A055140 -id:A055140 - cp's OEIS frontend

A231846 Polynomials for total Pontryagin classes. Refinement of double Pochhammer triangle.

1, 1, 2, 1, 8, 6, 1, 48, 32, 12, 12, 1, 384, 240, 160, 80, 60, 20, 1, 3840, 2304, 1440, 640, 720, 960, 120, 160, 180, 30, 1, 46080, 26880, 16128, 13440, 8064, 10080, 4480, 3360, 1680, 3360, 840, 280, 420, 42, 1, 645120, 368640, 215040, 172032, 80640, 107520, 129024, 107520, 40320, 35840, 21504, 40320, 17920, 26880, 1680, 3360, 8960, 3360, 448, 840, 56, 1

Offset: 0

Tom Copeland, Nov 14 2013

The W. Lang link in A036039 explicitly gives the first several cycle index polynomials for the symmetric group S_n, or the partition polynomials for the refined Stirling numbers of the first kind. In line with the discussion in the Fecko link, null the indeterminates with odd indices, divide the 2n-th partition polynomial by the double factorial of odd numbers given in A001147, and re-index. The sum of the resulting row coefficients are also equal to A001147.

			In terms of the trace of a curvature form Tr(F^n)={n} or indeterminates c_n=[n]:
P_0 = 1,
P_1 = Tr(F^2) = {2}
    = c_1 = [1],
P_2 = 2Tr(F^4)+Tr(F^2)^2 = 2{4}+{2}^2
    = 2c_2+ (c_1)^2 = 2[2]+[1]^2,
P_3 = 8Tr(F^6)+6Tr(F^2)Tr(F^4)+Tr(F^2)^3= 8{6}+6{2}{4}+{2}^3
    = 8c_3+6c_1 c_2+(c_1)^3 = 8[3]+6[1][2]+[1]^3,
P_4 = 48{8}+32{2}{6}+12{4}^2+12{2}^2{4}+{2}^4
    = 48[4]+32[1][3]+12[2]^2+12[1]^2[2]+[1]^4,
P_5 = 384{10}+240{2}{8}+160{4}{6}+80{2}^2{6}
      + 60{2}{4}^2+20{2}^3{4}+{2}^5
    = 384[5]+240[1][4]+160[2][3]+80[1]^2[3]
      + 60[1][2]^2+20[1]^3[2]+[1]^5
P_6 = 3840[6]+2304[1][5]+1440[2][4]+640[3]^2+720[1]^2[4]
  +960[1][2][3]+120[2]^3+160[1]^3[3]+180[1]^2[2]^2+30[1]^4[2]+[1]^6
P_7 = 46080[7]+26880[1][6]+16128[2][5]+13440[3][4]+8064[1]^2[5]
  +10080[1][2][4]+4480[1][3]^2+3360[2]^2[3]+1680[1]^3[4]
  +3360[1]^2[2][3]+840[1][2]^3+280[1]^4[3]+420[1]^3[2]^2+42[1]^5[2]+[1]^7
....
Summing over partitions with the same number of blocks gives the unsigned double Pochhammer triangle A039683. Row sums are A001147. Multiplying P_n by the row sum gives the 2n-th partition polynomial of A036039 with its odd-indexed indeterminates nulled.
For c_1 = c_2 = x and c_n = 0 otherwise, see A119275. Let Omega(t) = xi(1/2 + i*t)/xi(1/2) where xi is the Landau version of the Riemann xi function, t is real, and i^2 = -1. The Taylor series coefficients vanish for odd order derivatives and, for even, are c_(2n) = Omega^(2n)(0) = (-1)^n * xi^(2n)(1/2) / xi(1/2) = A001147(n) * P_n as in the Example section with F^(2n) = -2 * Sum(1/x_k^(2n)) = -2 * Tr_(2n) where x_k is the imaginary part of the k-th zero of the Riemann zeta function and k ranges over all the zeros above the real axis. E.g., (see the Mathematics Stack Exchange question) summing over the first several thousands of zeros, c_4 = A001147(2)*P_2 = 3*[2*(-2*Tr_4) + (-2*Tr_2)^2] = 12*[-(0.000372) + (0.02311)^2] = .005962 and c_4 = xi^(4)*(1/2)/xi(1/2) = 0.002963/0.497 = 0.005962 (rounding off). Conversely, the Tr_(2n) can be calculated from the c_n using the Faber polynomials (A263916), as indicated in A036039. See Coffey for Taylor coefficients of Omega(t) about t = 0 and the MSE question for Tr_(2n). The traces are convergent and any zeros in the critical strip off the critical line would have a slightly more complicated real contribution to the traces but negligible to any practical order. - _Tom Copeland_, May 27 2020

M. Coffey, Relations and positivity results for derivatives of the Riemann xi function, J. Comput. Appl. Math. 166(2) (2004), 525-534.
Tom Copeland, Appells and Roses: Newton, Leibniz, Euler, Riemann and Symmetric Polynomials, 2020.
M. Fecko, Selected topological concepts used in physics pp. 37-41 and 56-58.
Mathematics Stack Exchange, Sums of the reciprocals of powers of the imaginary part of the nontrivial Riemann zeros, a MSE question posed by T. Copeland and answered by G. Helms, 2020.
Wikipedia, Pontryagin class
Y. Zhang, A brief introduction to characteristic classes from the differentiable viewpoint p. 27.

Cf. A000165, A001147, A036039, A055140, A119275.
Cf. A263916.
The terms are indexed by partitions in the Abramowitz and Stegun order, A036036.

rows[n_] := {{1}}~Join~With[{s = Exp[Sum[b[k] t^k/(2 k), {k, n}] + O[t]^(n+1)]}, Table[Expand@Coefficient[(2 k)!! s, t^k Product[b[t], {t, p}]], {k, n}, {p, Sort[Sort /@ IntegerPartitions[k]]}]];
rows[8] // Flatten (* Andrey Zabolotskiy, Feb 19 2024 *)

From Tom Copeland, Oct 11 2016: (Start)
A generating function for the polynomials PB_n[b_2,b_4,..,b_(2n)] of this array is
exp[b_2 y^2/2 + b_4 y^4/4 + b_6 y^6/6 + ...] = Sum_{n >= 0} PB_n y^(2n) / A000165(n) = Sum_{n >= 0} St1[2n,0,b_2,0,b_4,0,..,b_(2n)] y^(2n) / (2n)! = Sum_{n >= 0} PB_n *(y/sqrt(2))^(2n) / n! with b_n = Tr(F^n), as in the examples, and St1(n,b_1,b_2,..,b_n), the partition polynomials of A036039. Then St1[2n,0,b_2,0,b_4,..,0,b_(2n)] = A001147(n) * PB_n.
The polynomials PC_n(c_1,c_2,..,c_n) of this array with c_k = b_(2k) are an Appell sequence in the indeterminate c_1 with lowering operator L = d/d(c_1), i.e., L*PC_n(c_1,..,c_n) = d(PC_n)/d(c_1) = n * PC_(n-1)[c_1,..,c_(n-1)].
With [PC.(c_1,c_2,..)]^n = PC_n(c_1,..,c_n), the e.g.f. is G(t,c_1,c_2,..) = exp[t*PC.(0,c_2,c_3,..)] * exp(t*c_1) = exp{t*[c_1 + PC.(0,c_2,c_3,..)]} = exp[t*PC.(c_1,c_2,..)] = exp[(1/2) * sum_{n > 0} c_n (2t)^n/n ] = exp[-log(1-2c.t) / 2], where, umbrally, (c.)^n = c_n.
The raising operator is R = d[log(G(L,c_1,c_2,..))]/dL = sum_{n >= 0} 2^n * c_(n+1) * (d/dc_1)^n = c./(1-2c.L), umbrally. R PC_n(c_1,..,c_n) = P_(n+1)[c_1,..,c_(n+1)].
Another generator: G(L,0,c_2,c_3,..) (c_1)^n = PC_n(c_1,c_2,..,c_n).
The Appell umbral compositional inverse sequence UPC_n to the PC_n sequence has e.g.f. UG(t,c_1,c_2,..) = [1 / G(t,0,c_2,c_3,..)] *  exp(t*c_1) with lowering operator L, as above, and raising operator RU = c_1 - sum_{n > 0} 2^n * c_(n+1) * (d/dc_1)^n. It follows that UPC_n(c_1,c_2,..,c_n) = PC_n(c_1,-c_2,..,-c_n) and PC_n(PC.(c_1,c_2,..),-c_2,-c_3,..) = PC_n(PC.(c_1,-c_2,-c_3,..),c_2,c_3,..) = (c_1)^n, e.g., PC_2(PC.(c_1,-c_2,..),c_2) = 2 c_2 + (PC.(c_1,-c_2,..))^2 = 2 c_2 + PC_2(c_1,-c_2) = 2 c_2 + 2 (-c_2) + (c_1)^2 = (c_1)^2.
Letting c_1 = x and all other c_n = 1 gives the row polynomials of A055140.
(End)

Polynomials P_6 and P_7 added by Tom Copeland, Oct 11 2016
Correction to P_3 in Example by Tom Copeland, May 27 2020
Terms in rows 6-7 reordered, row 8 added by Andrey Zabolotskiy, Feb 19 2024

A370347 Number T(n,k) of partitions of [3n] into n sets of size 3 having exactly k sets {3j-2,3j-1,3j} (1<=j<=n); triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 9, 0, 1, 252, 27, 0, 1, 14337, 1008, 54, 0, 1, 1327104, 71685, 2520, 90, 0, 1, 182407545, 7962624, 215055, 5040, 135, 0, 1, 34906943196, 1276852815, 27869184, 501795, 8820, 189, 0, 1, 8877242235393, 279255545568, 5107411260, 74317824, 1003590, 14112, 252, 0, 1

Offset: 0

Alois P. Heinz, Feb 15 2024

			T(2,0) = 9: 124|356, 125|346, 126|345, 134|256, 135|246, 136|245, 145|236, 146|235, 156|234.
T(2,2) = 1: 123|456.
Triangle T(n,k) begins:
            1;
            0,          1;
            9,          0,        1;
          252,         27,        0,      1;
        14337,       1008,       54,      0,    1;
      1327104,      71685,     2520,     90,    0,   1;
    182407545,    7962624,   215055,   5040,  135,   0, 1;
  34906943196, 1276852815, 27869184, 501795, 8820, 189, 0, 1;
  ...

Alois P. Heinz, Rows n = 0..140, flattened
Wikipedia, Partition of a set

Row sums give A025035.
Column k=0 gives A370357.
T(n+1,n-1) gives A027468.
T(n+2,n-1) gives 252*A000292.
Cf. A023531, A055140, A370358.

b:= proc(n) option remember; `if`(n<3, [1, 0, 9][n+1],
      9*(n*(n-1)/2*b(n-1)+(n-1)^2*b(n-2)+(n-1)*(n-2)/2*b(n-3)))
    end:
T:= (n, k)-> b(n-k)*binomial(n, k):
seq(seq(T(n, k), k=0..n), n=0..10);

T(n,k) = binomial(n,k) * A370357(n-k).
Sum_{k=1..n} T(n,k) = A370358(n).
T(n,k) mod 9 = A023531(n,k).

A155517 Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} for which the number of j < ceiling(n/2) such that p(j) + p(n+1-j) = n+1 is equal to k (n>=1; 0<=k <=ceiling(n/2)).

Original entry on oeis.org

0, 1, 0, 2, 4, 0, 2, 16, 0, 8, 64, 48, 0, 8, 384, 288, 0, 48, 2880, 1536, 576, 0, 48, 23040, 12288, 4608, 0, 384, 208896, 115200, 30720, 7680, 0, 384, 2088960, 1152000, 307200, 76800, 0, 3840, 23193600, 12533760, 3456000, 614400, 115200, 0, 3840, 278323200

Offset: 1

Emeric Deutsch, Jan 26 2009

For the permutation 31756284 of S_8 we have k=2 because p(2) + p(7) = 1+8 = 9 and p(3) + p(6) = 7+2 = 9; for the permutation 3214756 of S_7 we have k=2 because p(3) + p(5) = 1+7 = 8 and p(4) + p(4) = 4+4 = 8.
Row sums are the factorial numbers (A000142).
Row n contains 1 + ceiling(n/2)entries.
T(2n,n) = n!*2^n = A037223(2n) = number of centrosymmetric permutations in S[2n];
T(2n+1,n+1) = n!*2^n = A037223(2n+1) = number of centrosymmetric permutations in S[2n+1].
T(n,0) = A155518(n).
Sum_{k=0..ceiling(n/2)} k*T(n,k) = A155519(n).

			T(4,2)=8 because we have 1234, 4231, 1324, 4321, 2143, 3142, 2413 and 3412.
Triangle starts:
    0,   1;
    0,   2;
    4,   0,   2;
   16,   0,   8;
   64,  48,   0,   8;
  384, 288,   0,  48;

Cf. A000142, A037223, A055140, A155518, A155519.

g[0] := 1: g[1] := 0: for n from 2 to 20 do g[n] := (2*(n-1))*(g[n-1]+g[n-2]) end do: T := proc (n, k) if `mod`(n, 2) = 0 then 2^((1/2)*n)*factorial((1/2)*n)*g[(1/2)*n-k]*binomial((1/2)*n, k) else 2^((1/2)*n-1/2)*factorial((1/2)*n-1/2)*g[(1/2)*n+1/2-k]*binomial((1/2)*n+1/2, k) end if end proc: for n to 12 do seq(T(n, k), k = 0 .. ceil((1/2)*n)) end do;

T(2n,k) = n!*2^n*A055140(n,k);
T(2n-1,k) = (n-1)!*2^(n-1)*A055140(n,k);
here A055140(n,k) = A053871(n-k)*binomial(n,k), where g(n) = A053871(n) is defined by g(0)=1, g(1)=0, g(n) = 2(n-1)(g(n-1)+g(n-2)).

A370253 Number of deranged matchings of 2n people with partners (of either sex) such that at least one person is matched with their spouse.

Original entry on oeis.org

0, 1, 1, 7, 45, 401, 4355, 56127, 836353, 14144545, 267629139, 5601014255, 128455425593, 3203605245777, 86317343312395, 2498680706048191, 77336483434140705, 2548534969132415297, 89087730603300393443, 3292572900736818264015, 128281460895447809211529

Offset: 0

Sam Coutteau, Feb 13 2024

nonn

			For n=0, there is no matching which has at least one person matched with their original partner.
For n=1, there are only 2 people, so there is only one way to match them and it is with their original partner.
For n=2, we have two couples, A0 with A1, and B0 with B1. Of the three ways to match them [(A0,A1),(B0,B1)], [(A0,B0),(A1,B1)] and [(A0,B1),(A1,B0)], only the first matching has a person matched up with their original partner.

Alois P. Heinz, Table of n, a(n) for n = 0..404

Cf. A001147 (total number of matchings for 2n people).
Cf. A053871 (number of deranged matchings of 2n people with partners (of either sex) other than their spouse).
Cf. A055140, A057427.

a:= proc(n) option remember; `if`(n<3, signum(n),
      (4*n-7)*a(n-1)-2*(2*n^2-10*n+11)*a(n-2)-2*(n-2)*(2*n-5)*a(n-3))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Feb 14 2024

a[n_] := Sum[(-1)^(n-i+1)*Binomial[n, i]*(2i-1)!!, {i, 0, n-1}];
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Feb 29 2024 *)

import math
A001147 = lambda i: math.factorial(2*i) // ( 2 ** i * math.factorial(i) )
A370253 = lambda n: int( sum( (-1)**(i+1) * math.comb(n,n-i) * A001147(n-i) for i in range(1,n+1) ) )
print( ", ".join( str(A370253(i)) for i in range(0,21) ) )

a(n) = A001147(n) - A053871(n).
a(n) = Sum_{i=0..n-1} (-1)^(n - i + 1) * binomial(n,i)*A001147(i).
a(n) mod 2 = A057427(n).
a(n) = Sum_{k=1..n} A055140(n,k). - Alois P. Heinz, Feb 14 2024

A325754 Triangle read by rows giving the number of configurations of n indistinguishable pairs placed on the vertices of the ladder graph P_2 X P_n such that exactly k such pairs are joined by a horizontal edge.

Original entry on oeis.org

1, 1, 0, 2, 0, 1, 7, 4, 4, 0, 43, 38, 21, 2, 1, 372, 360, 168, 36, 9, 0, 4027, 3972, 1818, 478, 93, 6, 1, 51871, 51444, 23760, 6640, 1260, 144, 16, 0, 773186, 768732, 358723, 103154, 20205, 2734, 278, 12, 1, 13083385, 13027060, 6129670, 1796740, 363595, 52900, 5650, 400, 25, 0

Offset: 0

Donovan Young, May 19 2019

This is the number of "k-horizontal-domino" configurations in the game of memory played on a 2 X n rectangular array, see [Young].

			The first few rows of T(n,k) are:
  1;
  1,  0;
  2,  0,  1;
  7,  4,  4,  0;
  43, 38, 21, 2, 1;
  ...
For n=2, let the vertex set of P_2 X P_2 be {A,B,C,D} and the edge set be {AB, AC, BD, CD}, where AB and CD are horizontal edges. For k=0, we may place the pairs on A, C and B, D or on A, D and B, C, hence T(2,0) = 2. If we place a pair on one of the horizontal edges we are forced to place the other pair on the remaining horizontal edge, hence T(2,1)=0 and T(2,2)=1.

D. Young, The Number of Domino Matchings in the Game of Memory, Journal of Integer Sequences, Vol. 21 (2018), Article 18.8.1.
Donovan Young, Generating Functions for Domino Matchings in the 2 * k Game of Memory, arXiv:1905.13165 [math.CO], 2019. Also in J. Int. Seq., Vol. 22 (2019), Article 19.8.7.

Cf. A046741, A055140, A079267 A178523, A265167, A318243, A318244, A318267, A318268, A318269, A318270, A325753.

CoefficientList[Normal[Series[Sum[Factorial2[2*k-1]*y^k/(1-(1-z)*y)/(1+(1-z)*y)^(2*k+1), {k, 0, 20}], {y, 0, 20}]], {y, z}];

G.f.: Sum_{j>=0} (2*j-1)!! y^j/(1-(1-z)*y)/(1+(1-z)*y)^(2*j+1).

E.g.f.: exp((sqrt(1 - 2 y)-1) (1 - z))/sqrt(1 - 2 y) - exp((y - 2) (1 - z)) sqrt(Pi/2) sqrt(1 - z) (-erfi(sqrt(2) sqrt(1 - z)) + erfi(((1 + sqrt(1 - 2 y)) sqrt(1 - z))/sqrt(2))).

T is an example of the group of matrices outlined in the table in A132382--the associated matrix for aC(1,1). The e.g.f. for the row polynomials is exp(x*t) * exp(x) * (1-2*x)^(1/2). T(n,k) = Binomial(n,k)* s(n-k) where s = A055142 with an e.g.f. of exp(x) * (1-2*x)^(1/2) which is the reciprocal of the e.g.f. of A053871. The row polynomials form an Appell sequence. [From Tom Copeland, Sep 11 2008]

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A055141 Matrix inverse of triangle A055140.

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A231846 Polynomials for total Pontryagin classes. Refinement of double Pochhammer triangle.

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A370347 Number T(n,k) of partitions of [3n] into n sets of size 3 having exactly k sets {3j-2,3j-1,3j} (1<=j<=n); triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A155517 Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} for which the number of j < ceiling(n/2) such that p(j) + p(n+1-j) = n+1 is equal to k (n>=1; 0<=k <=ceiling(n/2)).

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A370253 Number of deranged matchings of 2n people with partners (of either sex) such that at least one person is matched with their spouse.

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A325754 Triangle read by rows giving the number of configurations of n indistinguishable pairs placed on the vertices of the ladder graph P_2 X P_n such that exactly k such pairs are joined by a horizontal edge.

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