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 1-4 of 4 results.

A361430 Multiplicative with a(p^e) = e - 1.

Original entry on oeis.org

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

Views

Author

Vaclav Kotesovec, Mar 11 2023

Keywords

Comments

a(n) is the number of coreful divisors d of n such that n/d is also a coreful divisor of n (a coreful divisor d of a number n is a divisor with the same set of distinct prime factors as n, see A307958). - Amiram Eldar, Aug 15 2023

Crossrefs

Cf. A001694, A298826, A307958, A335850 (indices of records).
Cf. A005361 (multiplicative with a(p^e) = e), A000005 (e+1), A343443 (e+2), A360997 (e+3), A360908 (2*e-1), A360910 (3*e-1), A360911 (3*e-2).

Programs

  • Mathematica
    g[p_, e_] := e-1; a[1] = 1; a[n_] := Times @@ g @@@ FactorInteger[n]; Array[a, 200]
  • PARI
    for(n=1, 100, print1(direuler(p=2, n, 1 + 1/(1 - 1/X)^2)[n], ", "))
    
  • PARI
    a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1] = f[k,2]-1; f[k,2] = 1); factorback(f); \\ Michel Marcus, Mar 13 2023
    
  • Python
    from math import prod
    from sympy import factorint
    def A361430(n): return prod(e-1 for e in factorint(n).values()) # Chai Wah Wu, Apr 15 2025

Formula

Dirichlet g.f.: Product_{primes p} (1 + 1/(p^s - 1)^2).
Dirichlet g.f.: zeta(2*s) * zeta(3*s)^2 * Product_{primes p} (1 + 2/p^(4*s) + 2/p^(5*s) - 1/p^(6*s) - 2/p^(7*s) - 2/p^(8*s)).
Let f(s) = Product_{primes p} (1 + 2/p^(4*s) + 2/p^(5*s) - 1/p^(6*s) - 2/p^(7*s) - 2/p^(8*s)), then
Sum_{k=1..n} a(k) ~ f(1/2) * zeta(3/2)^2 * sqrt(n) + zeta(2/3) * (f(1/3) * (log(n) + 6*gamma - 3 + 2*zeta'(2/3)/zeta(2/3)) + f'(1/3)) * n^(1/3) / 3, where
f(1/2) = Product_{primes p} (1 + 2/p^2 + 2/p^(5/2) - 1/p^3 - 2/p^(7/2) - 2/p^4) = 2.20286226691565931157047065666916419062717171359087693723221239...
f(1/3) = Product_{primes p} (1 + 2/p^(4/3) + 2/p^(5/3) - 1/p^2 - 2/p^(7/3) - 2/p^(8/3)) = 6.250573144372477079986352917664218040797528021629950408099536...
f'(1/3) = f(1/3) * Sum_{primes p} (-2*(-8 + p^(1/3) + 4*p^(2/3)) * log(p) / (-2 + p^(2/3) + p - p^(5/3) + p^2)) = -90.898558294301467740374653006294640945295... and gamma is the Euler-Mascheroni constant A001620.
Conjecture: a(n) = abs(A298826(n)).
a(n) > 0 if and only if n is powerful (A001694). - Amiram Eldar, Aug 15 2023

A298825 Row sums of A298824.

Original entry on oeis.org

1, 0, 0, 4, 0, 0, 0, 16, -9, 0, 0, 0, 0, 0, 0, 48, 0, 0, 0, 0, 0, 0, 0, 0, -25, 0, -54, 0, 0, 0, 0, 128, 0, 0, 0, -36, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -49, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 320, 0, 0, 0, 0, 0, 0, 0, -144, 0, 0, 0, 0, 0, 0, 0, 0, -243, 0, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Mats Granvik, Jan 27 2018

Keywords

Comments

Positions of nonzero entries appear to be given by A001694.

Crossrefs

Programs

  • Mathematica
    a[n_] := n*Sum[If[Mod[n, j] == 0,DivisorSigma[0, n/j]*1/j*Sum[Sum[If[Mod[j, k] == 0, If[Mod[m, j/k] == 0, MoebiusMu[j/k]*j/k, 0], 0]*If[Mod[m + 2, k/1] == 0, MoebiusMu[k/1]*k/1, 0], {k, 1, j}], {m, 1, j}], 0], {j, 1, n}]; a /@ Range[85] (* Mats Granvik, Mar 03 2019 *)
  • PARI
    up_to = 256;
    DirConv(ma,h) = { my(u = matsize(ma)[1], md = matrix(u,u)); for(n=1,u-h,for(k=1,u,md[n,k] = sumdiv(k,d,ma[n,d]*ma[n+h,k/d]))); (md); };
    A298825list(up_to) = { my(h=2, matA = matrix(up_to+h,up_to+h,n,k,!(n%k)), matB = matrix(up_to+h,up_to+h,n,k,(!(k%n))*moebius(n)*n), matT = matA*matB, matD = DirConv(matT,2)); vector(up_to,i,sum(j=1,i,matD[j,i])); };
    v298825 = A298825list(up_to);
    A298825(n) = v298825[n]; \\ Antti Karttunen, Sep 30 2018

Formula

From Mats Granvik, Mar 03 2019: (Start)
a(n) = n*Sum_{j=1..n} [j divides n]*A000005(n/j)*A306653(j).
a(n) = n*Sum_{j=1..n} [j divides n]*A000005(n/j)*Sum_{m=1..n} Sum_{k=1..n} [k divides j]*[j/k divides m]*A008683(j/k)*j/k*[k divides m + 2^p]*A008683(k)*k, p=1,2,3,4,5,...,infinity.
(End)

A298674 Square matrix read by antidiagonals up. Matrix of Dirichlet series associated with Sum_{n<=X} MangoldtLambda(n) * MangoldtLambda(n+2).

Original entry on oeis.org

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

Views

Author

Mats Granvik, Jan 24 2018

Keywords

Comments

For n > 1: Sum_{k>=1} T(n,k) = log(A014963(n))*log(A014963(n+2)).
Triangular submatrix of this matrix is A298824.
Row sums of A298824 are found in A298825. A298825(n)/n = A298826(n). A298826 appears to be relevant to the heuristic for the twin prime conjecture.
By varying the prime gap "h" in the program it appears that prime gaps that are powers of "h" have the same row sums of the triangular submatrix, which in turn seems to imply that prime gaps equal to powers of "h" have the same density.

Examples

			The square matrix starts:
{
  {1,  2, -1,  3,  2, -2,  2,  4, -3,  4,  2, -3},
  {1, -2,  2, -1,  2, -4,  2,  0,  3, -4,  2, -2},
  {1,  2, -1,  3, -3, -2,  2,  4, -3, -6,  2, -3},
  {1, -2, -1, -1,  2,  2,  2,  0, -3, -4,  2,  1},
  {1,  2,  2,  3, -3,  4, -5,  4,  3, -6,  2,  6},
  {1, -2, -1, -1,  2,  2,  2,  0, -3, -4,  2,  1},
  {1,  2, -1,  3,  2, -2, -5,  4, -3,  4,  2, -3},
  {1, -2,  2, -1, -3, -4,  2,  0,  3,  6,  2, -2},
  {1,  2, -1,  3,  2, -2,  2,  4, -3,  4, -9, -3},
  {1, -2, -1, -1, -3,  2,  2,  0, -3,  6,  2,  1}
}
		

Crossrefs

Programs

  • Mathematica
    h = 2; nn = 14;
    A = Table[If[Mod[n, k] == 0, 1, 0], {n, nn}, {k, nn}];
    B = Table[If[Mod[k, n] == 0, MoebiusMu[n]*n, 0], {n, nn}, {k, nn}];
    T = (A.B);
    TableForm[TwinMangoldt = Table[a = T[[All, kk]];
        F1 = Table[If[Mod[n, k] == 0, a[[n/k]], 0], {n, nn}, {k, nn}];
        b = T[[All, kk + h]];
        F2 = Table[If[Mod[n, k] == 0, b[[n/k]], 0], {n, nn}, {k, nn}];
        (F1.F2)[[All, 1]], {kk, nn - h}]];
    Flatten[Table[TwinMangoldt[[n - k + 1, k]], {n, nn - h}, {k, n}]]

Formula

Let h = 2.
Let A(n,k) = 1 if n mod k = 0, otherwise 0.
Let B(n,k) = A008683(n)*n if k mod n = 0, otherwise 0.
Let T = A.B (where "." is matrix multiplication).
Take the Dirichlet convolution of a row in T(n,k) and a row in T(n+h,k) for n=1,2,3,4,5,... infinity, and form this matrix from the first columns of the convolutions. See Mathematica program for more precise description.

A306653 a(n) = Sum_{m=1..n} Sum_{k=1..n} [k divides n]*[n/k divides m]*A008683(n/k)*n/k*[k divides m + 2^p]*A008683(k)*k, where p can be a positive integer: 1,2,3,4,5,...

Original entry on oeis.org

1, -2, -2, 2, -2, 4, -2, 0, 0, 4, -2, -4, -2, 4, 4, 0, -2, 0, -2, -4, 4, 4, -2, 0, 0, 4, 0, -4, -2, -8, -2, 0, 4, 4, 4, 0, -2, 4, 4, 0, -2, -8, -2, -4, 0, 4, -2, 0, 0, 0, 4, -4, -2, 0, 4, 0, 4, 4, -2, 8, -2, 4, 0, 0, 4, -8, -2, -4, 4, -8, -2, 0, -2, 4, 0, -4, 4, -8, -2, 0, 0, 4, -2, 8, 4
Offset: 1

Views

Author

Mats Granvik, Mar 03 2019

Keywords

Comments

It appears that for p=1 and p=2, a(n) is identically the same for all n except for n equal to multiples of 16. See A306652 for comparison.

Crossrefs

Programs

  • Mathematica
    (* Dirichlet Convolution. *)
    p=1;
    a[n_] := 1/n*Sum[Sum[If[Mod[n, k] == 0, 1, 0]*If[Mod[m, n/k] == 0, 1, 0]*If[Mod[m + 2^p, k] == 0, 1, 0]*MoebiusMu[n/k]*n/k*MoebiusMu[k]*k, {k, 1, n}], {m, 1, n}]; a /@ Range[85]
    (* conjectured faster program *)
    nn = 85;
    b = 4*Select[Range[1, nn, 2], SquareFreeQ];
    bb = Table[DivisorSigma[0, n]*(MoebiusMu[n] + Sum[If[b[[j]] == n, LiouvilleLambda[n]*2/3, 0], {j, 1, Length[b]}]), {n, 1, nn}]
  • PARI
    A306653(n) = (1/n)*sum(m=1, n, sumdiv(n, k, if( !(m%(n/k)) && !((m+(2^1))%k), n*moebius(n/k)*moebius(k),0))); \\ Antti Karttunen, Mar 15 2019

Formula

a(n) = 1/n * Sum_{m=1..n} Sum_{k=1..n} [k divides n]*[n/k divides m]*[k divides m + 2^p]*A008683(n/k)*n/k*A008683(k)*k, where p can be any positive integer: 1,2,3,4,5,...
a(n) = A000005(n)*(A008683(n) + Sum_{j=1..n} [4*A056911(j)=n]*A008836(n)*2/3) (conjectured formula verified for n=1..2500).
Dirichlet generating function, after Daniel Suteu in A298826 and Álvar Ibeas in A076479, appears to be: Sum_{n>=1} a(n)/n^s = (Product_{j>=1} (1 - 2*prime(j)^(-s)))*(1 + Sum_{n>=2} (1/2/2^(n*(s - 1)))). - Mats Granvik, Apr 07 2019
The conjectured Dirichlet generating function simplifies to: Sum_{n>=1} a(n)/n^s = (Product_{j>=1} (1 - 2*prime(j)^(-s)))*(1 + 2^(1 - s)/(2^s - 2)). - Steven Foster Clark, Sep 23 2022
Showing 1-4 of 4 results.