cp's OEIS Frontend

Number of permutations in S_{2n} in which all cycles have even length (cf. A087137).

Also number of permutations in S_{2n} in which all cycles have odd length. - Vladeta Jovovic, Aug 10 2007

a(n) is the sum over all multinomials M2(2*n,k), k from {1..p(2*n)} restricted to partitions with only even parts. p(2*n)= A000041(2*n) (partition numbers) and for the M2-multinomial numbers in A-St order see A036039(2*n,k). - Wolfdieter Lang, Aug 07 2007

From Zhi-Wei Sun, Jun 26 2022: (Start)

Conjecture 1: For any primitive 2n-th root zeta of unity, the permanent of the 2n X 2n matrix [m(j,k)]_{j,k=1..2n} coincides with a(n) = ((2n-1)!!)^2, where m(j,k) is (1+zeta^(j-k))/(1-zeta^(j-k)) if j is not equal to k, and 1 otherwise.

The determinant of [m(j,k)]_{j,k=1..2n} was shown to be (-1)^(n-1)*((2n-1)!!)^2/(2n-1) by Han Wang and Zhi-Wei Sun in 2022.

Conjecture 2: Let p be an odd prime. Then the permanent of (p-1) X (p-1) matrix [f(j,k)]_{j,k=1..p-1} is congruent to a((p-1)/2) = ((p-2)!!)^2 modulo p^2, where f(j,k) is (j+k)/(j-k) if j is not equal to k, and f(j,k) = 1 otherwise. (End)

From David W. Wilson: Numbers triangular, differences square.

To be precise, the differences are the squares of the powers of three with positive indices. Hence a(n+1) - a(n) = (A000244(n+1))^2 = A001019(n+1). [Added by Ant King, Jan 05 2011]

Partial sums of A001019. This is m-th triangular number, where m is partial sums of A000244. a(n) = A000217(A003462(n)). - Lekraj Beedassy, May 25 2004

With offset 0, binomial transform of A003951. - Philippe Deléham, Jul 22 2005

Numbers in base 9: 1, 11, 111, 1111, 11111, 111111, 1111111, etc. - Zerinvary Lajos, Apr 26 2009

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=9, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=10, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 2, a(n-1) = (-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010

From Hieronymus Fischer, Jan 30 2013: (Start)

Least index k such that A052382(k) >= 10^(n-1), for n > 0.

Also index k such that A052382(k) = (10^n-1)/9, n > 0.

A052382(a(n)) is the least zerofree number with n digits, for n > 0.

For n > 1: A052382(a(n)-1) is the greatest zerofree number with n-1 digits. (End)

For n > 0, 4*a(n) is the total number of holes in a certain triangle fractal (start with 9 triangles, 4 holes) after n iterations. See illustration in links. - Kival Ngaokrajang, Feb 21 2015

For n > 0, a(n) is the sum of the numerators and denominators of the reduced fractions 0 < (b/3^(n-1)) < 1 plus 1. Example for n=3 gives fractions 1/9, 2/9, 1/3, 4/9, 5/9, 2/3, 7/9, and 8/9 plus 1 has sum of numerators and denominators +1 = a(3) = 91. - J. M. Bergot, Jul 11 2015

Except for 0 and 1, all terms are Brazilian repunits numbers in base 9, so belong to A125134. All these terms are composite because a(n) is the ((3^n - 1)/2)-th triangular number. - Bernard Schott, Apr 23 2017

These are also the second steps after the junctions of the Collatz trajectories of 2^(2k-1)-1 and 2^2k-1. - David Rabahy, Nov 01 2017

4 times the variance of the area under an n-step random walk: e.g., with three steps, the area can be 9/2, 7/2, 3/2, 1/2, -1/2, -3/2, -7/2, or -9/2 each with probability 1/8, giving a variance of 35/4 or a(3)/4. - Henry Bottomley, Jul 14 2003

Number of standard tableaux of shape (2n-1,1,1,1) (n>=1). - Emeric Deutsch, May 30 2004

Also a(n) = (1/6)*(8*n^3-2*n), n>0: structured octagonal diamond numbers (vertex structure 9). Cf. A059722 = alternate vertex; A000447 = structured diamonds; and structured tetragonal anti-diamond numbers (vertex structure 9). Cf. A096000 = alternate vertex; A100188 = structured anti-diamonds. Cf. A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004

The n-th tetrahedral (or pyramidal) number is n(n+1)(n+2)/6. This sequence contains the tetrahedral numbers of A000292 obtained for n= 1,3,5,7,... (see A015219). - Valentin Bakoev, Mar 03 2009

Using three consecutive numbers u, v, w, (u+v+w)^3-(u^3+v^3+w^3) equals 18 times the numbers in this sequence. - J. M. Bergot, Aug 24 2011

This sequence is related to A070893 by A070893(2*n-1) = n*a(n)-sum(i=0..n-1, a(i)). - Bruno Berselli, Aug 26 2011

Number of integer solutions to 1-n <= x <= y <= z <= n-1. - Michael Somos, Dec 27 2011

Partial sums of A016754. - Reinhard Zumkeller, Apr 02 2012

Also the number of cubes in the n-th Haüy square pyramid. - Eric W. Weisstein, Sep 27 2017

Discussion of Central Factorial Numbers by N. J. A. Sloane, Feb 01 2011: (Start)

Here is Riordan's definition of the central factorial numbers t(n,k) given in Combinatorial Identities, Section 6.5:

For n >= 0, expand the polynomial

x^[n] = x*Product{i=1..n-1} (x+n/2-i) = Sum_{k=0..n} t(n,k)*x^k.

The t(n,k) are not always integers. The cases n even and n odd are best handled separately.

For n=2m, we have:

x^[2m] = Product_{i=0..m-1} (x^2-i^2) = Sum_{k=1..m} t(2m,2k)*x^(2k).

E.g. x^[8] = x^2(x^2-1^2)(x^2-2^2)(x^2-3^2) = x^8-14x^6+49x^4-36x^2,

which corresponds to row 4 of the present triangle.

So the m-th row of the present triangle gives the absolute values of the coefficients in the expansion of Product_{i=0..m-1} (x^2-i^2).

Equivalently, and simpler, the n-th row gives the coefficients in the expansion of Product_{i=1..n-1}(x+i^2), highest powers first.

For n odd, n=2m+1, we have:

x^[2m+1] = x*Product_{i=0..m-1}(x^2-((2i+1)/2)^2) = Sum_{k=0..m} t(2m+1,2k+1)*x^(2k+1).

E.g. x^[5] = x(x^2-(1/2)^2)(x^2-(3/2)^2) = x^5-10x^3/4+9x/16,

which corresponds to row 2 of the triangle in A008956.

We now rescale to get integers by replacing x by x/2 and multiplying by 2^(2m+1) (getting 1, -10, 9 from the example).

The result is that row m of triangle A008956 gives the coefficients in the expansion of x*Product_{i=0..m} (x^2-(2i+1)^2).

Equivalently, and simpler, the n-th row of A008956 gives the coefficients in the expansion of Product_{i=0..n-1} (x+(2i+1)^2), highest powers first.

Note that the n-th row of A182867 gives the coefficients in the expansion of Product_{i=1..n} (x+(2i)^2), highest powers first.

(End)

Contribution from Johannes W. Meijer, Jun 18 2009: (Start)

We define Beta(n-z,n+z)/Beta(n,n) = Gamma(n-z)*Gamma(n+z)/Gamma(n)^2 = sum(EG2[2m,n]*z^(2m), m = 0..infinity) with Beta(z,w) the Beta function. The EG2[2m,n] coefficients are quite interesting, see A161739. Our definition leads to EG2[2m,1] = 2*eta(2m) and the recurrence relation EG2[2m,n] = EG2[2m,n-1] - EG2[2m-2,n-1]/(n-1)^2 for m = -2, -1, 0, 1, 2, ... and n = 2, 3, ... , with eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. We found for the matrix coefficients EG2[2m,n] = sum((-1)^(k+n)*t1(n-1,k-1)*2*eta(2*m-2*n+2*k)/((n-1)!)^2,k=1..n) with the central factorial numbers t1(n,m) as defined above, see also the Maple program.

From the EG2 matrix we arrive at the ZG2 matrix, see A161739 for its odd counterpart, which is defined by ZG2[2m,1] = 2*zeta(2m) and the recurrence relation ZG2[2m,n] = ZG2[2m-2,n-1]/(n*(n-1))-(n-1)*ZG2[2m,n-1]/n for m = -2, -1, 0, 1, 2, ... and n = 2, 3, ... . We found for the ZG2[2m,n] = Sum_{k=1..n} (-1)^(k+1)*t1(n-1,k-1)* 2* zeta(2*m-2*n+2*k)/((n-1)!*(n)!), and we see that the central factorial numbers t1(n,m) once again play a crucial role.

(End)

Or, triangle of central factorial numbers T(2n,2k) (in Riordan's notation).

Can be used to calculate the Bernoulli numbers via the formula B_2n = (1/2)*Sum_{k = 1..n} (-1)^(k+1)*(k-1)!*k!*T(n,k)/(2*k+1). E.g., n = 1: B_2 = (1/2)*1/3 = 1/6. n = 2: B_4 = (1/2)*(1/3 - 2/5) = -1/30. n = 3: B_6 = (1/2)*(1/3 - 2*5/5 + 2*6/7) = 1/42. - Philippe Deléham, Nov 13 2003

From Peter Bala, Sep 27 2012: (Start)

Generalized Stirling numbers of the second kind. T(n,k) is equal to the number of partitions of the set {1,1',2,2',...,n,n'} into k disjoint nonempty subsets V1,...,Vk such that, for each 1 <= j <= k, if i is the least integer such that either i or i' belongs to Vj then {i,i'} is a subset of Vj. An example is given below.

Thus T(n,k) may be thought of as a two-colored Stirling number of the second kind. See Matsumoto and Novak, who also give another combinatorial interpretation of these numbers. (End)

The coefficients of the LS1 matrix are defined by LS1[2*m,n] = int(y^(2*m)/(sinh(y))^(2*n-1),y=0..infinity)/factorial(2*m) for m = 1, 2, 3, .. and n = 1, 2, 3, .. under the condition that n <= m.

This definition leads to LS1[2*m,n=1] = 2*lambda(2*m+1), for m = 1, 2, .. , and the recurrence relation LS1[2*m,n] = ((2*n-3)/(2*n-2))*(LS1[2*m-2,n-1]/(2*n-3)^2- LS1[2*m,n-1]). As usual lambda(m) = (1-2^(-m))*zeta(m) with zeta(m) the Riemann zeta function.

These two formulas enable us to determine the values of the LS1[2*m,n] coefficients, for all integers m and all positive integers n, but not for all n. If we choose, somewhat but not entirely arbitrarily, LS1[m=0,n=1] = gamma, with gamma the Euler-Mascheroni constant, we can determine them all.

The coefficients in the columns of the LS1 matrix, for m = 0, 1, 2, .. , and n = 2, 3, 4 .. , can be generated with the GL(z;n) polynomials for which we found the following general expression GL(z;n) = (h(n)*CFN2(z;n)*GL(z;n=1) + LAMBDA(z;n))/p(n).

The CFN2(z;n) polynomials depend on the central factorial numbers A008956.

The LAMBDA(z;n) are the Lambda polynomials which lead to the Lambda triangle.

The zero patterns of the Lambda polynomials resemble a UFO. These patterns resemble those of the Eta, Zeta and Beta polynomials, see A160464, A160474 and A160480.

The first Maple algorithm generates the coefficients of the Lambda triangle. The second Maple algorithm generates the LS1[2*m,n] coefficients for m= -1, -2, -3, .. .

Some of our results are conjectures based on numerical evidence.

The coefficients of the BS1 matrix are defined by BS1[2*m-1,n] = int(y^(2*m-1)/(cosh(y))^(2*n-1),y=0..infinity)/factorial(2*m-1) for m = 1, 2, ... and n = 1, 2, ... .

This definition leads to BS1[2*m-1,n=1] = 2*beta(2*m), for m = 1, 2, ..., and the recurrence relation BS1 [2*m-1,n] = (2*n-3)/(2*n-2)*(BS1[2*m-1,n-1] - BS1[2*m-3,n-1]/(2*n-3)^2) which we used to extend our definition of the BS1 matrix coefficients to m = 0, -1, -2, ... . We discovered that BS1[ -1,n] = 1 for n = 1, 2, ... . As usual beta(m) = sum((-1)^k/(1+2*k)^m, k=0..infinity).

The coefficients in the columns of the BS1 matrix, for m = 1, 2, 3, ..., and n = 2, 3, 4, ..., can be generated with the GK(z;n) polynomials for which we found the following general expression GK(z;n) = ((-1)^(n+1)*CFN2(z;n)*GK(z;n=1) + BETA(z;n))/p(n).

The CFN2(z;n) polynomials depend on the central factorial numbers A008956.

The BETA(z;n) are the Beta polynomials which lead to the Beta triangle.

The zero patterns of the Beta polynomials resemble a UFO. These patterns resemble those of the Eta, Zeta and Lambda polynomials, see A160464, A160474 and A160487.

The first Maple algorithm generates the coefficients of the Beta triangle. The second Maple algorithm generates the BS1[2*m-1,n] coefficients for m = 0, -1, -2, -3, ... .

Some of our results are conjectures based on numerical evidence, see especially A160481.