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.

Previous Showing 11-16 of 16 results.

A070183 Primes p such that x^6 = 2 has a solution mod p, but x^(6^2) = 2 has no solution mod p.

Original entry on oeis.org

17, 41, 137, 401, 433, 449, 457, 521, 569, 641, 761, 809, 857, 919, 929, 953, 977, 1361, 1409, 1423, 1657, 1697, 1999, 2017, 2081, 2143, 2153, 2287, 2297, 2417, 2609, 2633, 2729, 2753, 2777, 2791, 2801, 2897, 2953, 3041, 3209, 3329, 3457, 3593, 3617
Offset: 1

Views

Author

Klaus Brockhaus, Apr 29 2002

Keywords

Crossrefs

Programs

  • Magma
    [p: p in PrimesUpTo(5000) | not exists{x: x in ResidueClassRing(p) | x^36 eq 2} and exists{x: x in ResidueClassRing(p) | x^6 eq 2}]; // Vincenzo Librandi, Sep 21 2012
    
  • Maple
    select(p -> isprime(p) and [msolve(x^6=2,p)]<>[] and [msolve(x^36=2,p)]=[] , [seq(i,i=3..10^4,2)]); # Robert Israel, May 13 2018
  • PARI
    forprime(p=2,3700,x=0; while(x
    				
  • PARI
    ok(p, r, k1, k2)={
        if (  Mod(r,p)^((p-1)/gcd(k1,p-1))!=1, return(0) );
        if (  Mod(r,p)^((p-1)/gcd(k2,p-1))==1, return(0) );
        return(1);
    }
    forprime(p=2,10^4, if (ok(p,2,6,6^2),print1(p,", ")));
    /* Joerg Arndt, Sep 21 2012 */
    
  • Python
    from itertools import count, islice
    from sympy import nextprime, is_nthpow_residue
    def A070183_gen(startvalue=2): # generator of terms >= startvalue
        p = max(nextprime(startvalue-1),2)
        while True:
            if is_nthpow_residue(2,6,p) and not is_nthpow_residue(2,36,p):
                yield p
            p = nextprime(p)
    A070183_list = list(islice(A070183_gen(),20)) # Chai Wah Wu, May 02 2024

A070187 Primes p such that x^11 = 2 has a solution mod p, but x^(11^2) = 2 has no solution mod p.

Original entry on oeis.org

12101, 13553, 30493, 32429, 44771, 66067, 103577, 128987, 180533, 182711, 187793, 201829, 242243, 257489, 264749, 299113, 314359, 330331, 337349, 341947, 356467, 371471, 431729, 442619, 475289, 484243, 505781, 513767, 540871, 558053, 564103, 573299, 581527, 582011, 586367, 593869, 596047, 630169
Offset: 1

Views

Author

Klaus Brockhaus, Apr 29 2002

Keywords

Crossrefs

Programs

  • PARI
    forprime(p=2,550000,x=0; while(x
    				
  • PARI
    N=10^6;  default(primelimit,N);
    ok(p, r, k1, k2)={
        if (  Mod(r,p)^((p-1)/gcd(k1,p-1))!=1, return(0) );
        if (  Mod(r,p)^((p-1)/gcd(k2,p-1))==1, return(0) );
        return(1);
    }
    forprime(p=2,N, if (ok(p,2,11,11^2),print1(p,", ")));
    /* Joerg Arndt, Sep 21 2012 */

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

A070182 Primes p such that x^5 = 2 has a solution mod p, but x^(5^2) = 2 has no solution mod p.

Original entry on oeis.org

151, 251, 3251, 3301, 4751, 8501, 11251, 11701, 13751, 14251, 14951, 15551, 16451, 17401, 18401, 21401, 21601, 24251, 28351, 28901, 32251, 32401, 32801, 34301, 36151, 36451, 37201, 40351, 42451, 42701, 44201, 45751, 46051, 46451, 46901
Offset: 1

Views

Author

Klaus Brockhaus, Apr 29 2002

Keywords

Crossrefs

Programs

  • Magma
    [p: p in PrimesUpTo(50000) | not exists{x: x in ResidueClassRing(p) | x^25 eq 2} and exists{x: x in ResidueClassRing(p) | x^5 eq 2}]; // Vincenzo Librandi, Sep 21 2012
    
  • PARI
    forprime(p=2,47000,x=0; while(x
    				
  • PARI
    ok(p, r, k1, k2)={
        if (  Mod(r,p)^((p-1)/gcd(k1,p-1))!=1, return(0) );
        if (  Mod(r,p)^((p-1)/gcd(k2,p-1))==1, return(0) );
        return(1);
    }
    forprime(p=2,10^5, if (ok(p,2,5,5^2),print1(p,", ")));
    /* Joerg Arndt, Sep 21 2012 */

A294919 Numbers n such that 2^(n-1), (2*n-1)*(2^((n-1)/2)), (4*ceiling((1/4)*n)-2), and (2^((n+1)/2) + floor((3/4)*n)*2^(((n+1)/2)+1)) are all congruent to 1 (mod n).

Original entry on oeis.org

5, 13, 29, 37, 53, 61, 101, 109, 149, 157, 173, 181, 197, 229, 269, 277, 293, 317, 349, 373, 389, 397, 421, 461, 509, 541, 557, 613, 653, 661, 677, 701, 709, 733, 757, 773, 797, 821, 829, 853, 877, 941, 997, 1013, 1021, 1061, 1069, 1093, 1109, 1117, 1181, 1213
Offset: 1

Views

Author

Jonas Kaiser, Nov 10 2017

Keywords

Comments

It appears that A007521 is a subsequence.
a(118) = 3277 = 29*113 is the first nonprime term.

Crossrefs

Programs

  • Mathematica
    okQ[n_] := AllTrue[{2^(n-1), (2*n-1)*(2^((n-1)/2)), (4*Ceiling@(n/4) - 2), (2^((n+1)/2) + Floor@((3/4)*n)*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((1/4)*n)-2)), n) )== 1)&&lift((Mod(2, n)^((n+1)/2) +Mod(floor((3/4)*n),n)*Mod(2, n)^(((n+1)/2)+1 ))== 1)

Extensions

More terms from Alois P. Heinz, Nov 10 2017

A294993 Numbers n > 1 such that all of 2^(n-1), 3^(n-1), 5^(n-1), (2*n-1)*(2^((n-1)/2)), 4*ceiling((3/4)*n)-2, and (2^((n+1)/2) + floor(n/4)*2^((n+3)/2)) are congruent to 1 (mod n).

Original entry on oeis.org

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, 1283
Offset: 1

Views

Author

Jonas Kaiser, Nov 12 2017

Keywords

Comments

It appears that A007520 is a subsequence. Up to 10^7 there are no composites in this sequence.
The first composite is a(17465859) = 1397357851; there are probably infinitely many. - Charles R Greathouse IV, Nov 12 2017

Crossrefs

Programs

  • Mathematica
    Select[Range[2, 1300], Function[n, AllTrue[Join[Prime[Range@3]^(n - 1), {(2 n - 1) (2^((n - 1)/2)), 4 Ceiling[3 n/4] - 2, (2^((n + 1)/2) + Floor[n/4]*2^((n + 3)/2))}], Mod[#, n] == 1 &]]] (* Michael De Vlieger, Nov 15 2017 *)
  • PARI
    is(n) = n%2 && Mod(2, n)^(n-1)==1 && Mod(3, n)^(n-1)==1 && Mod(5, n)^(n-1)==1 && (2*n-1)*Mod(2, n)^((n-1)/2)== 1 && Mod(4*ceil((3/4)*n)-2, n)==1 && Mod(2, n)^((n+1)/2)+floor(n/4)*Mod(2, n)^((n+3)/2)==1
Previous Showing 11-16 of 16 results.