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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A375918 Composite numbers k == 5, 7 (mod 12) such that 3^((k-1)/2) == -1 (mod k).

Original entry on oeis.org

703, 1891, 3281, 8911, 12403, 16531, 44287, 63139, 79003, 97567, 105163, 152551, 182527, 188191, 211411, 218791, 288163, 313447, 320167, 364231, 385003, 432821, 453259, 497503, 563347, 638731, 655051, 658711, 801139, 859951, 867043, 973241, 994507, 1024651, 1097227
Offset: 1

Views

Author

Jianing Song, Sep 02 2024

Keywords

Comments

Odd composite numbers k such that 3^((k-1)/2) == (3/k) = -1 (mod k), where (3/k) is the Jacobi symbol (or Kronecker symbol).

Examples

			3281 is a term because 3281 = 17*193 is composite, 3281 == 5 (mod 12), and 3^((3281-1)/2) == -1 (mod 3281).
		

Crossrefs

| b=2 | b=3 | b=5 |
-----------------------------------+-------------------+----------+---------+
(b/k)=1, b^((k-1)/2)==1 (mod k) | A006971 | A375917 | A375915 |
-----------------------------------+-------------------+----------+---------+
(b/k)=-1, b^((k-1)/2)==-1 (mod k) | A244628 U A244626 | this seq | A375916 |
-----------------------------------+-------------------+----------+---------+
b^((k-1)/2)==-(b/k) (mod k), also | A306310 | A375490 | A375816 |
(b/k)=-1, b^((k-1)/2)==1 (mod k) | | | |
-----------------------------------+-------------------+----------+---------+
Euler-Jacobi pseudoprimes | A047713 | A048950 | A375914 |
(union of first two) | | | |
-----------------------------------+-------------------+----------+---------+
Euler pseudoprimes | A006970 | A262051 | A262052 |
(union of all three) | | | |

Programs

  • PARI
    isA375918(k) = !isprime(k) && (k%12==5 || k%12==7) && Mod(3,k)^((k-1)/2) == -1

A270697 Composite numbers k == 3 (mod 4) such that (1 + i)^k == 1 - i (mod k), where i = sqrt(-1).

Original entry on oeis.org

2047, 42799, 90751, 256999, 271951, 476971, 514447, 741751, 877099, 916327, 1302451, 1325843, 1397419, 1441091, 1507963, 1530787, 1907851, 2004403, 2205967, 2304167, 2748023, 2811271, 2953711, 2976487, 3090091, 3116107, 4469471, 4863127, 5016191
Offset: 1

Views

Author

Keywords

Comments

Composite k == 3 (mod 4) such that 2*(-4)^((k-3)/4) == -1 (mod k). - Robert Israel, Mar 21 2016
2*(-4)^((p-3)/4) == -1 (mod p) is satisfied by all primes p == 3 (mod 4), see A318908. - Jianing Song, Sep 05 2018
Numbers in A047713 that are congruent to 3 mod 4. Most terms are congruent to 7 mod 8. For terms congruent to 3 mod 8, see A244628. - Jianing Song, Sep 05 2018
Question: Is this a subsequence of A001262? I have verified that it contains all terms up to 2^64. - Joseph M. Shunia, Jul 02 2019

Crossrefs

Subsequence of A001567 and A047713.
A244628 is a proper subsequence.

Programs

  • Maple
    select(t -> not isprime(t) and 1 + 2*(-4) &^ ((t-3)/4) mod t = 0, [seq(i, i=7..10^7, 4)]); # Robert Israel, Mar 21 2016
  • Mathematica
    Select[3 + 4*Range[10000000], PrimeQ[#] == False && PowerMod[1 + I, #, #] == Mod[1 - I, #] &]
  • PARI
    forstep(n=3, 10^7, 4, if(Mod(2, n)^((n-1)/2)==kronecker(2, n) && !isprime(n), print1(n, ", ")))

A294912 Numbers n such that 2^(n-1), (2*n-1)*(2^((n-1)/2)), (4*ceiling((3/4)*n)-2), and (2^((n+1)/2) + floor((1/4)*n)*2^(((n+1)/2)+1)) are all congruent to 1 (mod n).

Original entry on oeis.org

3, 11, 19, 43, 59, 67, 83, 107, 131, 139, 163, 179, 211, 227, 251, 283, 307, 331, 347, 379, 419, 443, 467, 491, 499, 523, 547, 563, 571, 587, 619, 643, 659, 683, 691, 739, 787, 811, 827, 859, 883, 907, 947, 971, 1019, 1051, 1091, 1123, 1163, 1171, 1187, 1259
Offset: 1

Views

Author

Jonas Kaiser, Nov 10 2017

Keywords

Comments

It appears that A007520 is a subsequence.
The first composite term is a(9969) = 476971 = 11*131*331. - Alois P. Heinz, Nov 10 2017
From Hilko Koning, Dec 03 2019: (Start)
The next composite terms < 1999979 are
a(17428) = 877099 = 307*2857
a(25090) = 1302451 = 571*2281
a(25518) = 1325843 = 499*2657
a(26785) = 1397419 = 67*20857
a(27549) = 1441091 = 347*4153
a(28715) = 1507963 = 971*1553
a(29117) = 1530787 = 619*2473
a(35635) = 1907851 = 11*251*691
(End)
From Hilko Koning, Dec 05 2019: (Start)
The next composite terms < 24999971 are
a(37344) = 2004403 = 307*6529
a(55773) = 3090091 = 1163*2657
a(56189) = 3116107 = 883*3529
a(91332) = 5256091 = 811*6481
a(102027) = 5919187 = 1777*3331
a(133230) = 7883731 = 811*9721
a(156407) = 9371251 = 1531*6121
a(182911) = 11081459 = 227*48817
a(189922) = 11541307 = 1699*6793
a(201043) = 12263131 = 811*15121
a(213203) = 13057787 = 467*27961
a(217484) = 13338371 = 3163*4217
a(257526) = 15976747 = 3739*4273
a(274961) = 17134043 = 1097*15619
a(299096) = 18740971 = 1531*12241
a(308928) = 19404139 = 2011*9649
a(321676) = 20261251 = 2251*9001
a(341902) = 21623659 = 1163*18593
a(348622) = 22075579 = 163*135433
a(380162) = 24214051 = 281*86171
The composite terms < 25*10^6 match the terms of A244628.
(End)
It appears that composites of the form 2k+1 such that 3*(2k+1) divides 2^k+1 are the composite terms of this sequence. - Hilko Koning, Dec 09 2019

Crossrefs

Programs

  • Mathematica
    okQ[n_] := AllTrue[{2^(n-1), (2*n-1)*(2^((n-1)/2)), (4*Ceiling@((3/4)*n) - 2), (2^((n+1)/2) + Floor@(n/4)*2^(((n+1)/2)+1))}, Mod[#, n] == 1&];
    Select[Range[1300], okQ] (* Jean-François Alcover, Feb 18 2019 *)
  • PARI
    isok(n) = (n%2) && lift((Mod(2, n)^(n-1))==1)&&lift((Mod((2*n-1), n)*Mod(2, n)^((n-1)/2)) == 1)&&lift((Mod(((4*ceil((3/4)*n)-2)), n) )== 1)&&lift((Mod(2, n)^((n+1)/2) +Mod(floor((1/4)*n),n)*Mod(2, n)^(((n+1)/2)+1 ))== 1)

Extensions

More terms from Alois P. Heinz, Nov 10 2017

A356638 Odd composite numbers k such that 2^((k-1)/2) == -1 (mod k).

Original entry on oeis.org

3277, 29341, 49141, 80581, 88357, 104653, 196093, 314821, 458989, 476971, 489997, 800605, 838861, 873181, 877099, 1004653, 1251949, 1302451, 1325843, 1373653, 1397419, 1441091, 1507963, 1509709, 1530787, 1678541, 1811573, 1907851, 1987021, 2004403, 2269093
Offset: 1

Views

Author

Jeppe Stig Nielsen, Aug 19 2022

Keywords

Comments

Counterexamples (pseudoprimes) to the hypothesis that this congruence is sufficient to prove that an odd number is prime.
Compare with Proth's theorem.

Crossrefs

Programs

  • Mathematica
    Select[Range[1, 2.5*10^6, 2], CompositeQ[#] && PowerMod[2, (# - 1)/2, #] == # - 1 &] (* Amiram Eldar, Aug 19 2022 *)
  • PARI
    forstep(k=3,10^8,2,isprime(k)&&next();Mod(2,k)^((k-1)/2)==-1&&print1(k,", "))
Previous Showing 11-14 of 14 results.