A003060 -id:A003060 - cp's OEIS frontend

A059892 a(n) = |{m : multiplicative order of 10 mod m is equal to n}|.

3, 3, 5, 6, 9, 53, 9, 36, 12, 33, 9, 186, 21, 33, 111, 144, 9, 564, 3, 330, 239, 273, 3, 1756, 84, 165, 76, 714, 93, 16167, 21, 5952, 111, 177, 363, 4288, 21, 15, 99, 5724, 45, 48807, 45, 4314, 1140, 183, 9, 14192, 36, 2940, 495, 1338, 45, 11572, 747, 11484

Offset: 1

Vladeta Jovovic, Feb 06 2001

The multiplicative order of a mod m, gcd(a,m)=1, is the smallest natural number d for which a^d = 1 (mod m).
The number of unit fractions 1/k having a decimal expansion of period n and with k coprime to 10. - T. D. Noe, May 18 2007
Also, number of primitive factors of 10^n - 1 (cf. A003060). - Max Alekseyev, May 03 2022
a(n) is odd if and only if n is squarefree. Proof: Note that 10^d - 1 == 3 (mod 4) for d >= 2, so 10^d - 1 is a square if and only if d = 1. From the formula we can see that a(n) is odd if and only if mu(n) is nonzero, or n is squarefree. - Jianing Song, Jun 15 2021

Table of n, a(n) for n = 1..352

Number of primitive factors of b^n - 1: A059499 (b=2), A059885(b=3), A059886 (b=4), A059887 (b=5), A059888 (b=6), A059889 (b=7), A059890 (b=8), A059891 (b=9), this sequence (b=10).
Cf. A000005, A008683, A002329, A007138, A005422, A057951, A058946, A070528.
Column k=10 of A212957.

with(numtheory):
a:= n-> add(mobius(n/d)*tau(10^d-1), d=divisors(n)):
seq(a(n), n=1..30);  # Alois P. Heinz, Oct 12 2012

f[n_, d_] := MoebiusMu[n/d]*Length[Divisors[10^d - 1]]; a[n_] := Total[(f[n, #] & ) /@ Divisors[n]]; Table[a[n], {n, 1, 56}] (* Jean-François Alcover, Mar 21 2011 *)

j=[]; for(n=1,10,j=concat(j,sumdiv(n,d,moebius(n/d)*numdiv(10^d-1)))); j

from sympy import divisors, mobius, divisor_count
def a(n): return sum(mobius(n//d)*divisor_count(10**d - 1) for d in divisors(n)) # Indranil Ghosh, Apr 23 2017

a(n) = Sum_{d|n} mu(n/d)*tau(10^d-1), (mu(n) = Moebius function A008683, tau(n) = number of divisors of n A000005).

More terms from Jason Earls, Aug 06 2001.
Terms to a(280) in b-file from T. D. Noe, Oct 01 2013
a(281)-a(322) in b-file from Ray Chandler, May 03 2017
a(323)-a(352) in b-file from Max Alekseyev, May 03 2022

A070528 Number of divisors of 10^n-1 (999...999 with n digits).

Original entry on oeis.org

3, 6, 8, 12, 12, 64, 12, 48, 20, 48, 12, 256, 24, 48, 128, 192, 12, 640, 6, 384, 256, 288, 6, 2048, 96, 192, 96, 768, 96, 16384, 24, 6144, 128, 192, 384, 5120, 24, 24, 128, 6144, 48, 49152, 48, 4608, 1280, 192, 12, 16384, 48, 3072, 512, 1536, 48, 12288, 768

Offset: 1

Henry Bottomley, May 02 2002

			a(7)=12 since the divisors of 9999999 are 1, 3, 9, 239, 717, 2151, 4649, 13947, 41841, 1111111, 3333333, 9999999.

Table of n, a(n) for n = 1..352

Cf. A004023 (indices of 6's).
Cf. A018282, A018766, A027894, A027893, A027892, A027891, A027890, A027889, A027895.
Cf. A046801, A005422, A102347, A046053, A057951, A007138, A003060, A059892, A085035.

DivisorSigma[0,#]&/@(10^Range[60]-1) (* Harvey P. Dale, Jan 14 2011 *)
Table[DivisorSigma[0, 10^n - 1], {n, 60}] (* T. D. Noe, Aug 18 2011 *)

a(n) = numdiv(10^n - 1); \\ Michel Marcus, Sep 08 2015

a(n) = A000005(A002283(n)).
a(n) = Sum_{d|n} A059892(d).
a(n) = A070529(n)*(A007949(n)+3)/(A007949(n)+1).

Terms to a(280) in b-file from Hans Havermann, Aug 19 2011
a(281)-a(322) in b-file from Ray Chandler, Apr 22 2017
a(323)-a(352) in b-file from Max Alekseyev, May 04 2022

A085035 Number of prime factors of cyclotomic(n,10), which is A019328(n), the value of the n-th cyclotomic polynomial evaluated at x=10.

Original entry on oeis.org

2, 1, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 3, 1, 2, 2, 2, 2, 1, 2, 3, 4, 1, 1, 3, 2, 3, 3, 5, 3, 3, 5, 2, 3, 3, 1, 3, 1, 1, 2, 4, 4, 4, 3, 2, 4, 2, 1, 2, 3, 4, 2, 4, 2, 4, 2, 3, 2, 2, 3, 7, 1, 5, 4, 2, 2, 3, 3, 3, 2, 2, 3, 3, 3, 3, 2, 4, 5, 6, 2, 6, 2, 3, 2, 3, 3, 3

Offset: 1

T. D. Noe, Jun 19 2003

nonn

The Mobius transform of this sequence yields A057951, number of prime factors of 10^n-1.

See references at A085021.

Max Alekseyev, Table of n, a(n) for n = 1..352 (first 322 terms from Ray Chandler)
Makoto Kamada, Factorizations of Phi_n(10)

omega(Phi(n,x)): A085021 (x=2), A085028 (x=3), A085029 (x=4), A085030 (x=5), A085031 (x=6), A085032 (x=7), A085033 (x=8), A085034 (x=9), this sequence (x=10).

Cf. A001222, A003060, A005422, A007138, A019328, A057951, A070528, A059892.

Table[Plus@@Transpose[FactorInteger[Cyclotomic[n, 10]]][[2]], {n, 1, 100}]

a(n) = A001222(A019328(n)). - Ray Chandler, May 10 2017

A212953 Minimal order of degree-n irreducible polynomials over GF(2).

Original entry on oeis.org

1, 3, 7, 5, 31, 9, 127, 17, 73, 11, 23, 13, 8191, 43, 151, 257, 131071, 19, 524287, 25, 49, 69, 47, 119, 601, 2731, 262657, 29, 233, 77, 2147483647, 65537, 161, 43691, 71, 37, 223, 174763, 79, 187, 13367, 147, 431, 115, 631, 141, 2351, 97, 4432676798593, 251

Offset: 1

Alois P. Heinz, Jun 01 2012

nonn

a(n) = smallest odd m such that A002326((m-1)/2) = n. - Thomas Ordowski, Feb 04 2014

For n > 1; n < a(n) < 2^n, wherein a(n) = n+1 iff n+1 is A001122 a prime with primitive root 2, or a(n) = 2^n-1 iff n is a Mersenne exponent A000043. - Thomas Ordowski, Feb 08 2014

			For n=4 the degree-4 irreducible polynomials p over GF(2) are 1+x+x^2+x^3+x^4, 1+x+x^4, 1+x^3+x^4. Their orders (i.e., the smallest integer e for which p divides x^e+1) are 5, 15, 15. (Example: (1+x+x^2+x^3+x^4) * (1+x) == x^5+1 (mod 2)). Thus the minimal order is 5 and a(4) = 5.

W. Narkiewicz, Elementary and Analytic Theory of Algebraic Numbers, Springer 2004, Third Edition, 4.3 Factorization of Prime Ideals in Extensions. More About the Class Group (Theorem 4.33), 4.4 Notes to Chapter 4 (Theorem 4.40). - Regarding the first comment.

Max Alekseyev, Table of n, a(n) for n = 1..1236 (first 179 terms from Alois P. Heinz)
Eric Weisstein's World of Mathematics, Irreducible Polynomial
Eric Weisstein's World of Mathematics, Polynomial Order

Cf. A003060, A059912, A213224.

with(numtheory):
M:= proc(n) option remember;
      divisors(2^n-1) minus U(n-1)
    end:
U:= proc(n) option remember;
      `if`(n=0, {}, M(n) union U(n-1))
    end:
a:= n-> min(M(n)[]):
seq(a(n), n=1..50);

M[n_] := M[n] = Divisors[2^n-1] ~Complement~ U[n-1];
U[n_] := U[n] = If[n == 0, {}, M[n] ~Union~ U[n-1]];
a[n_] := Min[M[n]];
Array[a, 50] (* Jean-François Alcover, Mar 22 2017, translated from Maple *)

a(n) = min(M(n)) with M(n) = {d : d|(2^n-1)} \ U(n-1) and U(n) = M(n) union U(n-1) for n>0, U(0) = {}.

a(n) = A059912(n,1) = A213224(n,1).

A170945 Least number k such that the decimal representation of 1/k has period Fibonacci(n).

Original entry on oeis.org

3, 11, 27, 41, 73, 53, 43, 103, 1321, 497867, 323, 467, 11311, 20141, 12169, 232159532264041847249

Offset: 2

Michel Lagneau, Feb 19 2010

The period of 1/k is the least integer p such that 10^p = 1 (mod k). The integer p is also known as the multiplicative order of 10 (mod k).

			p(k) is the period of 1/k, we obtain with k=3,11,27,41,73,53,43,103 p(3)=1,p(11)=2,p(27)=3,p(41)=5,p(73)=8, p(53)=13,p(43)=21, p(103)=34

Mohammad K. Azarian, The Generating Function for the Fibonacci Sequence, Missouri Journal of Mathematical Sciences, Vol. 2, No. 2, Spring 1990, pp. 78-79. Zentralblatt MATH, Zbl 1097.11516.
Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, 2001.
S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989.

Dario Alejandro Alpern, Factorization using the Elliptic Curve Method
C. K. Caldwell, Fibonacci Numbers

Cf. A000045, A039834 (signed Fibonacci numbers), A002329, A072859 (periodic sequences), A003060

For the great numbers (p > 70), the maple program is very slow. That's what we use an process of two steps: factoring 10^p-1 with elliptic curve method (see the first web site), and then, for each factor q(k), k=1,2,...,r computation the periods of 1/q(k) and keep the period q(i) such that q(i) = Fibonacci number. The 17th term required 3h 2m for the computing of (10^1597) -1 T:=array(0..100);U:=array(0..100); n0:=0:n1:=1:T[1] = 1:for i from 2 to 30 do: n2:=n0+n1:T[i]:=n2:n0:=n1:n1:=n2:od:U[1]:=3:U[2]:=3:for q from 3 to 10 do: p0:=T[q]: indic:=0:for n from 1 to 2000 do:for p from 1 to 150 while(irem(10^p,n)<>1 or gcd(n,10)<>1 ) do:od: if irem(10^p,n) = 1 and gcd(n,10) = 1 and p=p0 and indic=0 then U[q]:=n:indic:=1:else fi:od: od: for n from 1 to 10 do:print( U[n]):od:

Edited by T. D. Noe, Apr 14 2010

A291943 a(0)=0; for n>0, a(n) = (2n)-th digit after the decimal point in the decimal expansion of 1/(2n+1).

Original entry on oeis.org

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

Offset: 0

Marco Matosic, Sep 06 2017

			a(3)=7 since we want the sixth decimal digit of 1/7.

John H. Conway & Richard K. Guy, The Book of Numbers; Springer 1996.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A002371, A003060, A006883, A036275, A051626, A061480, A114205, A121090, A270392.

f:= proc(n) floor(10^(2*n)/(2*n+1)) mod 10 end proc:
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Oct 31 2017

f[n_] := Mod[Floor[10^(2n)/(2n +1)], 10]; f[0] = 0; Array[f, 105, 0] (* Robert G. Wilson v, Oct 31 2017 *)

Edited by N. J. A. Sloane, Oct 30 2017

a(82) corrected by Robert Israel, Oct 31 2017

A381370 Smallest number with reciprocal of period length n in base 9.

Original entry on oeis.org

1, 2, 5, 7, 32, 11, 35, 547, 17, 19, 25, 23, 224, 398581, 29, 31, 128, 103, 95, 1597, 352, 43, 115, 47, 97, 151, 53, 109, 928, 59, 155, 683, 256, 161, 515, 71, 608, 18427, 7985, 79, 187, 83, 203, 431, 89, 181, 235, 1223, 896, 491, 101

Offset: 0

Erich Friedman, Feb 25 2025

For n > 1, a(n) is the smallest positive d such that d divides 9^n - 1 and does not divide any of 9^k - 1 for 0 < k < n.

			a(3)=7 because 1/7 has period 3 in base 9 (.125125125...) and no smaller number has this property.

J. Brillhart et al., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.

Cf. A003060, A379642.

a[n_] := First[Select[Divisors[9^n - 1], MultiplicativeOrder[9, #] == n &, 1]];
a[0] = 1; a[1] = 2; Table[a[n], {n, 0, 50}]

from sympy import divisors
def A381370(n): return next(d for d in divisors(9**n-1) if d>1 and all(pow(9,k,d)!=1 for k in range(1,n))) if n else 1 # Chai Wah Wu, Feb 28 2025

A381493 Smallest number with reciprocal of period length n in base 8.

Original entry on oeis.org

1, 7, 3, 73, 5, 31, 19, 49, 17, 262657, 11, 23, 37, 79, 43, 631, 97, 103, 81, 32377, 25, 3577, 67, 47, 323, 601, 237, 2593, 29, 233, 209, 2147483647, 193, 199, 307, 71, 405, 223, 571, 937, 187, 13367, 817, 431, 115, 271, 139, 2351, 577, 343, 251

Offset: 0

Erich Friedman, Feb 25 2025

For n > 1, a(n) is the smallest positive d such that d divides 8^n - 1 and does not divide any of 8^k - 1 for 0 < k < n.

			a(3)=73 because 1/73 has period 3 in base 8 (.007007007...) and no smaller number has this property.

J. Brillhart et al., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.

Cf. A003060, A379641.

a[n_] := First[Select[Divisors[8^n - 1], MultiplicativeOrder[8, #] == n &, 1]];
a[0] = 1; a[1] = 7; Table[a[n], {n, 0, 50}]

from sympy import divisors
def A381493(n):
    if n == 0: return 1
    for d in divisors(8**n-1):
        if d>1 and all(pow(8,k,d)!=1 for k in range(1,n)):
            return d # Chai Wah Wu, Feb 28 2025

A381494 Smallest number with reciprocal of period length n in base 7.

Original entry on oeis.org

1, 2, 4, 9, 5, 2801, 36, 29, 64, 27, 11, 1123, 13, 16148168401, 113, 31, 17, 14009, 108, 419, 55, 261, 23, 47, 73, 2551, 53, 81, 145, 59, 99, 311, 256, 3631, 56036, 81229, 135, 223, 1676, 486643, 41, 83, 1017, 166003607842448777, 115, 837, 188, 13722816749522711, 153, 3529, 10204

Offset: 0

Erich Friedman, Feb 25 2025

For n > 1, a(n) is the smallest positive d such that d divides 7^n - 1 and does not divide any of 7^k - 1 for 0 < k < n.

			a(3)=9 since 1/9 has period 3 in base 7 (.053053053...) and no smaller number has this property.

J. Brillhart et al., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.

Cf. A003060, A379640.

f:= proc(n) local d,k;
      for d in sort(convert(numtheory:-divisors(7^n-1),list)) do
        if andmap(k -> 7^k-1 mod d <> 0, [$1 .. n-1]) then return d fi
      od
end proc:
f(0):= 1: f(1):= 2:
map(f, [$0..80]); # Robert Israel, Feb 28 2025

a[n_] := First[Select[Divisors[7^n - 1], MultiplicativeOrder[7, #] == n &, 1]];
a[0] = 1; a[1] = 2; Table[a[n], {n, 0, 50}]

from sympy import divisors
def A381494(n): return next(d for d in divisors(7**n-1) if d>1 and all(pow(7,k,d)!=1 for k in range(1,n))) if n else 1 # Chai Wah Wu, Feb 28 2025

Conjecture: a(n) = A218358(n) for n>=2. - R. J. Mathar, Mar 03 2025

A176351 Numbers n such that 2*3^n + 1 is a primitive prime factor of 10^3^n - 1.

Original entry on oeis.org

4, 180, 320, 5480, 12096, 17720, 82780, 1175232

Offset: 1

T. D. Noe, Apr 15 2010

Consider the problem of finding the smallest number k such that the decimal representation of 1/k has period 3^e for a given e. The number k is usually 3^(e+2). However, if e is one of the n in this sequence, then the prime 2*3^n+1 is a smaller k. The first instance of these exceptions is 1/163, which has a period of 81.
Subsequence of A003306.
10 must be a square residue modulo 2*3^n + 1, implying that n must be a multiple of 4.

Cf. A003306 (primes of the form 2*3^n+1), A003060 (least k such that 1/k has period n).

Select[Range[10000], PrimeQ[1+2*3^# ] && MultiplicativeOrder[10,1+2*3^# ] == 3^# &]

Two more terms from Max Alekseyev, May 03 2010

cp's OEIS Frontend

A059892 a(n) = |{m : multiplicative order of 10 mod m is equal to n}|.

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A070528 Number of divisors of 10^n-1 (999...999 with n digits).

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A085035 Number of prime factors of cyclotomic(n,10), which is A019328(n), the value of the n-th cyclotomic polynomial evaluated at x=10.

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A212953 Minimal order of degree-n irreducible polynomials over GF(2).

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A170945 Least number k such that the decimal representation of 1/k has period Fibonacci(n).

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A291943 a(0)=0; for n>0, a(n) = (2n)-th digit after the decimal point in the decimal expansion of 1/(2n+1).

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A381370 Smallest number with reciprocal of period length n in base 9.

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