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.

Showing 1-5 of 5 results.

A047713 Euler-Jacobi pseudoprimes: 2^((n-1)/2) == (2 / n) mod n, where (2 / n) is a Jacobi symbol.

Original entry on oeis.org

561, 1105, 1729, 1905, 2047, 2465, 3277, 4033, 4681, 6601, 8321, 8481, 10585, 12801, 15841, 16705, 18705, 25761, 29341, 30121, 33153, 34945, 41041, 42799, 46657, 49141, 52633, 62745, 65281, 74665, 75361, 80581, 85489, 87249, 88357, 90751, 104653
Offset: 1

Views

Author

N. J. A. Sloane, Richard Pinch and Robert G. Wilson v

Keywords

Comments

Odd composite numbers n such that 2^((n-1)/2) == (-1)^((n^2-1)/8) mod n. - Thomas Ordowski, Dec 21 2013
Most terms are congruent to 1 mod 8 (cf. A006971). Among the first 1000 terms, the number of terms congruent to 1, 3, 5 and 7 mod 8 are 764, 47, 125 and 64, respectively. - Jianing Song, Sep 05 2018

References

  • R. K. Guy, Unsolved Problems in Number Theory, A12.
  • H. Riesel, Prime numbers and computer methods for factorization, Progress in Mathematics, Vol. 57, Birkhauser, Boston, 1985.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes the subsequence A006971).

Links

Crossrefs

Cf. A002997, A001567, A048950.
Terms in this sequence satisfying certain congruence: A270698 (congruent to 1 mod 4), A270697 (congruent to 3 mod 4), A006971 (congruent to +-1 mod 8), A244628 (congruent to 3 mod 8), A244626 (congruent to 5 mod 8).

Programs

  • Mathematica
    Select[ Range[ 3, 105000, 2 ], Mod[ 2^((# - 1)/2) - JacobiSymbol[ 2, # ], # ] == 0 && ! PrimeQ[ # ] & ]
  • PARI
    is(n)=n%2 && Mod(2,n)^(n\2)==kronecker(2,n) && !isprime(n) \\ Charles R Greathouse IV, Dec 20 2013

Extensions

Corrected by Eric W. Weisstein; more terms from David W. Wilson

A006971 Composite numbers k such that k == +-1 (mod 8) and 2^((k-1)/2) == 1 (mod k).

Original entry on oeis.org

561, 1105, 1729, 1905, 2047, 2465, 4033, 4681, 6601, 8321, 8481, 10585, 12801, 15841, 16705, 18705, 25761, 30121, 33153, 34945, 41041, 42799, 46657, 52633, 62745, 65281, 74665, 75361, 85489, 87249, 90751, 113201, 115921, 126217, 129921, 130561, 149281, 158369
Offset: 1

Views

Author

Richard Pinch

Keywords

Comments

Previous name was "Terms of A047713 that are congruent to +-1 mod 8".
Complement of (A244626 union A244628) with respect to A047713. - Jianing Song, Sep 18 2018

References

  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004, Section A12, pp. 44-50.
  • Hans Riesel, Prime numbers and computer methods for factorization, Progress in Mathematics, Vol. 57, Birkhäuser, Boston, 1985.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Subsequence of A001567 and A047713.
Cf. A244626, A244628, A270697, A270698.

Programs

  • Mathematica
    Select[Range[10^5], MemberQ[{1, 7}, Mod[#, 8]] && CompositeQ[#] && PowerMod[2, (# - 1)/2, #] == 1 &] (* Amiram Eldar, Nov 06 2023 *)

Extensions

This sequence appeared as M5461 in Sloane-Plouffe (1995), but was later mistakenly declared to be an erroneous form of A047713. Thanks to Jianing Song for providing the correct definition. - N. J. A. Sloane, Sep 17 2018
Formal definition by Jianing Song, Sep 18 2018

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

José María Grau Ribas, Mar 21 2016

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

Links

Crossrefs

Subsequence of A001567 and A047713.
A244628 is a proper subsequence.
Cf. A001262, A006971, A270698, A318908.

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, ", ")))

A318898 a(n) = ((-4)^((p-1)/4) - 1)/p, where p is the n-th prime congruent to 1 mod 4.

Original entry on oeis.org

-1, -5, 15, -565, -7085, 25575, -1266205, -17602325, 941362695, 197665011735, 2901803883615, -11147523830125, -165269711096165, 637677823344495, 2154364271382137415, -126774939137440139965, -1925041114036033717685, -447232673152232758272805, -6839447730858454557453725, 410508614063545790640124095, -1608693655111966245554191885
Offset: 1

Views

Author

Jianing Song, Sep 05 2018

Keywords

Comments

a(n) is always an integer. If p == 1 (mod 8), then (-4)^((p-1)/4) == 4^((p-1)/4) == 2^((p-1)/2) (mod p). 2 is a quadratic residue modulo p so 2^((p-1)/2) == 1 (mod p). If p == 5 (mod 8), then (-4)^((p-1)/4) == -4^((p-1)/4) == -2^((p-1)/2) (mod p). 2 is a quadratic nonresidue modulo p so 2^((p-1)/2) == -1 (mod p). Furthermore, for n > 1, a(n) is always an odd multiple of 5.
(-4)^((p-1)/4) == 1 (mod p) implies -4 is always a quartic residue modulo p. Note that x^4 + 4 = (x^2 + 2)^2 - (2x)^2 = (x^2 + 2x + 2)(x^2 - 2x + 2) = ((x + 1)^2 + 1)*((x - 1)^2 + 1), so the solutions to x^4 == -4 (mod p) are x == ((p - 1)/2)! + 1, ((p - 1)/2)! - 1, -((p - 1)/2)! + 1 and -((p - 1)/2)! - 1 (mod p).

Examples

			The second prime congruent to 1 mod 4 is 13, so a(2) = ((-4)^3 - 1)/13 = (-65)/13 = -5. Also, the four solutions to x^4 == -4 (mod 13) are x == 4, 6, 7 and 9 (mod 13).

Crossrefs

Cf. A002144 (primes of the form 4n + 1).
Cf. A270698 (composite k == 1 (mod 4) that divides (-4)^((k-1)/4) - 1).

Programs

  • PARI
    forstep(p=5, 100, 4, if(isprime(p), print1(((-4)^((p-1)/4)-1)/p, ", ")))

A329705 Composite numbers k such that (1 - w)^(k-1) == 1 (mod k) in the ring of Eisenstein integers (w = (-1 + sqrt(3)*i)/2).

Original entry on oeis.org

121, 703, 1729, 1891, 2821, 7381, 8401, 8911, 10585, 12403, 15457, 15841, 16531, 18721, 19345, 23521, 24661, 28009, 29341, 31621, 41041, 44287, 46657, 47197, 49141, 50881, 52633, 55969, 63139, 63973, 74593, 75361, 79003, 82513, 87913, 88573, 93961, 97567, 105163
Offset: 1

Views

Author

Amiram Eldar, Feb 28 2020

Keywords

Comments

w = exp(2*Pi*i/3) = (-1 + sqrt(3)*i)/2, where i is the imaginary unit, is a unit in the ring of Eisenstein integers (usually denoted by the Greek letter omega).
Also Euler-Jacobi pseudoprimes to base 3 that are congruent to 1 (mod 6).

Links

Crossrefs

Intersection of A016921 and A048950.
Cf. A066408, A270698.

Programs

  • Mathematica
    eisProd[z1_, z2_] := {z1[[1]]*z2[[1]] - z1[[2]]*z2[[2]], z1[[1]]*z2[[2]] + z1[[2]]*z2[[1]] - z1[[2]]*z2[[2]]}; seq = {}; z = {1, 0}; Do[z = eisProd[{1, -1}, z]; If[CompositeQ[n] && And @@ Divisible[z - {1, 0}, n], AppendTo[seq, n]], {n, 2, 10^4}]; seq
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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