cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A307627 Primes p such that 2 is a primitive root modulo p while 8 is not.

Original entry on oeis.org

13, 19, 37, 61, 67, 139, 163, 181, 211, 349, 373, 379, 421, 523, 541, 547, 613, 619, 661, 709, 757, 787, 829, 853, 859, 877, 883, 907, 1117, 1123, 1171, 1213, 1237, 1291, 1381, 1453, 1483, 1531, 1549, 1621, 1669, 1693, 1741, 1747, 1861, 1867, 1987, 2029, 2053
Offset: 1

Views

Author

Jianing Song, Apr 19 2019

Keywords

Comments

Primes p such that 2 is a primitive root modulo p (i.e., p is in A001122) and that p == 1 (mod 3).
According to Artin's conjecture, the number of terms <= N is roughly ((2/5)*C)*PrimePi(N), where C is the Artin's constant = A005596, PrimePi = A000720. Compare: the number of terms of A001122 that are no greater than N is roughly C*PrimePi(N).

Examples

			For p = 67, the multiplicative order of 2 modulo 67 is 66, while 8^22 == 2^(3*22) == 1 (mod 67), so 67 is a term.
		

Crossrefs

Complement of A019338 with respect to A001122.
Cf. also A005596, A000720, A307628.

Programs

  • Maple
    select(p -> isprime(p) and numtheory:-order(2,p) = p-1,
    [seq(i,i=1..10000,6)]); # Robert Israel, Apr 23 2019
  • Mathematica
    Select[Prime@ Range[5, 310], And[FreeQ[#, 8], ! FreeQ[#, 2]] &@ PrimitiveRootList@ # &] (* Michael De Vlieger, Apr 23 2019 *)
  • PARI
    forprime(p=3, 2000, if(znorder(Mod(2, p))==(p-1) && p%3==1, print1(p, ", ")))

A307628 Primes p such that 2 is a primitive root modulo p while 32 is not.

Original entry on oeis.org

11, 61, 101, 131, 181, 211, 421, 461, 491, 541, 661, 701, 821, 941, 1061, 1091, 1171, 1291, 1301, 1381, 1451, 1531, 1571, 1621, 1741, 1861, 1901, 1931, 2131, 2141, 2221, 2371, 2531, 2621, 2741, 2851, 2861, 3011, 3371, 3461, 3491, 3571, 3581, 3691, 3701, 3851, 3931
Offset: 1

Views

Author

Jianing Song, Apr 19 2019

Keywords

Comments

Primes p such that 2 is a primitive root modulo p (i.e., p is in A001122) and that p == 1 (mod 5).
By Artin's conjecture, the number of terms <= N is roughly ((4/19)*C)*PrimePi(N), where C is the Artin's constant = A005596, PrimePi = A000720. Compare: the number of terms of A001122 that are no greater than N is roughly C*PrimePi(N).

Examples

			For p = 61, the multiplicative order of 2 modulo 61 is 60, while 32^12 == 2^(5*12) == 1 (mod 61), so 61 is a term.
		

Crossrefs

Complement of A019358 with respect to A001122.
Cf. also A005596, A000720, A307627.

Programs

  • Maple
    select(p -> isprime(p) and numtheory:-order(2,p) = p-1,
    [seq(i,i=1..10000,10)]); # Robert Israel, Apr 23 2019
  • Mathematica
    {11}~Join~Select[Prime@ Range[11, 550], And[FreeQ[#, 32], ! FreeQ[#, 2]] &@ PrimitiveRootList@ # &] (* Michael De Vlieger, Apr 23 2019 *)
  • PARI
    forprime(p=3, 4000, if(znorder(Mod(2, p))==(p-1) && p%5==1, print1(p, ", ")))

A348971 a(n) = Product(p*(p+1)^(e-1)) - Product((p-1)*p^(e-1)), when n = Product(p^e), with p primes, and e their exponents.

Original entry on oeis.org

0, 1, 1, 4, 1, 4, 1, 14, 6, 6, 1, 14, 1, 8, 7, 46, 1, 18, 1, 22, 9, 12, 1, 46, 10, 14, 30, 30, 1, 22, 1, 146, 13, 18, 11, 60, 1, 20, 15, 74, 1, 30, 1, 46, 36, 24, 1, 146, 14, 40, 19, 54, 1, 78, 15, 102, 21, 30, 1, 74, 1, 32, 48, 454, 17, 46, 1, 70, 25, 46, 1, 192, 1, 38, 50, 78, 17, 54, 1, 238, 138, 42, 1, 102, 21
Offset: 1

Views

Author

Antti Karttunen, Nov 05 2021

Keywords

Comments

Möbius transform of A348507.

Crossrefs

Programs

  • Mathematica
    f1[p_, e_] := p*(p + 1)^(e - 1); f2[p_, e_] := (p - 1)*p^(e - 1); a[1] = 0; a[n_] := Times @@ f1 @@@ (f = FactorInteger[n]) - Times @@ f2 @@@ f; Array[a, 100] (* Amiram Eldar, Nov 05 2021 *)
  • PARI
    A348971(n) = { my(f=factor(n),m1=1,m2=1,p); for(i=1, #f~, p = f[i, 1]; m1 *= p*(p+1)^(f[i, 2]-1); m2 *= (p-1)*p^(f[i, 2]-1)); (m1-m2); };
    
  • PARI
    A348971(n) = { my(f=factor(n),p); for (i=1, #f~, p = f[i, 1]; f[i, 1] = p*(p+1)^(f[i, 2]-1); f[i, 2] = 1); factorback(f)-eulerphi(n); }

Formula

a(n) = A003968(n) - A000010(n).
a(n) = Sum_{d|n} A008683(n/d) * A348507(d).
Sum_{k=1..n} a(k) ~ c * n^2, where c = A104141 * (1/A005596 - 1) = 0.5088692487... . - Amiram Eldar, Oct 05 2023

A049232 Primes p such that p+2 is divisible by a square.

Original entry on oeis.org

2, 7, 23, 43, 47, 61, 73, 79, 97, 151, 167, 173, 223, 241, 277, 313, 331, 349, 359, 367, 373, 421, 439, 457, 523, 547, 601, 619, 673, 691, 709, 727, 733, 773, 823, 839, 853, 907, 929, 997, 1033, 1051, 1069, 1087, 1123, 1181, 1213, 1223, 1231, 1249, 1303
Offset: 1

Views

Author

Keywords

Comments

This sequence is infinite and its relative density in the sequence of the primes is equal to 1 - 2 * Product_{p prime} (1-1/(p*(p-1))) = 1 - 2 * A005596 = 0.252088... - Amiram Eldar, Feb 27 2021

Examples

			47 is a term since 47+2 = 49 = 7^2 is a square.
523 is a term since 523+2 = 525 = 5^2*21 is divisible by a square.
		

Crossrefs

A091880 gives prime indices.

Programs

  • Mathematica
    Select[Prime[Range[100]], ! SquareFreeQ[ # + 2] &] (* Zak Seidov, Oct 28 2008 *)
  • PARI
    powerfreep3(n,p,k) = { c=0; pc=0; forprime(x=2,n, pc++; if(ispowerfree(x+k,p)==0, c++; print1(x","); ) ); print(); print(c","pc","c/pc+.0) }
    ispowerfree(m,p1) = { flag=1; y=component(factor(m),2); for(i=1,length(y), if(y[i] >= p1,flag=0;break); ); return(flag) }

Formula

Primes p such that mu(p+2) = 0.

Extensions

Corrected by Cino Hilliard and Ray Chandler, Dec 08 2003
Edited by N. J. A. Sloane, Oct 27 2008 at the suggestion of R. J. Mathar

A105880 Primes for which -8 is a primitive root.

Original entry on oeis.org

5, 23, 29, 47, 53, 71, 101, 149, 167, 173, 191, 197, 239, 263, 269, 293, 311, 317, 359, 383, 389, 461, 479, 503, 509, 557, 599, 647, 653, 677, 701, 719, 743, 773, 797, 821, 839, 863, 887, 941, 983, 1031, 1061, 1109, 1151, 1223, 1229, 1277, 1301, 1319, 1367, 1373, 1439
Offset: 1

Views

Author

N. J. A. Sloane, Apr 24 2005

Keywords

Comments

From Jianing Song, May 12 2024: (Start)
Members of A105874 that are not congruent to 1 mod 3. Terms are congruent to 5 or 23 modulo 24.
According to Artin's conjecture, the number of terms <= N is roughly ((3/5)*C)*PrimePi(N), where C is the Artin's constant = A005596, PrimePi = A000720. Compare: the number of terms of A001122 that are no greater than N is roughly C*PrimePi(N). (End)

Crossrefs

Programs

  • Mathematica
    pr=-8; Select[Prime[Range[400]], MultiplicativeOrder[pr, # ] == #-1 &] (* N. J. A. Sloane, Jun 01 2010 *)
    a[p_, q_]:= Sum[2 Cos[2^n Pi/((2 q+1)(2 p+1))],{n,1,2 q p}]
    2 Select[Range[800], Rationalize[N[a[#, 9], 20]] == 1 &] + 1
    (* Gerry Martens, Apr 28 2015 *)
  • PARI
    is(n)=isprime(n) && n>3 && znorder(Mod(-8,n))==n-1 \\ Charles R Greathouse IV, May 21 2015

Formula

Let a(p,q)=sum(n=1,2*p*q,2*cos(2^n*Pi/((2*q+1)*(2*p+1)))). Then 2*p+1 is a prime of this sequence when a(p,9)==1. - Gerry Martens , May 21 2015

A192862 Flat primes: odd primes p such that p+1 is a squarefree number times a power of two.

Original entry on oeis.org

3, 5, 7, 11, 13, 19, 23, 29, 31, 37, 41, 43, 47, 59, 61, 67, 73, 79, 83, 101, 103, 109, 113, 127, 131, 137, 139, 151, 157, 163, 167, 173, 181, 191, 193, 211, 223, 227, 229, 239, 257, 263, 271, 277, 281, 283, 307, 311, 313, 317, 331, 347, 353, 367, 373, 379
Offset: 1

Views

Author

Keywords

Comments

Broughan & Qizhi show that this sequence has relative density 2A in the primes, where A = A005596 is Artin's constant. Consequently, there exists a flat number between x and (1+e)x for every e > 0 and large enough x.

Crossrefs

Subsequence of A192861.

Programs

Formula

a(n) ~ k * n * log(n) with k = 1/(2A) = 1.3370563...

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

Views

Author

Jonathan Sondow, Feb 02 2013

Keywords

Comments

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

Examples

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

References

  • 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.

Crossrefs

Programs

  • Maple
    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
  • Mathematica
    Select[ Prime[ Range[4000]], Mod[(# - 1)/4, 5] == 2 && PrimeQ[(# - 1)/4] &]
  • PARI
    is(n)=n%20==9 && isprime(n) && isprime(n\4) \\ Charles R Greathouse IV, Apr 18 2013

Formula

a(n) = 4*A221982(n) + 1.
a(n) >> n log^2 n. - Charles R Greathouse IV, Dec 30 2024

A222008 Primes of the form 4^k + 1 for some k > 0.

Original entry on oeis.org

5, 17, 257, 65537
Offset: 1

Views

Author

Jonathan Sondow, Feb 04 2013

Keywords

Comments

Same as Fermat primes 2^(2^m) + 1 for m >= 1. See A019434 for comments, etc.
Chebyshev showed that 3 is a primitive root of all primes of the form 2^(2*k) + 1 with k > 0. 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 = 3.
As a(n) is a Fermat prime > 3, by Pépin's test a(n) has primitive root 3.
Conjecture: let p a prime number, a(n) is not congruent to p mod (p^2-3)/2. - Vincenzo Librandi, Jun 15 2014
This conjecture is false when p = a(n), but may be true for primes p != a(n). - Jonathan Sondow, Jun 15 2014
Primes p with the property that k-th smallest divisor of its squares p^2, for all 1 <= k <= tau(p^2), contains exactly k "1" digits in its binary representation. Corresponding values of squares p^2: 25, 289, 66049, 4295098369. Example: p = 257, set of divisors of p^2 = 66049 in binary representation: {1, 100000001, 10000001000000001}. Subsequence of A255401. - Jaroslav Krizek, Dec 21 2016

Examples

			4^1 + 1 = 5 is prime, so a(1) = 5. Also, 3^k == 3, 4, 2, 1 (mod 5) for k = 1, 2, 3, 4, resp., so 3 is a primitive root for a(1).
		

References

  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966, p. 102, nr. 3.
  • P. L. Chebyshev, Theory of congruences. Elements of number theory, Chelsea, 1972, p. 306.

Crossrefs

Programs

  • Mathematica
    Select[Table[4^k + 1, {k, 10^3}], PrimeQ] (* Michael De Vlieger, Dec 22 2016 *)

Formula

a(n) = A019434(n+1) for n > 0.

A307410 Numerators of the product in the singular series.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 5, 1, 1, 3, 9, 1, 11, 5, 3, 1, 15, 1, 17, 3, 5, 9, 21, 1, 3, 11, 1, 5, 27, 3, 29, 1, 9, 15, 5, 1, 35, 17, 11, 3, 39, 5, 41, 9, 3, 21, 45, 1, 5, 3, 15, 11, 51, 1, 27, 5, 17, 27, 57, 3, 59, 29, 5, 1, 11, 9, 65, 15, 21, 5, 69, 1, 71, 35, 3, 17, 3, 11, 77, 3, 1, 39, 81, 5, 45
Offset: 1

Views

Author

Mats Granvik, Apr 07 2019

Keywords

Comments

Differs from A305444 at n = 35, 65, 70, ...

Crossrefs

Cf. A005596, A005597, A305444, A380839 (denominators).

Programs

  • Maple
    f:= proc(n) numer(mul((p-2)/(p-1),p=select(type,numtheory:-factorset(n),odd))) end proc:
    map(f, [$1..100]); # Robert Israel, Apr 07 2019
  • Mathematica
    Table[Times@@(DeleteDuplicates[DeleteCases[DeleteCases[Exp[MangoldtLambda[Divisors[h]]], 1],2]] - 2)/Times@@(DeleteDuplicates[DeleteCases[DeleteCases[Exp[MangoldtLambda[Divisors[h]]], 1], 2]] - 1), {h, 1, 85}]
    Numerator[%]
    f[p_, e_] := If[p == 2, 1, (p-2)/(p-1)]; a[n_] := Numerator[Times @@ f @@@ FactorInteger[n]]; a[1] = 1; Array[a, 100] (* Amiram Eldar, Mar 03 2025 *)
  • PARI
    a(n) = my(f=factor(n)[,1]~); numerator(prod(k=1, #f, if (f[k]>2, (f[k]-2)/(f[k]-1), 1))); \\ Michel Marcus, Apr 07 2019

Formula

a(n) = numerator of Product_{p|n;p>2}(p-2)/(p-1) where p is a prime number.
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k)/A380839(k) = 2 * Product_{p prime} (1-1/(p^2-p)) = 2 * A005596 = 0.7479116272384045761094... . - Amiram Eldar, Mar 03 2025

A323594 Primes p such that 3 is a primitive root modulo p while 27 is not.

Original entry on oeis.org

7, 19, 31, 43, 79, 127, 139, 163, 199, 211, 223, 283, 331, 379, 463, 487, 571, 607, 631, 691, 739, 751, 811, 823, 859, 907, 1039, 1063, 1087, 1123, 1231, 1279, 1291, 1327, 1423, 1447, 1459, 1483, 1567, 1579, 1627, 1663, 1699, 1723, 1747, 1831, 1951, 1987, 1999
Offset: 1

Views

Author

Jianing Song, Aug 30 2019

Keywords

Comments

Primes p such that 3 is a primitive root modulo p (i.e., p is in A019334) and that p == 1 (mod 3).
According to Artin's conjecture, the number of terms <= N is roughly ((2/5)*C)*PrimePi(N), where C is the Artin's constant = A005596, PrimePi = A000720. Compare: the number of terms of A001122 that are no greater than N is roughly C*PrimePi(N).

Crossrefs

Complement of A019353 with respect to A019334.
Cf. also A005596, A000720.
Primes p such that 3 is a primitive root modulo p and that p == 1 (mod q): this sequence (q=3), A323617 (q=5), A323628 (q=7).

Programs

  • PARI
    forprime(p=5, 2000, if(znorder(Mod(3, p))==(p-1) && p%3==1, print1(p, ", ")))
Previous Showing 51-60 of 99 results. Next