oeis.org

2, 3, 5, 29, 179, 293, 317, 467, 509, 659, 797, 1427, 1949, 2213, 2339, 2579, 2909, 3677, 4157, 4229, 4253, 4349, 5309, 5573, 5693, 5843, 5939, 6173, 6269, 6653, 6899, 6947, 7043, 7517, 7589, 8387, 8573, 8819, 9059, 9533, 10067, 10163, 10259, 10589, 11069, 11549, 11939, 13763, 14627, 15443

Offset: 1

N. J. A. Sloane, Dec 14 2016

nonn

Also, union of {2} and the primes p from A001122 such that gcd(p-1,A007733(p-1)) = 1. - Max Alekseyev, Feb 02 2024

Haifeng Xu, The largest cycles consist by the quadratic residues and Fermat primes, arXiv:1601.06509 [math.NT], 2016.

Excluding a(1)=2, forms a subsequence of A001122.

Cf. A279185, A279186, A279187, A279188.

T[n_, k_] := Module[{g, y, r}, If[k == 0, Return[1]]; y = n; g = GCD[k, y]; While[g > 1, y = y/g; g = GCD[k, y]]; If[y == 1, Return[1]]; r = MultiplicativeOrder[k, y]; r = r/2^IntegerExponent[r, 2]; If[r == 1, Return[1]]; MultiplicativeOrder[2, r]];
L[n_] := L[n] = Table[T[n, k], {k, 0, n - 1}] // Max;
For[p = 2, p < 1000, p = NextPrime[p], If[L[p^2] == (p-1) L[p], Print[p]]] (* Jean-François Alcover, Oct 07 2018, after Robert Israel in A279186 *)

a(8)-a(11) from Jean-François Alcover, Oct 07 2018

Terms a(12) onward from Max Alekseyev, Feb 02 2024

A074203 Odd numbers k such that the number of 1's in the binary representation of k divides 2^k-1.

Original entry on oeis.org

1, 351, 375, 381, 471, 477, 501, 687, 699, 747, 855, 861, 885, 939, 981, 1119, 1143, 1149, 1239, 1245, 1269, 1311, 1335, 1341, 1359, 1371, 1383, 1389, 1395, 1401, 1431, 1437, 1461, 1479, 1485, 1491, 1497, 1509, 1521, 1623, 1629, 1653, 1707, 1749, 1815

Offset: 1

Benoit Cloitre, Sep 17 2002

Except for 1, terms seem always divisible by 3.
From Robert Israel, Jan 14 2019: (Start)
An odd number k is in the sequence if and only if A000120(k) is in A036259 and k is divisible by A007733(A000120(k)).  In particular, there are infinitely many of these for every member of A036259 except 1.
Thus a(2) to a(28842) have A000120(k)=7 and are divisible by 3, but a(28843) = 12582911 has A000120(12582911) = 23 and is divisible by A007733(23) = 11 but not by 3. (End)

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

Cf. A000120, A007733, A036259, A074202.

filter:= n -> 2 &^ n - 1 mod convert(convert(n,base,2),`+`) = 0:
select(filter, [seq(i,i=1..2000,2)]); # Robert Israel, Jan 13 2019

Join[{1}, Select[Range[3, 2000, 2], PowerMod[2, #, DigitCount[#, 2, 1]] == 1 &]] (* Amiram Eldar, Jun 08 2022 *)

isok(n) = (n % 2) && !((2^n-1) % hammingweight(n)); \\ Michel Marcus, Nov 29 2013

A104013 First digit-cycle of binary expansion of 1/n. Any initial 0's are to be placed at end of cycle.

Original entry on oeis.org

0, 0, 10, 0, 1100, 10, 100, 0, 111000, 1100, 1011101000, 10, 100111011000, 100, 1000, 0, 11110000, 111000, 110101111001010000, 1100, 110000, 1011101000, 10110010000, 10, 10100011110101110000, 100111011000, 100101111011010000

Offset: 1

Bryan Jacobs (bryanjj(AT)gmail.com), Feb 25 2005

			1/5 = 0.00110011001100... in binary, so a(5) = 1100.

Andrey Zabolotskiy, Table of n, a(n) for n = 1..1018

Cf. A007733, A036275, A104014.

f[n_] := If[IntegerQ@ Log2@ n, 0, FromDigits[ RealDigits[1/n, 2][[1, 1]]]]; Array[f, 27] (* Robert G. Wilson v, Sep 01 2015 *)

A137332 Primes which are equal to the order of 2 modulo a prime q, sorted with respect to the value of q.

Original entry on oeis.org

2, 3, 11, 5, 23, 11, 7, 83, 37, 29, 131, 179, 191, 43, 73, 239, 251, 359, 419, 431, 443, 491, 29, 659, 683, 233, 179, 719, 743, 911, 239, 1019, 1031, 29, 1103, 47, 397, 1223, 79, 461, 1439, 1451, 1499, 1511, 1559, 1583, 557, 113, 431, 577, 601, 1811, 1931

Offset: 1

Joerg Arndt, Apr 07 2008

nonn

This is a multipermutation of the primes A000040 with every prime p appearing exactly A001221(2^p-1) times. - Max Alekseyev, May 01 2008

			The k-th term of the sequence is ord(2 mod A122094(k)).
For example, 223 is the 9th term of A122094 and ord(2 mod 223)=37, so 37 is the 9th term of this sequence.
11 is both the third term because ord(2 mod 23) == 11 and the sixth term because ord(2 mod 89) == 11.
Note both 23 and 89 divide 2^11-1; the third and sixth terms of A122094 are 23 and 89.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..106 from Joerg Arndt)

Cf. A002326, A122094.

Select[MultiplicativeOrder[2, #] & /@ Select[Range[3, 4000, 2], PrimeQ], PrimeQ] (* Amiram Eldar, Apr 04 2020 *)

forprime (p=3, 10^4, r = znorder( Mod(2,p) ); if ( isprime(r), print1(r, ", "); ); );

a(n) = A007733(A122094(n)) = A002326((A122094(n)-1)/2). - Max Alekseyev, May 01 2008

Definition revised by Max Alekseyev, May 01 2008

cp's OEIS Frontend

A119513 a(n) = A119957(n) / n.

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A163956 Multiplicative order of 2 in Z/mZ with m = A002997(n).

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A204983 a(n) = 2^(k-1)-2^(j-1), where (2^(k-1),2^(j-1)) is the least pair of distinct positive powers of 2 for which n divides 2^(k-1)-2^(j-1).

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A226181 Primes p such that p-1 divided by the period of the binary expansion of 1/p equals 2^x for some nonnegative integer x.

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A246503 Numbers m such that m^2 divides 2^k - 1 for some k, 0 < k <= m.

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A250211 Square array read by antidiagonals: A(m,n) = multiplicative order of m mod n, or 0 if m and n are not coprime.

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A279189 Primes p such that L(p^2) = (p-1)*L(p), where L(i) = A279186(i).

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A074203 Odd numbers k such that the number of 1's in the binary representation of k divides 2^k-1.

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