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 1-6 of 6 results.

A378008 a(n) = b(5*n+1), with the sequence {b(n)} having Dirichlet g.f. Product_{chi} L(chi,s), where chi runs through all Dirichlet characters modulo 5; 5th column of A378007.

Original entry on oeis.org

1, 0, 4, 1, 0, 0, 4, 0, 4, 0, 0, 0, 4, 0, 4, 0, 1, 0, 0, 0, 4, 0, 0, 0, 10, 0, 4, 0, 0, 0, 4, 0, 0, 0, 0, 4, 4, 0, 4, 0, 0, 0, 4, 0, 0, 0, 0, 0, 4, 0, 4, 1, 0, 0, 4, 0, 4, 0, 0, 0, 0, 0, 4, 0, 0, 0, 4, 0, 16, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 4, 0, 4, 0, 0, 0, 16
Offset: 0

Views

Author

Jianing Song, Nov 14 2024

Keywords

Examples

			(1 + 1/2^s + 1/3^s + 1/4^s + ...)*(1 + i/2^s - i/3^s - 1/4^s + ...)*(1 - 1/2^s - 1/3^s + 1/4^s + ...)*(1 - i/2^s + i/3^s - 1/4^s + ...) = 1 + 4/11^s + 1/16^s + 4/31^s + 4/41^s + ...

Links

Crossrefs

Cf. A378007.

Programs

  • PARI
    A378008(n) = {
my(f = factor(5*n+1), res = 1); for(i=1, #f~,
if(f[i,1] % 5 == 1, res *= binomial(f[i,2]+3, 3));
if(f[i,1] % 5 == 4, if(f[i,2] % 2 == 0, res *= f[i,2]/2+1, return(0)));
if(f[i,1] % 5 == 2 || f[i,1] % 5 == 3, if(f[i,2] % 4 != 0, return(0))));
res; }

Formula

a(n) = b(5*n+1), where {b(n)} is multiplicative with:
- b(5^e) = 0;
- for p == 1 (mod 5), b(p^e) = binomial(e+3,3) = (e+3)*(e+2)*(e+1)/6;
- for p == 4 (mod 5), b(p^e) = e/2 + 1 if e is even, and 0 otherwise;
- for p == 2, 3 (mod 5), b(p^e) = 1 if 4 divides e, and 0 otherwise.

A378009 a(n) = b(7*n+1), with the sequence {b(n)} having Dirichlet g.f. Product_{chi} L(chi,s), where chi runs through all Dirichlet characters modulo 7; 7th column of A378007.

Original entry on oeis.org

1, 2, 0, 0, 6, 0, 6, 0, 0, 3, 6, 0, 0, 0, 0, 0, 6, 0, 6, 0, 0, 0, 0, 0, 3, 0, 0, 0, 6, 0, 6, 0, 0, 12, 6, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 6, 12, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 6, 0, 0, 0, 6, 0, 6, 0, 0, 0, 6, 0, 0, 4, 0, 0, 0, 0, 6, 0, 0, 12, 0, 0, 0, 0, 0, 0, 6, 0, 6
Offset: 0

Views

Author

Jianing Song, Nov 14 2024

Keywords

Examples

			Write w = exp(2*Pi*i/3) = (-1 + sqrt(3)*i)/2, then (1 + 1/2^s + 1/3^s + 1/4^s + 1/5^s + 1/6^s + ...)*(1 + w/2^s + (w+1)/3^s - (w+1)/4^s - w/5^s - 1/6^s + ...)*(1 - (w+1)/2^s + w/3^s + w/4^s - (w+1)/5^s + 1/6^s + ...)*(1 + 1/2^s - 1/3^s + 1/4^s - 1/5^s - 1/6^s + ...)*(1 + w/2^s - (w+1)/3^s - (w+1)/4^s + w/5^s + 1/6^s + ...)*(1 - (w+1)/2^s - w/3^s + w/4^s + (w+1)/5^s - 1/6^s + ...) = 1 + 2/8^s + 6/29^s + 6/43^s + 3/64^s + 6/71^s + ...

Links

Crossrefs

Cf. A378007.

Programs

  • PARI
    A378009(n) = {
my(f = factor(7*n+1), res = 1); for(i=1, #f~,
if(f[i,1] % 7 == 1, res *= binomial(f[i,2]+5, 5));
if(f[i,1] % 7 == 6, if(f[i,2] % 2 == 0, res *= binomial(f[i,2]/2+2, 2), return(0)));
if(f[i,1] % 7 == 2 || f[i,1] % 7 == 4, if(f[i,2] % 3 == 0, res *= f[i,2]/3+1, return(0)));
if(f[i,1] % 7 == 3 || f[i,1] % 7 == 5, if(f[i,2] % 6 != 0, return(0))));
res; }

Formula

a(n) = b(7*n+1), where {b(n)} is multiplicative with:
- b(7^e) = 0;
- for p == 1 (mod 7), b(p^e) = binomial(e+5,5) = (e+5)*(e+4)*(e+3)*(e+2)*(e+1)/120;
- for p == 6 (mod 7), b(p^e) = binomial(e/2+2,2) = (e/2+2)*(e/2+1)/2 if e is even, and 0 otherwise;
- for p == 2, 4 (mod 7), b(p^e) = e/3 + 1 if 3 divides e, and 0 otherwise;
- for p == 3, 5 (mod 7), b(p^e) = 1 if 6 divides e, and 0 otherwise.

A378010 a(n) = b(8*n+1), with the sequence {b(n)} having Dirichlet g.f. Product_{chi} L(chi,s), where chi runs through all Dirichlet characters modulo 8; 8th column of A378007.

Original entry on oeis.org

1, 2, 4, 2, 0, 4, 2, 0, 0, 4, 3, 4, 4, 0, 4, 2, 0, 4, 0, 8, 0, 2, 0, 0, 4, 0, 0, 0, 4, 4, 4, 0, 4, 0, 0, 4, 10, 0, 0, 4, 0, 0, 4, 0, 4, 2, 8, 0, 0, 0, 4, 4, 0, 8, 4, 4, 4, 4, 0, 0, 0, 0, 0, 0, 0, 4, 2, 0, 0, 0, 0, 4, 4, 0, 4, 4, 0, 4, 3, 0, 4, 0, 8, 0, 4, 0, 0, 16, 0, 0, 0
Offset: 0

Views

Author

Jianing Song, Nov 14 2024

Keywords

Examples

			(1 + 1/3^s + 1/5^s + 1/7^s + ...)*(1 + 1/3^s - 1/5^s - 1/7^s + ...)*(1 - 1/3^s + 1/5^s - 1/7^s + ...)*(1 - 1/3^s - 1/5^s + 1/7^s + ...) = 1 + 2/9^s + 4/17^s + 2/25^s + 4/41^s + 2/49^s + 4/73^s + 3/81^s + ...

Links

Crossrefs

Cf. A378007.

Programs

  • PARI
    A378010(n) = {
my(f = factor(8*n+1), res = 1); for(i=1, #f~,
if(f[i,1] % 8 == 1, res *= binomial(f[i,2]+3, 3));
if(f[i,1] % 8 == 3 || f[i,1] % 8 == 5 || f[i,1] % 8 == 7, if(f[i,2] % 2 == 0, res *= f[i,2]/2+1, return(0))));
res; }

Formula

a(n) = b(8*n+1), where {b(n)} is multiplicative with:
- b(2^e) = 0;
- for p == 1 (mod 8), b(p^e) = binomial(e+3,3) = (e+3)*(e+2)*(e+1)/6;
- for p == 3, 5, 7 (mod 8), b(p^e) = e/2 + 1 if e is even, and 0 otherwise.

A378011 a(n) = b(9*n+1), with the sequence {b(n)} having Dirichlet g.f. Product_{chi} L(chi,s), where chi runs through all Dirichlet characters modulo 9; 9th column of A378007.

Original entry on oeis.org

1, 0, 6, 0, 6, 0, 0, 1, 6, 0, 0, 0, 6, 0, 6, 0, 0, 0, 6, 0, 6, 0, 6, 0, 0, 0, 0, 0, 0, 0, 6, 0, 3, 0, 6, 0, 0, 0, 2, 0, 21, 0, 6, 0, 6, 0, 0, 0, 6, 0, 0, 0, 0, 0, 6, 0, 0, 0, 6, 0, 6, 0, 0, 0, 6, 0, 0, 0, 6, 0, 6, 0, 0, 0, 0, 0, 0, 0, 36, 0, 0, 0, 6, 0, 6, 0, 0, 0, 0, 0, 6
Offset: 0

Views

Author

Jianing Song, Nov 14 2024

Keywords

Examples

			Write w = exp(2*Pi*i/3) = (-1 + sqrt(3)*i)/2, then (1 + 1/2^s + 1/4^s + 1/5^s + 1/7^s + 1/8^s + ...)*(1 + (w+1)/2^s + w/4^s - w/5^s - (w+1)/7^s - 1/8^s + ...)*(1 + w/2^s - (w+1)/4^s - (w+1)/5^s + w/7^s + 1/8^s + ...)*(1 - 1/2^s + 1/4^s - 1/5^s + 1/7^s - 1/8^s + ...)*(1 - (w+1)/2^s + w/4^s + w/5^s - (w+1)/7^s + 1/8^s + ...)*(1 - w/2^s - (w+1)/4^s + (w+1)/5^s + w/7^s - 1/8^s + ...) = 1 + 6/19^s + 6/37^s + 1/64^s + 6/73^s + ...

Links

Crossrefs

Cf. A378007.

Programs

  • PARI
    A378011(n) = {
my(f = factor(9*n+1), res = 1); for(i=1, #f~,
if(f[i,1] % 9 == 1, res *= binomial(f[i,2]+5, 5));
if(f[i,1] % 9 == 8, if(f[i,2] % 2 == 0, res *= binomial(f[i,2]/2+2, 2), return(0)));
if(f[i,1] % 9 == 4 || f[i,1] % 9 == 7, if(f[i,2] % 3 == 0, res *= f[i,2]/3+1, return(0)));
if(f[i,1] % 9 == 2 || f[i,1] % 9 == 5, if(f[i,2] % 6 != 0, return(0))));
res; }

Formula

a(n) = b(9*n+1), where {b(n)} is multiplicative with:
- b(3^e) = 0;
- for p == 1 (mod 9), b(p^e) = binomial(e+5,5) = (e+5)*(e+4)*(e+3)*(e+2)*(e+1)/120;
- for p == 8 (mod 9), b(p^e) = binomial(e/2+2,2) = (e/2+2)*(e/2+1)/2 if e is even, and 0 otherwise;
- for p == 4, 7 (mod 9), b(p^e) = e/3 + 1 if 3 divides e, and 0 otherwise;
- for p == 2, 5 (mod 9), b(p^e) = 1 if 6 divides e, and 0 otherwise.

A378012 a(n) = b(10*n+1), with the sequence {b(n)} having Dirichlet g.f. Product_{chi} L(chi,s), where chi runs through all Dirichlet characters modulo 10; 10th column of A378007.

Original entry on oeis.org

1, 4, 0, 4, 4, 0, 4, 4, 1, 0, 4, 0, 10, 4, 0, 4, 0, 0, 4, 4, 0, 4, 0, 0, 4, 4, 0, 4, 4, 0, 0, 4, 0, 4, 16, 0, 2, 0, 0, 0, 4, 0, 4, 4, 0, 16, 4, 0, 0, 4, 0, 0, 4, 0, 4, 0, 0, 4, 0, 0, 4, 0, 0, 4, 4, 0, 4, 16, 0, 4, 4, 0, 0, 0, 0, 4, 4, 0, 16, 0, 0, 4, 4, 0, 2, 0, 0, 0, 4, 4, 0
Offset: 0

Views

Author

Jianing Song, Nov 14 2024

Keywords

Examples

			(1 + 1/3^s + 1/7^s + 1/9^s + ...)*(1 + i/3^s - i/7^s - 1/9^s + ...)*(1 - 1/3^s - 1/7^s + 1/9^s + ...)*(1 - i/3^s + i/7^s - 1/9^s + ...) = 1 + 4/11^s + 4/31^s + 4/41^s + 4/61^s + 4/71^s + 1/81^s + 4/101^s + ...

Links

Programs

  • PARI
    A378012(n) = {
my(f = factor(10*n+1), res = 1); for(i=1, #f~,
if(f[i,1] % 10 == 1, res *= binomial(f[i,2]+3, 3));
if(f[i,1] % 10 == 9, if(f[i,2] % 2 == 0, res *= f[i,2]/2+1, return(0)));
if(f[i,1] % 10 == 3 || f[i,1] % 10 == 7, if(f[i,2] % 4 != 0, return(0))));
res; }

Formula

a(n) = b(10*n+1), where {b(n)} is multiplicative with:
- b(2^e) = b(5^e) = 0;
- for p == 1 (mod 10), b(p^e) = binomial(e+3,3) = (e+3)*(e+2)*(e+1)/6;
- for p == 9 (mod 10), b(p^e) = e/2 + 1 if e is even, and 0 otherwise;
- for p == 3, 7 (mod 10), b(p^e) = 1 if 4 divides e, and 0 otherwise.
a(n) = A378008(2*n).

A378006 Square table read by descending antidiagonals: the k-th column has Dirichlet g.f. Product_{chi} L(chi,s), where chi runs through all Dirichlet characters modulo k.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 2, 0, 1, 1, 1, 0, 0, 0, 0, 0, 2, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 2, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
Offset: 1

Views

Author

Jianing Song, Nov 14 2024

Keywords

Comments

For fixed k, we have Product_{chi} L(chi,s) = Product_{p not dividing k} 1/(1 - 1/p^(ord(p,k)*s))^(phi(k)/ord(p,k)), where phi = A000010 is the Euler totient function and ord(a,k) is the multiplicative order of a modulo k; see Section 3.4 of Chapter VI, Proposition 13, page 72 of J.-P. Serre, A Course in Arithmetic. Using the series expansion of 1/(1-x)^r, we get Product_{chi} L(chi,s) = Product_{p not dividing k} (Sum_{n>=0} binomial(n+phi(k)/ord(p,k)-1,phi(k)/ord(p,k)-1)/p^(ord(p,k)*s)), giving us the formula to calculate T(n,k).
From the formula we can wee that T(n,k) = 0 unless n == 1 (mod k). A378007 is the condensed version giving only {T(k*n+1,k)}.

Examples

			Table starts
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
  1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
  1, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...
  1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...
  1, 1, 0, 2, 0, 0, 0, 0, 0, 0, ...
  1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
  1, 1, 2, 0, 0, 2, 0, 0, 0, 0, ...
  1, 0, 0, 0, 0, 0, 2, 0, 0, 0, ...
  1, 1, 0, 1, 0, 0, 0, 2, 0, 0, ...
  1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
See A378007 for more details.

Links

Crossrefs

Columns: A000012 (k=1), A000035 (k=2), A045833 (k=3), A008442 (k=4).
Cf. A378007.

Programs

  • PARI
    A378006(n,k) = {
my(f = factor(n), res = 1); for(i=1, #f~, if(k % f[i,1] == 0, return(0));
my(d = znorder(Mod(f[i,1],k))); if(f[i,2] % d != 0, return(0), my(m = f[i,2]/d, r = eulerphi(k)/d); res *= binomial(m+r-1,r-1)));
res;}

Formula

Each column is multiplicative: T(p^e,k) = 0 if p divides k; 0 if e is not divisible by ord(p,k); binomial(e/ord(p,k)+phi(k)/ord(p,k)-1,phi(k)/ord(p,k)-1) otherwise.
For odd k, T(2*k,n) = T(k,n) for odd n, 0 for even n.
Showing 1-6 of 6 results.

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