A001913 -id:A001913 - cp's OEIS frontend

A060259 Denoting 4 consecutive primes by p, q, r and s, these are the values of q such that q and r have 10 as a primitive root, but p and s do not.

59, 109, 179, 229, 571, 701, 937, 1019, 1171, 1429, 1619, 1777, 1811, 1847, 2063, 2269, 2297, 2339, 2383, 2447, 2731, 2819, 2927, 3257, 3299, 3331, 3461, 3571, 3593, 3617, 3701, 3833, 3967, 4139, 4259, 4421, 4567, 4691, 4937, 5087, 5153, 5179, 5417

Offset: 1

Jeff Burch, Mar 23 2001

nonn

A prime p has 10 as a primitive root iff the length of the period of the decimal expansion of 1/p is p-1.

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

Cf. A001913, A002371, A060260, A060261, A060262.

test[p_] := MultiplicativeOrder[10, p]===p-1; Prime/@Select[Range[2, 800], test[Prime[ # ]]&&test[Prime[ #+1]]&&!test[Prime[ #-1]]&&!test[Prime[ #+2]]&]
Prime[#+1]&/@SequencePosition[Table[If[MultiplicativeOrder[10,p]===p-1,1,0],{p,Prime[Range[ 800]]}],{0,1,1,0}][[;;,1]] (* Harvey P. Dale, Nov 29 2023 *)

Edited by Dean Hickerson, Jun 17 2002

Offset corrected by Amiram Eldar, Oct 03 2021

A060260 Numbers k such that prime(k), prime(k+1) and prime(k+2) have 10 as a primitive root, but prime(k-1) and prime(k+3) do not.

Original entry on oeis.org

55, 75, 141, 164, 184, 199, 358, 371, 380, 432, 559, 702, 745, 808, 825, 858, 882, 1077, 1097, 1279, 1299, 1303, 1328, 1408, 1431, 1486, 1502, 1558, 1654, 1702, 1724, 1744, 1768, 1820, 1835, 1873, 1901, 1905, 1953, 1977, 2050, 2148, 2216, 2220, 2267

Offset: 1

Jeff Burch, Mar 23 2001

nonn

A prime p has 10 as a primitive root iff the length of the period of the decimal expansion of 1/p is p-1.

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

The corresponding primes are in A060261.

Cf. A001913, A002371, A060259, A060262.

test[p_] := MultiplicativeOrder[10, p]===p-1; Select[Range[2, 2500], test[Prime[ # ]]&&test[Prime[ #+1]]&&test[Prime[ #+2]]&&!test[Prime[ #-1]]&&!test[Prime[ #+3]]&]

Edited by Dean Hickerson, Jun 17 2002

A060261 Denoting 5 consecutive primes by p, q, r, s and t, these are the values of q such that q, r and s have 10 as a primitive root, but p and t do not.

Original entry on oeis.org

257, 379, 811, 971, 1097, 1217, 2411, 2539, 2617, 3011, 4051, 5297, 5657, 6211, 6337, 6659, 6857, 8647, 8807, 10457, 10651, 10687, 10937, 11731, 11939, 12451, 12577, 13099, 14011, 14537, 14731, 14887, 15137, 15607, 15737, 16091, 16411

Offset: 1

Jeff Burch, Mar 23 2001

nonn

A prime p has 10 as a primitive root iff the length of the period of the decimal expansion of 1/p is p-1.

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

The indices of these primes are in A060260.

Cf. A001913, A002371, A060259, A060262.

test[p_] := MultiplicativeOrder[10, p]===p-1; Prime/@Select[Range[2, 2500], test[Prime[ # ]]&&test[Prime[ #+1]]&&test[Prime[ #+2]]&&!test[Prime[ #-1]]&&!test[Prime[ #+3]]&]

Edited by Dean Hickerson, Jun 17 2002

Offset corrected by Amiram Eldar, Oct 03 2021

A060262 a(n) is the smallest k such that prime(k), prime(k+1), ..., prime(k+n-1) all have 10 as a primitive root, but prime(k-1) and prime(k+n) do not.

Original entry on oeis.org

4, 17, 55, 7, 93, 754, 2611, 31092, 55207, 301252, 955428, 805428, 3651249, 3686621, 5510710, 42337888, 109670084, 590903433, 1010572448

Offset: 1

Jeff Burch, Mar 23 2001

A prime p has 10 as a primitive root iff the length of the period of the decimal expansion of 1/p is p-1.

a(21) = 9774718453 and a(23) = 9525468065. - Amiram Eldar, Oct 03 2021

Cf. A001913, A002371, A060259, A060260, A060261.

test[p_] := MultiplicativeOrder[10, p]===p-1; For[n=1, n<100, n++, a[n]=0]; v=4; While[True, For[n=1, test[Prime[v+n]], n++, Null]; If[a[n]==0, a[n]=v; Print["a(", n, ") = ", v]]; For[v+=n+1, !test[Prime[v]], v++, Null]]

Edited by Dean Hickerson, Jun 17 2002

a(13)-a(19) from Amiram Eldar, Oct 03 2021

A060282 Periodic part of decimal expansion of reciprocal of n-th prime (leading 0's omitted).

Original entry on oeis.org

0, 3, 0, 142857, 9, 76923, 588235294117647, 52631578947368421, 434782608695652173913, 344827586206896551724137931, 32258064516129, 27, 2439, 23255813953488372093, 212765957446808510638297872340425531914893617

Offset: 1

N. J. A. Sloane, Mar 30 2001

			1/7 = 0.142857142..., so a(4) = 142857.
1/11 = 0.09090909..., so a(5) = 9.

T. D. Noe, Table of n, a(n) for n=1..100

Cf. A002371, A036275, A060283, A060251, A036275, A001913.

primePer[1] = primePer[3] = 0; primePer[n_] := FromDigits[(d = RealDigits[1/Prime[n]])[[1, 1]]] * 10^d[[2]]; Array[ primePer, 15] (* Amiram Eldar, Apr 28 2020 *)

f(n)=if(n<4,n==2,znorder(Mod(10, prime(n)))) \\ A002371
for(n=1,100,print1(floor(10^f(n)/prime(n)),","))

a(n) = floor(10^A002371(n)/prime(n)).
a(n) = 0 if and only if n = 1 or 3, corresponding to the primes 2 and 5, which are factors of 10. - Alonso del Arte, Apr 03 2020
ceiling(log_10(a(n))) = prime(n) - 1 if prime(n) is a full reptend prime (A001913). - Alonso del Arte, Apr 14 2020

More terms from Klaus Brockhaus, Mar 30 2001

A087023 Maximal exponent in prime factorization of n-th cyclic number.

Original entry on oeis.org

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

Offset: 1

Reinhard Zumkeller, Jul 30 2003

A004042(n) factorized with Dario Alpern's ECM.

Extended using factors of 10^(A001913(n)-1)-1, see Kamada link.

			A004042(2) = 142857 = 37*13*11*3^3, therefore a(1) = 3.

Dario A. Alpern, Factorization using the Elliptic Curve Method.
Makoto Kamada, Factorizations of 11...11 (Repunit)
Eric Weisstein's World of Mathematics, Cyclic Number

Cf. A001913, A087020, A087021, A087022, A087024, A087025, A087026.

a(n) = A051903(A004042(n+1)).

a(12)-a(42) from Ray Chandler, Nov 16 2011

a(43)-a(51) from Max Alekseyev, May 14 2022

A180340 Numbers with x digits such that the first x multiples are cyclic permutations of the number, leading 0's omitted (or cyclic numbers).

Original entry on oeis.org

142857, 588235294117647, 52631578947368421, 434782608695652173913, 344827586206896551724137931, 212765957446808510638297872340425531914893617, 169491525423728813559322033898305084745762711864406779661

Offset: 1

Ralph Kerchner (daxkerchner(AT)hotmail.com), Aug 28 2010

Periodic part of decimal expansion of 1/A001913(n). The number of digits in each term (including leading zeros), plus one, makes the sequence A001913.

			142857 is in the sequence because it has 6 digits and the first 6 multiples of 142857 are 142857, 285714, 428571, 571428, 714285, and 857142, all cyclic permutations of the number. Also the first term of A001913 is 7, and 1/7 = 0.142857142857... .
588235294117647 is the next number because 0588235294117647 has 16 digits and the first 16 multiples are cyclic permutations of the number; the second term of A001913 is 17, and 1/17 = 0.05882352941176470588235294117647... .

Ray Chandler, Table of n, a(n) for n = 1..60
Edwin E. Freed, Binary Magic Numbers, Dr. Dobb's Journal, Vol. 78 (April 1983), pp. 24-37.
OEIS Wiki, Cyclic numbers
Eric Weisstein's World of Mathematics, Cyclic number
Wikipedia, Cyclic number

A006883 starting from the second term of A006883, omitting ending 0's.

The n-th terms of A060284 where n is a member of A001913.

Map[(10^(# - 1) - 1)/# &, Select[Prime@ Range@ 17, MultiplicativeOrder[10, #] == # - 1 &]] (* Michael De Vlieger, Apr 03 2017 *)

a(n) = (10^(A001913(n)-1) - 1) / A001913(n).

A221981 Primes q = 4*p+1, where p == 2 (mod 5) is also prime.

Original entry on oeis.org

29, 149, 269, 389, 509, 1109, 1229, 1949, 2309, 2909, 3989, 4349, 5189, 5309, 6269, 6389, 7109, 7949, 8069, 9749, 10589, 10709, 11069, 11549, 12149, 12269, 13229, 13829, 14549, 15629, 16229, 17189, 17789, 18269, 19949, 20789, 22109, 22229, 24029, 24989, 25349, 25469, 25589, 26189, 26309, 28109, 28229, 28949, 29669, 30029, 30869, 31469, 32069, 33149, 34589, 34949, 36269, 36629, 36749, 37589

Offset: 1

Jonathan Sondow, Feb 02 2013

nonn

Moree (2012) says that Chebyshev observed that if q = 4p + 1 is prime, with prime p == 2 (mod 5), then 10 is a primitive root modulo q.
If the sequence is infinite, then Artin's conjecture ("every nonsquare integer n != -1 is a primitive root of infinitely many primes q") is true for n = 10.
The corresponding primes p are A221982.
The sequence is infinite under Dickson's conjecture, thus Dickson's conjecture implies Artin's conjecture for n = 10. - Charles R Greathouse IV, Apr 18 2013
Two conjectures: (a) These primes have primitive root 40; (b) if a(n)*8 + 1 is prime then it has primitive root 10. - Davide Rotondo, Dec 31 2024

			29 is a member because 29 = 4*7 + 1 and 7 == 2 (mod 5) are prime.

P. L. Chebyshev, Theory of congruences, Elements of number theory, Chelsea, 1972, p. 306.
Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section F9, pp. 377-380.

Amiram Eldar, Table of n, a(n) for n = 1..10000
P. L. Chebyshev, Theorie der Congruenzen, Mayer & Mueller, 1889, pp. 306-313.
Pieter Moree, Artin's primitive root conjecture - a survey, arXiv:math/0412262 [math.NT], 2004, revised 2012, p. 1.
Index entries for sequences related to Artin's conjecture
Index entries for primes by primitive root

Cf. A001913, A005596, A006883, A045380, A106849, A221982, A222008.

A221981:=n->`if`(isprime(4*n+1) and isprime(n) and n mod 5 = 2, 4*n+1, NULL): seq(A221981(n), n=1..10^4); # Wesley Ivan Hurt, Dec 11 2015

Select[ Prime[ Range[4000]], Mod[(# - 1)/4, 5] == 2 && PrimeQ[(# - 1)/4] &]

is(n)=n%20==9 && isprime(n) && isprime(n\4) \\ Charles R Greathouse IV, Apr 18 2013

a(n) = 4*A221982(n) + 1.

a(n) >> n log^2 n. - Charles R Greathouse IV, Dec 30 2024

A000353 Primes p == 7, 19, 23 (mod 40) such that (p-1)/2 is also prime.

Original entry on oeis.org

7, 23, 47, 59, 167, 179, 263, 383, 503, 863, 887, 983, 1019, 1367, 1487, 1619, 1823, 2063, 2099, 2207, 2447, 2459, 2579, 2819, 2903, 3023, 3167, 3623, 3779, 3863, 4007, 4127, 4139, 4259, 4703, 5087, 5099, 5807, 5927, 5939, 6047, 6659, 6779, 6899, 6983, 7247

Offset: 1

Robert A. J. Matthews

nonn

The decimal expansion of 1/a(n) will produce a stream of a(n)-1 pseudo-random digits. - Reinhard Zumkeller, Feb 10 2009

The condition in the name is sufficient for primes p such that the decimal expansion of 1/p recurs after p-1 digits, which is the maximum-possible cycle length. - Robert A. J. Matthews, Oct 31 2023

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Robert A. J. Matthews, Maximally periodic reciprocals, Bull. Institute of Mathematics and Its Applications, vol. 28, p. 147-148, 1992.
Wikipedia, Sophie Germain prime
Index entries for sequences related to decimal expansion of 1/n.

Subset of A005385.

Subsequence of A001913, A006883.

q:= p-> irem(p, 40) in {7, 19, 23} and andmap(isprime, [p, (p-1)/2]):
select(q, [$1..10000])[];  # Alois P. Heinz, Oct 31 2023

Select[Prime[Range[1000]], MatchQ[Mod[#, 40], 7|19|23] && PrimeQ[(#-1)/2]&] (* Jean-François Alcover, Feb 07 2016 *)

is(n)=my(k=n%40); (k==7||k==19||k==23) && isprime(n\2) && isprime(n) \\ Charles R Greathouse IV, Nov 20 2014

a(n) = 2*A000355(n)+1. - Reinhard Zumkeller, Feb 10 2009

More terms from Reinhard Zumkeller, Feb 10 2009

A069531 Smallest positive k such that 10^k + 1 is divisible by n, or 0 if no such number exists.

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 3, 0, 0, 0, 1, 0, 3, 0, 0, 0, 8, 0, 9, 0, 0, 0, 11, 0, 0, 0, 0, 0, 14, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 23, 0, 21, 0, 0, 0, 0, 0, 0, 0, 0, 0, 29, 0, 30, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 22, 0, 3, 0, 0, 0, 0, 0, 48, 0, 0, 0, 2

Offset: 1

Amarnath Murthy, Apr 01 2002

nonn

When a(n) is not zero, it is a divisor of phi(n). If n is a prime with primitive root 10 (cf. A001913) then a(n) = (n-1)/2.

			a(7) = a(13) = 3 as 1001 is divisible by 7 and 13. a(17) = 8 as 17 divides 100000001 = 10^8 + 1.

Antti Karttunen, Table of n, a(n) for n = 1..10001

Cf. A069521 to A069530.

Cf. A000010, A001913, A002329.

A069531(n) = { fordiv(eulerphi(n),k,if(!((1+(10^k))%n),return(k))); (0); }; \\ Antti Karttunen, Aug 23 2019

More terms from Vladeta Jovovic, Apr 03 2002

cp's OEIS Frontend

A060259 Denoting 4 consecutive primes by p, q, r and s, these are the values of q such that q and r have 10 as a primitive root, but p and s do not.

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A060260 Numbers k such that prime(k), prime(k+1) and prime(k+2) have 10 as a primitive root, but prime(k-1) and prime(k+3) do not.

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A060261 Denoting 5 consecutive primes by p, q, r, s and t, these are the values of q such that q, r and s have 10 as a primitive root, but p and t do not.

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A060262 a(n) is the smallest k such that prime(k), prime(k+1), ..., prime(k+n-1) all have 10 as a primitive root, but prime(k-1) and prime(k+n) do not.

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A060282 Periodic part of decimal expansion of reciprocal of n-th prime (leading 0's omitted).

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A087023 Maximal exponent in prime factorization of n-th cyclic number.

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A180340 Numbers with x digits such that the first x multiples are cyclic permutations of the number, leading 0's omitted (or cyclic numbers).

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A221981 Primes q = 4*p+1, where p == 2 (mod 5) is also prime.

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