A333119 -id:A333119 - cp's OEIS frontend

A062011 a(n) = 2tau(n) = 2A000005(n).

2, 4, 4, 6, 4, 8, 4, 8, 6, 8, 4, 12, 4, 8, 8, 10, 4, 12, 4, 12, 8, 8, 4, 16, 6, 8, 8, 12, 4, 16, 4, 12, 8, 8, 8, 18, 4, 8, 8, 16, 4, 16, 4, 12, 12, 8, 4, 20, 6, 12, 8, 12, 4, 16, 8, 16, 8, 8, 4, 24, 4, 8, 12, 14, 8, 16, 4, 12, 8, 16, 4, 24, 4, 8, 12, 12, 8, 16, 4, 20, 10, 8, 4, 24, 8, 8, 8, 16

Offset: 1

Ahmed Fares (ahmedfares(AT)my-deja.com), Jul 12 2001

Old definition was "Number of cyclic subgroups of the group C_n X C_2 (where C_n is the cyclic group with n elements)."
More generally, the number of cyclic subgroups of the group C_n X C_m is Sum_{i|n, j|m} phi(i)*phi(j)/phi(lcm(i,j)), where phi=Euler totient function, cf. A000010. - Vladeta Jovovic, Jul 15 2001
Number of divisors of p*n, where p is any prime not dividing n. - Reinhard Zumkeller, May 17 2006
From Enrique Pérez Herrero, Jul 21 2011: (Start)
If p(x) is a polynomial with integer coefficients, and if r is an integer zero of p(x), then r is a divisor of the constant term c_0 of p(x). Under this theorem, p(x) can have a(c_0) possible integer roots.
a(n) is the number of integer divisors of n, while A000005(n) is the number of positive divisors. (End)
Number of solutions to the Diophantine equation i*j = n*i + j. - Robert G. Wilson v, Apr 10 2019
a(n) is also the number of times n appears in the triangle A333119, or equivalently, the number of positive integer solutions of the equation A333119(x, y) = n for y < x. - Stefano Spezia, Oct 05 2022

Harry J. Smith, Table of n, a(n) for n = 1..1000
Omar E. Pol, Illustration of initial terms
Index entries for sequences related to groups

Cf. A000005, A060648, A060710.
Cf. A087560, A119416.
Cf. A333119.

A062011[n_] := 2*DivisorSigma[0,n]; Array[A062011,50] (* Enrique Pérez Herrero, Jul 15 2011 *)

for (n=1, 1000, write("b062011.txt", n, " ", 2*numdiv(n))) \\ Harry J. Smith, Jul 29 2009

a(n) = A000005(A087560(n)) = A000005(A119416(n)). - Reinhard Zumkeller, May 17 2006

L.g.f.: -log(Product_{k>=1} (1 - x^k)^(2/k)) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, Mar 18 2018

More terms from Vladeta Jovovic, Jul 14 2001

Edited by N. J. A. Sloane, Sep 20 2018, replacing old definition (which was of course correct) with a simple formula.

A338456 a(n) is the hafnian of a symmetric Toeplitz matrix M(2n) whose first row consists of a single zero followed by successive positive integers repeated (A004526).

Original entry on oeis.org

1, 1, 4, 45, 968, 34265, 1799748, 131572357, 12770710096, 1589142683313, 246658484353100

Offset: 0

Stefano Spezia, Oct 28 2020

			a(2) = 4 because the hafnian of
0  1  1  2
1  0  1  1
1  1  0  1
2  1  1  0
equals M_{1,2}*M_{3,4} + M_{1,3}*M_{2,4} + M_{1,4}*M_{2,3} = 4.

Wikipedia, Hafnian
Wikipedia, Symmetric matrix
Wikipedia, Toeplitz Matrix

Cf. A004526.
Cf. A002378 (conjectured determinant of M(2n+1)), A083392 (conjectured determinant of M(n+1)), A332566 (permanent of M(n)), A333119 (k-th super- and subdiagonal sums of the matrix M(n)).
Cf. A202038, A336114, A336286, A336400.

k[i_]:=Floor[i/2]; M[i_, j_, n_]:=Part[Part[ToeplitzMatrix[Array[k, n]], i], j]; a[n_]:=Sum[Product[M[Part[PermutationList[s, 2n], 2i-1], Part[PermutationList[s, 2n], 2i], 2n], {i, n}], {s, SymmetricGroup[2n]//GroupElements}]/(n!*2^n); Array[a, 5, 0]

tm(n) = {my(m = matrix(n, n, i, j, if (i==1, j\2, if (j==1, i\2)))); for (i=2, n, for (j=2, n, m[i, j] = m[i-1, j-1]; ); ); m; }
a(n) = {my(m = tm(2*n), s=0); forperm([1..2*n], p, s += prod(j=1, n, m[p[2*j-1], p[2*j]]);); s/(n!*2^n);} \\ Michel Marcus, Nov 11 2020

a(5) from Michel Marcus, Nov 11 2020

a(6)-a(10) from Pontus von Brömssen, Oct 14 2023

A332566 a(n) is the permanent of an n X n symmetric Toeplitz matrix M(n) whose first row consists of a single zero followed by successive positive integers repeated (A004526).

Original entry on oeis.org

1, 0, 1, 2, 16, 150, 2333, 45840, 1227816, 40715300, 1701223409, 84902728550, 5108474886424, 357837483830570, 29336856811970045, 2745407159100236484, 294324995624694053072, 35473014438701226021416, 4818705384665419284918401, 727012502373285844943278058, 122057159014198483893887865744

Offset: 0

Stefano Spezia, Feb 16 2020

nonn

			For n = 4 the matrix M(4) is
    0 1 1 2
    1 0 1 1
    1 1 0 1
    2 1 1 0
with permanent a(4) = 16.

Vaclav Kotesovec, Table of n, a(n) for n = 0..35
Wikipedia, Symmetric matrix
Wikipedia, Toeplitz Matrix

Cf. A004526, A083392, A331491, A333119.

nmax:=20; k[i_]:=Floor[i/2];a[n_]:=If[n==0,1,Permanent[ToeplitzMatrix[Array[k, n], Array[k, n]]]]; Table[a[n],{n,0,nmax}]

tm(n) = {my(m = matrix(n, n, i, j, if (i==1, floor(j/2), if (j==1, floor(i/2))))); for (i=2, n, for (j=2, n, m[i, j] = m[i-1, j-1]; ); ); m; }
a(n) = matpermanent(tm(n));

A374708 Triangle T read by rows: T(n,k) = (n - k)n(4n^2 - 4nk + 2k^2 - 1 + (-1)^k)/4, with 0 <= k < n.

Original entry on oeis.org

1, 16, 4, 81, 36, 15, 256, 144, 80, 32, 625, 400, 255, 140, 65, 1296, 900, 624, 396, 240, 108, 2401, 1764, 1295, 896, 609, 364, 175, 4096, 3136, 2400, 1760, 1280, 864, 544, 256, 6561, 5184, 4095, 3132, 2385, 1728, 1215, 756, 369, 10000, 8100, 6560, 5180, 4080, 3100, 2320, 1620, 1040, 500

Offset: 1

Stefano Spezia, Jul 17 2024

T(n, k) is the k-th super- and subdiagonal sum of the Hankel matrix M(n) whose permanent is A374668(n).

			n\k|    0    1    2    3    4    5
---+------------------------------
1  |    1
2  |   16    4
3  |   81   36   15
4  |  256  144   80   32
5  |  625  400  255  140   65
6  | 1296  900  624  396  240  108
      ...
For n = 3 the matrix M is
  [ 1,  4, 15]
  [ 4, 15, 32]
  [15, 32, 65]
and therefore T(3, 0) = 1 + 15 + 65 = 81, T(3, 1) = 4 + 32 = 36, and T(3, 2) = 15.

Cf. A317614 (diagonal), A374668.
Cf. A333119, A330613.
Cf. A000583 (k=0), A035287 (k=1), A123865, A374709 (row sums).

T[n_,k_]:=(n-k)*n*(4*n^2 - 4*n*k+2*k^2-1+(-1)^k)/4; Table[T[n,k],{n,10},{k,0,n-1}]//Flatten

O.g.f.: x*(1 - 4*x^8*y^5 + x*(11 + 2*y) - x^7*y^4*(7 + 16*y) - x^2*(-11 + 6*y - 6*y^2) - x^5*y^2*(2 - 46*y - 3*y^2) - x^6*y^3*(-2 - 27*y + 4*y^2) - x^3*(-1 + 18*y + 38*y^2 - 2*y^3) - x^4*y*(2 + 14*y + 2*y^2 - y^3))/((1 - x)^5*(1 - x*y)^4*(1 + x*y)^2).

T(n,2) = A123865(n-1) for n > 1.

A338796 Triangle T read by rows: T(n, k) is the k-th row sum of the symmetric Toeplitz matrix M(n) whose first row consists of a single zero followed by successive positive integers repeated (A004526).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 4, 3, 3, 4, 6, 5, 4, 5, 6, 9, 7, 6, 6, 7, 9, 12, 10, 8, 8, 8, 10, 12, 16, 13, 11, 10, 10, 11, 13, 16, 20, 17, 14, 13, 12, 13, 14, 17, 20, 25, 21, 18, 16, 15, 15, 16, 18, 21, 25, 30, 26, 22, 20, 18, 18, 18, 20, 22, 26, 30, 36, 31, 27, 24, 22, 21, 21, 22, 24, 27, 31, 36

Offset: 1

Stefano Spezia, Nov 12 2020

			n\k| 1 2 3 4 5 6
---+------------
1  | 0
2  | 1 1
3  | 2 2 2
4  | 4 3 3 4
5  | 6 5 4 5 6
6  | 9 7 6 6 7 9
...
For n = 4 the matrix M(4) is
        0 1 1 2
        1 0 1 1
        1 1 0 1
        2 1 1 0
and therefore T(4, 1) = 4, T(4, 2) = 3, T(4, 3) = 3 and T(4, 4) = 4.

Cf. A004526.
Cf. A002620, A033638, A212964, A290743.
Cf. A002378 (conjectured determinant of M(2n+1)), A083392 (conjectured determinant of M(n+1)), A332566 (permanent of M(n)), A333119 (k-th super- and subdiagonal sums of the matrix M(n)), A338456 (hafnian of M(n)).

T[n_,k_]:=((-1)^k+(-1)^(n-k+1)+4k^2+4n+2n^2-4k(n+1))/8; Flatten[Table[T[n,k],{n,12},{k,n}]] (* or *)
r[n_]:=Table[SeriesCoefficient[(2x^3y^2+y^2(1+y)+x^2(y-3y^2)-x(-1+2y+y^2))/((1-x)^3(1+x)(1-y)^3(1+y)),{x,0,i},{y,0,j}],{i,n,n},{j, n}]; Flatten[Array[r,12]] (* or *)
r[n_]:=Table[SeriesCoefficient[1/8 E^(-x-y)(-1+E^(2 x)+2 E^(2 (x+y))(x (3+x)-2 x y+2 y^2)),{x, 0, i},{y, 0, j}]i!j!,{i, n, n},{j, n}]; Flatten[Array[r, 12]]

tm(n) = {my(m = matrix(n, n, i, j, if (i==1, j\2, if (j==1, i\2)))); for (i=2, n, for (j=2, n, m[i, j] = m[i-1, j-1]; ); ); m; }
T(n, k) = my(m = tm(n)); sum(i=1, n, m[i, k]);
matrix(10, 10, n, k, if (n>=k, T(n,k), 0)) \\ Michel Marcus, Nov 12 2020

O.g.f.: (2*x^3*y^2 + y^2*(1 + y) + x^2*(y - 3*y^2) - x*(-1 + 2*y + y^2))/((1 - x)^3*(1 + x) *(1 - y)^3*(1 + y)).
E.g.f.: exp(-x-y)*(exp(2*x) + 2*exp(2*(x+y))*(x*(3 + x) - 2*x*y + 2*y^2 - 1))/8.
T(n, k) = ((-1)^k + (-1)^(n-k+1) + 4*k^2 + 4*n + 2*n^2 - 4*k*(n + 1))/8.
T(n, 1) = T(n, n) = A002620(n).
T(n, 2) = A033638(n-1).
T(n, 3) = A290743(n-2).
Sum_{k=1..n} T(n, k) = A212964(n+1).

cp's OEIS Frontend

A062011 a(n) = 2*tau(n) = 2*A000005(n).

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A338456 a(n) is the hafnian of a symmetric Toeplitz matrix M(2n) whose first row consists of a single zero followed by successive positive integers repeated (A004526).

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A332566 a(n) is the permanent of an n X n symmetric Toeplitz matrix M(n) whose first row consists of a single zero followed by successive positive integers repeated (A004526).

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A374708 Triangle T read by rows: T(n,k) = (n - k)*n*(4*n^2 - 4*n*k + 2*k^2 - 1 + (-1)^k)/4, with 0 <= k < n.

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A338796 Triangle T read by rows: T(n, k) is the k-th row sum of the symmetric Toeplitz matrix M(n) whose first row consists of a single zero followed by successive positive integers repeated (A004526).

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Programs

Mathematica

PARI

Formula

A062011 a(n) = 2tau(n) = 2A000005(n).

A374708 Triangle T read by rows: T(n,k) = (n - k)n(4n^2 - 4nk + 2k^2 - 1 + (-1)^k)/4, with 0 <= k < n.