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.

A328895 Decimal expansion of Sum_{k>=1} Kronecker(8,k)/k^2.

Original entry on oeis.org

8, 7, 2, 3, 5, 8, 0, 2, 4, 9, 5, 4, 8, 5, 9, 9, 4, 1, 7, 6, 9, 6, 9, 5, 1, 1, 7, 0, 2, 1, 1, 7, 5, 6, 6, 1, 2, 3, 9, 9, 8, 3, 2, 8, 3, 8, 6, 8, 5, 0, 5, 2, 9, 5, 7, 6, 9, 1, 8, 7, 0, 8, 3, 4, 3, 9, 9, 8, 8, 4, 7, 0, 3, 5, 4, 1, 3, 4, 6, 5, 1, 8, 3, 3, 4, 2, 5, 1, 6, 7, 1
Offset: 0

Views

Author

Jianing Song, Nov 19 2019

Keywords

Comments

Let Chi() be a primitive character modulo d, the so-called Dirichlet L-series L(s,Chi) is the analytic continuation (see the functional equations involving L(s,Chi) in the MathWorld link entitled Dirichlet L-Series) of the sum Sum_{k>=1} Chi(k)/k^s, Re(s)>0 (if d = 1, the sum converges requires Re(s)>1).
If s != 1, we can represent L(s,Chi) in terms of the Hurwitz zeta function by L(s,Chi) = (Sum_{k=1..d} Chi(k)*zeta(s,k/d))/d^s.
L(s,Chi) can also be represented in terms of the polylog function by L(s,Chi) = (Sum_{k=1..d} Chi'(k)*polylog(s,u^k))/(Sum_{k=1..d} Chi'(k)*u^k), where Chi' is the complex conjugate of Chi, u is any primitive d-th root of unity.
If m is a positive integer, we have L(m,Chi) = (Sum_{k=1..d} Chi(k)*polygamma(m-1,k/d))/((-d)^m*(m-1)!).
In this sequence we have Chi = A091337 and s = 2.

Examples

			1 - 1/3^2 - 1/5^2 + 1/7^2 + 1/9^2 - 1/11^2 - 1/13^2 + 1/15^2 + ... = Pi^2/(8*sqrt(2)) = 0.8723580249...
		

Crossrefs

Decimal expansion of Sum_{k>=1} Kronecker(d,k)/k^2, where d is a fundamental discriminant: A309710 (d=-8), A103133 (d=-7), A006752 (d=-4), A086724 (d=-3), A013661 (d=1), A328717 (d=5), this sequence (d=8), A258414 (d=12).
Decimal expansion of Sum_{k>=1} Kronecker(8,k)/k^s: A196525 (s=1), this sequence (s=2), A329715 (s=3).

Programs

  • Mathematica
    RealDigits[Pi^2/(8*Sqrt[2]), 10, 102] // First
  • PARI
    default(realprecision, 100); Pi^2/(8*sqrt(2))

Formula

Equals Pi^2/(8*sqrt(2)).
Equals (zeta(2,1/8) - zeta(2,3/8) - zeta(2,5/8) + zeta(2,7/8))/64, where zeta(s,a) is the Hurwitz zeta function.
Equals (polylog(2,u) - polylog(2,u^3) - polylog(2,-u) + polylog(2,-u^3))/sqrt(8), where u = sqrt(2)/2 + i*sqrt(2)/2 is an 8th primitive root of unity, i = sqrt(-1).
Equals (polygamma(1,1/8) - polygamma(1,3/8) - polygamma(1,5/8) + polygamma(1,7/8))/64.
Equals -Integral_{x=0..oo} log(x)/(x^4 + 1) dx. - Amiram Eldar, Jul 17 2020
Equals 1/(Product_{p prime == 1 or 7 (mod 8)} (1 - 1/p^2) * Product_{p prime == 3 or 5 (mod 8)} (1 + 1/p^2)). - Amiram Eldar, Dec 17 2023

A328717 Decimal expansion of Sum_{k>=1} Kronecker(5,k)/k^2.

Original entry on oeis.org

7, 0, 6, 2, 1, 1, 4, 0, 3, 2, 5, 9, 7, 4, 0, 9, 6, 9, 9, 3, 1, 0, 0, 3, 1, 7, 5, 7, 6, 2, 5, 6, 4, 0, 2, 7, 6, 6, 0, 2, 4, 6, 4, 7, 1, 8, 5, 2, 9, 4, 6, 8, 6, 3, 9, 4, 2, 1, 1, 7, 4, 0, 2, 1, 6, 5, 6, 7, 7, 6, 0, 4, 4, 3, 8, 3, 8, 3, 0, 0, 7, 6, 8, 3, 3, 7, 4, 5, 6, 6, 4
Offset: 0

Views

Author

Jianing Song, Nov 19 2019

Keywords

Comments

Let Chi() be a primitive character modulo d, the so-called Dirichlet L-series L(s,Chi) is the analytic continuation (see the functional equations involving L(s,Chi) in the MathWorld link entitled Dirichlet L-Series) of the sum Sum_{k>=1} Chi(k)/k^s, Re(s)>0 (if d = 1, the sum converges requires Re(s)>1).
If s != 1, we can represent L(s,Chi) in terms of the Hurwitz zeta function by L(s,Chi) = (Sum_{k=1..d} Chi(k)*zeta(s,k/d))/d^s.
L(s,Chi) can also be represented in terms of the polylog function by L(s,Chi) = (Sum_{k=1..d} Chi'(k)*polylog(s,u^k))/(Sum_{k=1..d} Chi'(k)*u^k), where Chi' is the complex conjugate of Chi, u is any primitive d-th root of unity.
If m is a positive integer, we have L(m,Chi) = (Sum_{k=1..d} Chi(k)*polygamma(m-1,k/d))/((-d)^m*(m-1)!).
In this sequence we have Chi = A080891 and s = 2.

Examples

			1 - 1/2^2 - 1/3^2 + 1/4^2 + 1/6^2 - 1/7^2 - 1/8^2 + 1/9^2 + ... = 4*Pi^2/(25*sqrt(5)) = 0.70621140325974096993100317576256402766024647185294...
		

Crossrefs

Decimal expansion of Sum_{k>=1} Kronecker(d,k)/k^2, where d is a fundamental discriminant: A309710 (d=-8), A103133 (d=-7), A006752 (d=-4), A086724 (d=-3), A013661 (d=1), this sequence (d=5), A328895 (d=8), A258414 (d=12).
Decimal expansion of Sum_{k>=1} Kronecker(5,k)/k^s: A086466 (s=1), this sequence (s=2), A328723 (s=3).

Programs

  • Mathematica
    RealDigits[4*Pi^2/(25*Sqrt[5]), 10, 102] // First
  • PARI
    default(realprecision, 100); 4*Pi^2/(25*sqrt(5))

Formula

Equals 4*Pi^2/(25*sqrt(5)).
Equals (zeta(2,1/5) - zeta(2,2/5) - zeta(2,3/5) + zeta(2,4/5))/25, where zeta(s,a) is the Hurwitz zeta function.
Equals (polylog(2,u) - polylog(2,u^2) - polylog(2,u^3) + polylog(2,u^4))/sqrt(5), where u = exp(2*Pi*i/5) is a 5th primitive root of unity, i = sqrt(-1).
Equals (polygamma(1,1/5) - polygamma(1,2/5) - polygamma(1,3/5) - polygamma(1,4/5))/25.
Equals Sum_{k>=1} Fibonacci(2*k)/(k^2*binomial(2*k,k)) = Sum_{k>=1} A001906(k)/A002736(k) (Seiffert, 1991). - Amiram Eldar, Jan 17 2022
Equals 1/(Product_{p prime == 1 or 4 (mod 5)} (1 - 1/p^2) * Product_{p prime == 2 or 3 (mod 5)} (1 + 1/p^2)). - Amiram Eldar, Dec 17 2023

A088965 Number of solutions to x^2 + 2y^2 == 1 (mod n).

Original entry on oeis.org

1, 2, 2, 4, 6, 4, 8, 16, 6, 12, 10, 8, 14, 16, 12, 32, 16, 12, 18, 24, 16, 20, 24, 32, 30, 28, 18, 32, 30, 24, 32, 64, 20, 32, 48, 24, 38, 36, 28, 96, 40, 32, 42, 40, 36, 48, 48, 64, 56, 60, 32, 56, 54, 36, 60, 128, 36, 60, 58, 48, 62, 64, 48, 128, 84, 40, 66, 64, 48, 96
Offset: 1

Views

Author

Yuval Dekel (dekelyuval(AT)hotmail.com), Oct 28 2003

Keywords

Crossrefs

Programs

  • Magma
    [n eq 1 select 1 else #[x: x in [1..n], y in [1..n] | (x^2+2*y^2) mod n eq 1]: n in [1..80]]; // Vincenzo Librandi, Jul 16 2018
  • Maple
    A088965 := proc(n) local a,x,y ; a := 0 ; for x from 0 to n-1 do for y from 0 to n-1 do if (x^2+2*y^2) mod n = 1 mod n then a := a+1 ; end if; end do; end do ; a ; end proc:
    seq(A088965(n),n=1..70) ; # R. J. Mathar, Jan 07 2011
  • Mathematica
    a[1]=1; a[n_]:=Length@Rest@Union@Flatten@Table[If[Mod[i^2 + 2 j^2, n]==1, i+I j, 0], {i, 0, n-1}, {j, 0, n-1}]; Table[a[n], {n, 1, 80}] (* Vincenzo Librandi, Jul 16 2018 *)
  • PARI
    a(n)={my(v=vector(n)); for(i=0, n-1, v[i^2%n + 1]++); sum(i=0, n-1, v[i+1]*v[(1-2*i)%n + 1])} \\ Andrew Howroyd, Jul 09 2018
    
  • PARI
    a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i,1], e=f[i,2]); if(p==2, 2^e*if(e>2, 2, 1), p^(e-1)*if(abs(p%8-2)==1, p-1, p+1)))} \\ Andrew Howroyd, Jul 09 2018
    

Formula

Multiplicative with a(2^e) = 2^e for e <= 2, a(2^e) = 2^(e + 1) for e > 2, a(p^e) = (p-1)*p^(e-1) for p-2 mod 8 = +-1, a(p^e) = (p+1)*p^(e-1) for p-2 mod 8 = +-3. - Andrew Howroyd, Jul 13 2018
Sum_{k=1..n} a(k) ~ c * n^2, where c = (9/(16*A309710)) = 0.528300880442971272... . - Amiram Eldar, Nov 21 2023

A087561 Number of solutions to x^2 + 2y^2 == 0 (mod n).

Original entry on oeis.org

1, 2, 5, 4, 1, 10, 1, 8, 21, 2, 21, 20, 1, 2, 5, 16, 33, 42, 37, 4, 5, 42, 1, 40, 25, 2, 81, 4, 1, 10, 1, 32, 105, 66, 1, 84, 1, 74, 5, 8, 81, 10, 85, 84, 21, 2, 1, 80, 49, 50, 165, 4, 1, 162, 21, 8, 185, 2, 117, 20, 1, 2, 21, 64, 1, 210, 133, 132, 5, 2, 1, 168, 145, 2, 125, 148, 21
Offset: 1

Views

Author

Yuval Dekel (dekelyuval(AT)hotmail.com), Oct 24 2003

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := If[n == 1, 1, Product[{p, e} = pe; Which[p == 2, 2^e, Abs[Mod[p, 8] - 2] != 1, (p^2)^Quotient[e, 2], True, (p + e (p-1)) p^(e-1)], {pe, FactorInteger[n]}]];
    a /@ Range[1, 100] (* Jean-François Alcover, Sep 20 2019, from PARI *)
  • PARI
    a(n)={my(v=vector(n)); for(i=0, n-1, v[i^2%n + 1]++); sum(i=0, n-1, v[i+1]*v[(-2*i)%n + 1])} \\ Andrew Howroyd, Jul 16 2018
    
  • PARI
    a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i, 1], e=f[i, 2]); if(p==2, 2^e, if(abs(p%8-2)<>1, (p^2)^(e\2), (p+e*(p-1))*p^(e-1))))} \\ Andrew Howroyd, Jul 16 2018

Formula

Multiplicative with a(2^e) = 2^e, a(p^e) = p^(2*floor(e/2)) for p - 2 == +-3 (mod 8), a(p^e) = ((p-1)*e+p)*p^(e-1) for p - 2 == +-1 (mod 8). - Andrew Howroyd, Jul 16 2018
Sum_{k=1..n} a(k) ~ c * n^2, where c = Pi/(4*sqrt(2)*A309710) = 0.521595326207... . - Amiram Eldar, Nov 21 2023

Extensions

More terms from David Wasserman, Jun 07 2005

A089002 Number of non-congruent solutions to x^2 + 2y^2 == -1 (mod n).

Original entry on oeis.org

1, 2, 2, 4, 6, 4, 8, 0, 6, 12, 10, 8, 14, 16, 12, 0, 16, 12, 18, 24, 16, 20, 24, 0, 30, 28, 18, 32, 30, 24, 32, 0, 20, 32, 48, 24, 38, 36, 28, 0, 40, 32, 42, 40, 36, 48, 48, 0, 56, 60, 32, 56, 54, 36, 60, 0, 36, 60, 58, 48, 62, 64, 48, 0, 84, 40, 66, 64, 48, 96
Offset: 1

Views

Author

Yuval Dekel (dekelyuval(AT)hotmail.com), Nov 02 2003

Keywords

Crossrefs

Programs

  • Mathematica
    f[2, e_] := If[e < 3, 2^e, 0]; f[p_, e_] := If[MemberQ[{1, 7}, Mod[p - 2, 8]], (p - 1), (p + 1)] * p^(e - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 11 2020 *)
  • PARI
    a(n)={my(v=vector(n)); for(i=0, n-1, v[i^2%n + 1]++); sum(i=0, n-1, v[i+1]*v[(-1-2*i)%n + 1])} \\ Andrew Howroyd, Jul 09 2018
    
  • PARI
    a(n)={my(f=factor(n)); prod(i=1, #f~, my(p=f[i,1], e=f[i,2]); if(p==2, if(e>2, 0, 2^e), p^(e-1)*if(abs(p%8-2)==1, p-1, p+1)))} \\ Andrew Howroyd, Jul 09 2018

Formula

Multiplicative with a(2^e) = 2^e for e <= 2, a(2^e) = 0 for e > 2, a(p^e) = (p-1)*p^(e-1) for p-2 mod 8 = +-1, a(p^e) = (p+1)*p^(e-1) for p-2 mod 8 = +-3. - Andrew Howroyd, Jul 15 2018
Sum_{k=1..n} a(k) ~ c * n^2, where c = 7/(16*A309710) = 0.410900684788977656... . - Amiram Eldar, Nov 21 2023
Showing 1-5 of 5 results.