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 1-10 of 15 results. Next

A091727 Norms of prime ideals of Z[sqrt(-5)].

Original entry on oeis.org

2, 3, 5, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89, 101, 103, 107, 109, 121, 127, 149, 163, 167, 169, 181, 223, 227, 229, 241, 263, 269, 281, 283, 289, 307, 347, 349, 361, 367, 383, 389, 401, 409, 421, 443, 449, 461, 463, 467, 487
Offset: 1

Views

Author

Paul Boddington, Feb 02 2004

Keywords

Comments

Consists of primes congruent to 1, 2, 3, 5, 7, 9 (mod 20) together with the squares of all other primes.
From Jianing Song, Feb 20 2021: (Start)
The norm of a nonzero ideal I in a ring R is defined as the size of the quotient ring R/I.
Note that Z[sqrt(-5)] has class number 2.
For primes p == 1, 9 (mod 20), there are two distinct ideals with norm p in Z[sqrt(-5)], namely (x + y*sqrt(-5)) and (x - y*sqrt(-5)), where (x,y) is a solution to x^2 + 5*y^2 = p.
For p == 3, 7 (mod 20), there are also two distinct ideals with norm p, namely (p, x+y*sqrt(-5)) and (p, x-y*sqrt(-5)), where (x,y) is a solution to x^2 + 5*y^2 = p^2 with y != 0; (2, 1+sqrt(-5)) and (sqrt(-5)) are respectively the unique ideal with norm 2 and 5.
For p == 11, 13, 17, 19 (mod 20), (p) is the only ideal with norm p^2. (End)

Examples

			From _Jianing Song_, Feb 20 2021: (Start)
Let |I| be the norm of an ideal I, then:
|(2, 1+sqrt(-5))| = 2;
|(3, 2+sqrt(-5))| = |(3, 2-sqrt(-5))| = 3;
|(sqrt(-5))| = 5;
|(7, 1+3*sqrt(-5))| = |(7, 1-3*sqrt(-5))| = 7;
|(23, 22+3*sqrt(-5))| = |(23, 22-3*sqrt(-5))| = 23;
|(3 + 2*sqrt(-5))| = |(3 - 2*sqrt(-5))| = 29;
|(6 + sqrt(-5))| = |(6 - sqrt(-5))| = 41. (End)
		

References

  • David A. Cox, Primes of the form x^2+ny^2, Wiley, 1989.
  • A. Frohlich and M. J. Taylor, Algebraic number theory, Cambridge university press, 1991.

Crossrefs

Cf. A091728.
The number of distinct ideals with norm n is given by A035170.
Norms of prime ideals in O_K, where K is the quadratic field with discriminant D and O_K be the ring of integers of K: A055673 (D=8), A341783 (D=5), A055664 (D=-3), A055025 (D=-4), A090348 (D=-7), A341784 (D=-8), A341785 (D=-11), A341786 (D=-15*), A341787 (D=-19), this sequence (D=-20*), A341788 (D=-43), A341789 (D=-67), A341790 (D=-163). Here a "*" indicates the cases where O_K is not a unique factorization domain.

Programs

  • PARI
    isA091727(n) = { my(ms = [1, 2, 3, 5, 7, 9], p, e=isprimepower(n,&p)); if(!e || e>2, 0, bitxor(e-1,!!vecsearch(ms,p%20))); }; \\ Antti Karttunen, Feb 24 2020

Extensions

Offset corrected by Jianing Song, Feb 20 2021

A363416 a(n) = 1/sqrt(5) * the imaginary part of Product_{k = 0..n} 1 + k*sqrt(-5).

Original entry on oeis.org

0, 1, 3, -24, -240, 1890, 40446, -311472, -12038544, 86898420, 5614173180, -36099955584, -3786960576672, 20307572439336, 3492389655843480, -14110473458954880, -4223754447793582464, 10493742733654512528, 6488421280167604253616, -4618066393756887442560, -12344309538368967592151040
Offset: 0

Views

Author

Peter Bala, Jun 01 2023

Keywords

Comments

Compare with A105751(n) = the imaginary part of Product_{k = 0..n} 1 + k*sqrt(-1).
Moll (2012) studied the prime divisors of the terms of A105750 - the real part of Product_{k = 0..n} 1 + k*sqrt(-1) - and divided the primes into three classes. Numerical calculation suggests that a similar division holds in this case.
Type 1: primes p that do not divide any element of the sequence {a(n)}.
We conjecture that in this case, unlike in A105750, the set of type 1 primes is empty; that is, every prime p divides some term of this sequence.
Type 2: primes p such that the p-adic valuation v_p(a(n)) has asymptotically linear behavior. An example is given below.
We conjecture that the set of type 2 primes consists of primes p == 1, 3, 7 or 9 (mod 20), equivalently, rational primes that split in the field extension Q(sqrt(-5)) of Q, together with the prime p = 2. See A139513.
Moll's conjecture 5.5 extends to this sequence and takes the form:
(i) the 2-adic valuation v_2(a(n)) ~ n/4 as n -> oo.
(ii) for the other primes of type 2, the p-adic valuation v_p(a(n)) ~ n/(p - 1) as n -> oo.
Type 3: primes p such that the sequence of p-adic valuations {v_p(a(n)) : n >= 0} exhibits an oscillatory behavior (this phrase is not precisely defined). An example is given below.
We conjecture that the set of type 3 primes consists of primes p == 11, 13, 17 or 19 (mod 20), equivalently, primes that remain inert in the field extension Q(sqrt(-5)) of Q, together with the prime p = 5, which ramifies in Q(sqrt(-5)). See A003626.

Examples

			Type 2 prime p = 3: the sequence of 3-adic valuations [v_3(a(n)) : n = 1..80] = [0, 1, 1, 1, 3, 3, 3, 4, 4, 4, 5, 5, 5, 7, 7, 7, 8, 8, 8, 9, 9, 9, 12, 12, 12, 13, 13, 13, 14, 14, 14, 16, 16, 16, 17, 17, 17, 18, 18, 18, 20, 20, 20, 21, 21, 21, 22, 22, 22, 25, 25, 25, 26, 26, 26, 27, 27, 27, 29, 29, 29, 30, 30, 30, 31, 31, 31, 33, 33, 33, 34, 34, 34, 35, 35, 35, 39, 39, 41, 40, 40].
Note that v_3(a(80)) = 40 = 80/(3 - 1), in agreement with the asymptotic behavior for type 2 primes conjectured above.
Type 3 prime p = 11: the sequence of 11-adic valuations [v_11(a(n)) : n = 1..121] = [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2], showing the oscillatory behavior for type 3 primes conjectured above.
		

Crossrefs

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 0 elif n = 1 then 1 else (
    (2*n - 1)*a(n-1) - n*(5*n^2 - 10*n + 6)*a(n-2) )/(n - 1) end if; end:
    seq(a(n), n = 0..20);

Formula

P-recursive: (n - 1)*a(n) = (2*n - 1)*a(n-1) - n*(5*n^2 - 10*n + 6)*a(n-2) with
a(0) = 0 and a(1) = 1.
a(n) = Sum_{k = 0..floor((n+1)/2)} (-5)^k*Stirling1(n+1,n-2*k).

A385225 Primes p such that multiplicative order of -5 modulo p is odd.

Original entry on oeis.org

2, 3, 7, 23, 29, 43, 47, 61, 67, 83, 103, 107, 127, 163, 167, 223, 227, 229, 263, 283, 307, 347, 349, 367, 383, 421, 443, 449, 463, 467, 487, 503, 509, 521, 523, 547, 563, 587, 607, 643, 647, 661, 683, 701, 709, 727, 743, 761, 787, 821, 823, 827, 863, 883, 887, 907, 947, 967, 983
Offset: 1

Views

Author

Jianing Song, Jun 22 2025

Keywords

Comments

The multiplicative order of -5 modulo a(n) is A385231(n).
Contained in primes congruent to 1, 3, 7, 9 modulo 20 (primes p such that -5 is a quadratic residue modulo p, A139513), and contains primes congruent to 3, 7 modulo 20 (A122870).
Conjecture: this sequence has density 1/3 among the primes.

Crossrefs

Subsequence of A139513. Contains A122870 as a subsequence.
Cf. A385231 (the actual multiplicative orders).
Cf. other bases: A014663 (base 2), A385220 (base 3), A385221 (base 4), A385192 (base 5), A163183 (base -2), A385223 (base -3), A385224 (base -4), this sequence (base -5).

Programs

  • Mathematica
    Select[Prime[Range[200]], OddQ[MultiplicativeOrder[-5, #]] &] (* Paolo Xausa, Jun 28 2025 *)
  • PARI
    isA385225(p) = isprime(p) && (p!=5) && znorder(Mod(-5,p))%2

A003626 Inert rational primes in Q(sqrt(-5)).

Original entry on oeis.org

11, 13, 17, 19, 31, 37, 53, 59, 71, 73, 79, 97, 113, 131, 137, 139, 151, 157, 173, 179, 191, 193, 197, 199, 211, 233, 239, 251, 257, 271, 277, 293, 311, 313, 317, 331, 337, 353, 359, 373, 379, 397, 419, 431, 433, 439, 457, 479, 491, 499, 557, 571, 577, 593, 599
Offset: 1

Views

Author

Keywords

Comments

Primes congruent to 11, 13, 17, 19 (mod 20). - Michael Somos, Aug 14 2012
Legendre symbol (-5, a(n)) = -1. For prime 5 this symbol is set to 0, and for other odd primes (-5, prime) = +1, given in A139513. - Wolfdieter Lang, Mar 05 2021

References

  • H. Hasse, Number Theory, Springer-Verlag, NY, 1980, p. 498.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A139513.

Programs

  • Mathematica
    Select[Prime[Range[1000]],MemberQ[{11,13,17,19},Mod[#,20]]&] (* Vincenzo Librandi, Aug 20 2012 *)
  • PARI
    {a(n) = local( cnt, m ); if( n<1, return( 0 )); while( cnt < n, if( isprime( m++) && kronecker( -20, m )==-1, cnt++ )); m} /* Michael Somos, Aug 14 2012 */

A343238 All positive integer moduli a(n) for which the congruence x^2 == -5 (mod a(n)) is solvable for integer x (representatives from {0, 1, ..., a(n)-1}); ordered increasingly.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 9, 10, 14, 15, 18, 21, 23, 27, 29, 30, 35, 41, 42, 43, 45, 46, 47, 49, 54, 58, 61, 63, 67, 69, 70, 81, 82, 83, 86, 87, 89, 90, 94, 98, 101, 103, 105, 107, 109, 115, 122, 123, 126, 127, 129, 134, 135, 138, 141, 145, 147, 149, 161, 162, 163, 166, 167, 174, 178, 181, 183, 189, 201, 202
Offset: 1

Views

Author

Wolfdieter Lang, May 16 2021

Keywords

Comments

This sequence includes A139513, that is, Legendre(-5, p) = +1 for odd primes not 5, that is, primes congruent to {1, 3, 7, 9} mod 20. Here 5 is a member of the sequence with solution x = 0.
The primes of this sequence are given in A240920.
The present sequence gives the numbers of the form 2^a*5^b*Product_{j=1..m} (p_j)^e(j), with a and b from {0, 1}, p_j a prime from {1, 3, 7, 9} (mod 20), i.e., from A139513, m >= 0 and e(j) >= 0 (this includes the number 1). These numbers are ordered increasingly.
This follows from the Legendre-symbol(-5, p)= +1 and the lifting theorem (see, e.g., Apostol, Theorem 5.30, p. 121-2) for p = 2 and 5 (no lifting for the solutions for p = 2 and p = 5), and the unique lifting for the primes satisfying Legendre-symbol(-5, p) = +1.
Therefore the number of representative solutions x from {0, 1, ..., a(n)-1}, denoted by M(a(n)), is 1 for precisely four cases: a(1) = 1 (x = 0), a(2) = 2 (x = 1), a(4) = 5 (x = 0) and a(8) = 10 = 2*5 (x = 5). For each of the mentioned prime powers there are just 2 solutions. This implies that for the number of solutions in the general a(n) case, n not 1, 2, 4, 8, only the primes p_j are of interest: M(a(n)) = 2^m(n).
For these solutions x see A343239, and for the multiplicity M(a(n)) see A343240.
This congruence is needed to find all proper solutions of the positive definite binary quadratic form of discriminant Disc = -20 = -4*5 representing k = a(n). The solutions x lead to the so-called representative parallel primitive forms (rpapfs). See A344231 for more details.
For a bisection see A344231 and A344232, related to integer solutions of X^2 + 5*Y^2 = A344231(k) and 2*X^2 + 2*X*Y + 3*Y^2 = A344232(k).

Examples

			a(3) = 3: two solutions 1 and 2.
a(7) = 3^2 = 9: two solutions 2 and 7.
a(8) = 10 = 2*5 only one solution 5.
a(53) = 135 = 5*3^3: two solutions 20 and 115.
		

References

  • Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, pp 121, 122.

Crossrefs

Programs

  • PARI
    isok(k) = issquare(Mod(-5, k)); \\ Michel Marcus, Sep 17 2023

Formula

There exists at least one x from {0, 1, ..., m-1} satisfying x^2 + 5 == 0 (mod m), for positive integer m. These m values are then ordered increasingly as (a(n))_{n>=1}.

A343240 The number of solutions x from {0, 1, ..., A343238(n)-1} of the congruence x^2 + 5 == 0 (mod A343238(n)) is given by a(n).

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 4, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 4, 4, 2, 4, 2, 2, 4, 4, 2, 4, 2, 4, 2, 2, 2, 2, 4, 2, 2, 4, 4, 4, 2, 4, 2, 2, 4
Offset: 1

Views

Author

Wolfdieter Lang, May 16 2021

Keywords

Comments

Row length of irregular triangle A343239.

Examples

			a(19) = 4 because A343238(19) = 42 = 2*3*7 has 2^(1+1) = 4 solutions from the primes 3 and 7.
		

Crossrefs

Programs

  • PARI
    isok(k) = issquare(Mod(-5, k)); \\ A343238
    lista(nn) = my(list = List()); for (n=1, nn, if (issquare(Mod(-5, n)), listput(list, sum(i=0, n-1, Mod(i,n)^2 + 5 == 0)););); Vec(list); \\ Michel Marcus, Sep 17 2023

Formula

a(n) = row length of A343239(n), for n >= 1.
a(1) = a(2) = a(4) = a(8) = 1, and otherwise a(n) = 2^{number of distinct primes from A139513}, that is, primes congruent to {1, 3, 7, 9} (mod 20), appearing in the prime factorization of A343238(n).

A035183 Coefficients in expansion of Dirichlet series Product_p (1-(Kronecker(m,p)+1)*p^(-s)+Kronecker(m,p)*p^(-2s))^(-1) for m = -5.

Original entry on oeis.org

1, 0, 2, 1, 1, 0, 2, 0, 3, 0, 0, 2, 0, 0, 2, 1, 0, 0, 0, 1, 4, 0, 2, 0, 1, 0, 4, 2, 2, 0, 0, 0, 0, 0, 2, 3, 0, 0, 0, 0, 2, 0, 2, 0, 3, 0, 2, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 6, 1, 0, 0, 2, 0, 4, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 5
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := If[n < 0, 0, DivisorSum[n, KroneckerSymbol[-5, #] &]]; Table[ a[n], {n, 1, 100}] (* G. C. Greubel, Apr 27 2018 *)
  • PARI
    my(m=-5); direuler(p=2,101,1/(1-(kronecker(m,p)*(X-X^2))-X))
    
  • PARI
    a(n) = sumdiv(n, d, kronecker(-5, d)); \\ Michel Marcus, Oct 07 2023

Formula

From Amiram Eldar, Oct 17 2022: (Start)
a(n) = Sum_{d|n} Kronecker(-5, d).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2*Pi/(3*sqrt(5)) = 0.936641... . (End)
Multiplicative with a(5^e) = 1, a(p^e) = (1+(-1)^e)/2 if Kronecker(-5, p) = -1 (p is in A296923), and a(p^e) = e+1 if Kronecker(-5, p) = 1 (p is in A139513). - Amiram Eldar, Nov 20 2023

A296937 Rational primes that decompose in the field Q(sqrt(13)).

Original entry on oeis.org

3, 17, 23, 29, 43, 53, 61, 79, 101, 103, 107, 113, 127, 131, 139, 157, 173, 179, 181, 191, 199, 211, 233, 251, 257, 263, 269, 277, 283, 311, 313, 337, 347, 367, 373, 389, 419, 433, 439, 443, 467, 491, 503, 521, 523, 547, 563, 569, 571, 599, 601, 607, 641
Offset: 1

Views

Author

N. J. A. Sloane, Dec 26 2017

Keywords

Comments

Is this the same sequence as A141188 or A038883? - R. J. Mathar, Jan 02 2018
From Jianing Song, Apr 21 2022: (Start)
Primes p such that Kronecker(13, p) = Kronecker(p, 13) = 1, where Kronecker() is the Kronecker symbol. That is to say, primes p that are quadratic residues modulo 13.
Primes p such that p^6 == 1 (mod 13).
Primes p == 1, 3, 4, 9, 10, 12 (mod 13). (End)

Crossrefs

Cf. A011583 (kronecker symbol modulo 13), A038883.
Rational primes that decompose in the quadratic field with discriminant D: A139513 (D=-20), A191019 (D=-19), A191018 (D=-15), A296920 (D=-11), A033200 (D=-8), A045386 (D=-7), A002144 (D=-4), A002476 (D=-3), A045468 (D=5), A001132 (D=8), A097933 (D=12), this sequence (D=13), A296938 (D=17).
Cf. A038884 (inert rational primes in the field Q(sqrt(13))).

Programs

Formula

Equals A038883 \ {13}. - Jianing Song, Apr 21 2022

A240920 Prime numbers that occur as divisors of numbers of the form m^2 + 5.

Original entry on oeis.org

2, 3, 5, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89, 101, 103, 107, 109, 127, 149, 163, 167, 181, 223, 227, 229, 241, 263, 269, 281, 283, 307, 347, 349, 367, 383, 389, 401, 409, 421, 443, 449, 461, 463, 467, 487, 503, 509, 521, 523, 541, 547, 563
Offset: 1

Views

Author

J. Lowell, Aug 02 2014

Keywords

Comments

Conjecture: a prime number is in this sequence if and only if its next-to-last digit is even.
The law of quadratic reciprocity shows an odd prime is in the sequence if and only if it is 1, 3, 5, 7 or 9 (mod 20). This proves the above conjecture, so the sequence is the union of {2, 5} and A139513. - Jens Kruse Andersen, Aug 09 2014

Examples

			23 is in the sequence because it divides 8^2+5=69 with m=8.
		

Crossrefs

Cf. A002313 (k=1 or k=4), A033203 (k=2), A045331 (k=3), A139513.

Programs

  • Maple
    isA240920 := proc(p)
        local n;
        if isprime(p) then
            for n from 0 to p do
                if modp(n^2+5,p) = 0 then
                    return true;
                end if;
            end do:
            false;
        else
            false;
        end if;
    end proc:
    for i from 1 to 600 do
        p := ithprime(i) ;
        if isA240920(p) then
            printf("%d,",p);
        end if;
    end do:
  • PARI
    select(p->issquare(Mod(-5,p)), primes(100)) \\ Charles R Greathouse IV, Nov 29 2016

Formula

a(n) ~ 2n log n. - Charles R Greathouse IV, Nov 29 2016

A296922 Primes p such that Legendre(-5,p) = 0 or 1.

Original entry on oeis.org

3, 5, 7, 23, 29, 41, 43, 47, 61, 67, 83, 89, 101, 103, 107, 109, 127, 149, 163, 167, 181, 223, 227, 229, 241, 263, 269, 281, 283, 307, 347, 349, 367, 383, 389, 401, 409, 421, 443, 449, 461, 463, 467, 487, 503, 509, 521, 523, 541, 547, 563, 569, 587, 601, 607, 641, 643
Offset: 1

Views

Author

N. J. A. Sloane, Dec 25 2017

Keywords

Comments

Primes == 1, 3, 5, 7, or 9 (mod 20). Primes whose 10's digit is even. - Robert Israel, Dec 27 2017

Crossrefs

Programs

  • Maple
    Load the Maple program HH given in A296920. Then run HH(-5,200);
    select(isprime, {seq(seq(20*i+j,j=[1,3,5,7,9]),i=0..100)}); # Robert Israel, Dec 27 2017
  • Mathematica
    Select[Prime@ Range@ 120, MemberQ[{0, 1}, KroneckerSymbol[-5, #]] &] (* or *)
    Select[Prime@ Range@ 120, MemberQ[Range[1, 9, 2], Mod[#, 20]] &] (* Michael De Vlieger, Jan 02 2018 *)
  • PARI
    lista(nn) = forprime(p=2, nn, if (kronecker(-5,p) >= 0, print1(p, ", "))); \\ Michel Marcus, Dec 26 2017

Formula

a(n) = A240920(n+1) for n >= 1. - Georg Fischer, Oct 30 2018
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