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.

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

Original entry on oeis.org

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

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Author

Tom Copeland, Nov 14 2013

Keywords

Comments

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.

Examples

			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
		

Crossrefs

Cf. A263916.
The terms are indexed by partitions in the Abramowitz and Stegun order, A036036.

Programs

  • Mathematica
    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 *)

Formula

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)

Extensions

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

A113279 Triangle T(n,k) of coefficients of r_1^n+r_2^n in terms of p and q, where r_1,r_2 are the roots of x^2+px+q=0.

Original entry on oeis.org

2, -1, 1, -2, -1, 3, 1, -4, 2, -1, 5, -5, 1, -6, 9, -2, -1, 7, -14, 7, 1, -8, 20, -16, 2, -1, 9, -27, 30, -9, 1, -10, 35, -50, 25, -2, -1, 11, -44, 77, -55, 11, 1, -12, 54, -112, 105, -36, 2, -1, 13, -65, 156, -182, 91, -13, 1, -14, 77, -210, 294, -196, 49, -2, -1, 15, -90, 275, -450, 378, -140, 15
Offset: 0

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Author

Floor van Lamoen, Oct 22 2005

Keywords

Comments

The triangle begins: 2 .-1 1..-2 .-1..3 1..-4.2
From Tom Copeland, Nov 07 2015: (Start)
The row polynomials are the power sums p_n = x1^n + x2^n of the solutions of 0 = (x-x1)(x-x2) = x^2 - (x1+x2) x + x1x2 = x^2 + px + q, so -p = x1+x2 = e1(x1,x2), the first order elementary symmetric polynomial for two variables, or indeterminates, and q = x1*x2 = e2(x1,x2), the second order elementary symmetric polynomial.
The Girard-Newton-Waring identities (Wikipedia) express the power sums in terms of the elementary symmetric polynomials, giving
p_0 = x1^0 + x0^0 = 2
p_1 = x1 + x2 = e1 = -p = F(1,p,q,0,..)
p_2 = x1^2 + x2^2 = e1^2 - 2 e2 = p^2 - 2 q = F(2,p,q,0,..)
p_3 = e1^3 - 3 e2 e1 = -p^3 + 3 pq = F(3,p,q,0,..)
p_4 = p^4 - 4 p^2 q + 2 q^2 = F(4,p,q,0,..)
... .
These bivariate partition polynomials are the Faber polynomials F(n,b1,b2,..,bn) of A263916 with b1 = -e1 = p, b2 = e2 = q, and all other indeterminates set to zero.
Let p = q = t, then
F(1,t,t,0,..)/t = -1
F(2,t,t,0,..)/t = -2 + t
F(3,t,t,0,..)/t^2 = 3 - t
F(4,t,t,0,..)/t^2 = 2 - 4 t + t^2
... .
Or,
t * F(1,1/t,1/t,0,..) = -1
t^2 * F(2,1/t,1/t,0,..) = 1 -2 t
t^3 * F(3,1/t,1/t,0,..) = -1 + 3 t
t^4 * F(4,1/t,1/t,0,..) = 1 - 4 t + 2 t^2
... .
The sequence of Faber polynomials F(n,1/t,1/t,0,..) is obtained from the logarithmic generator -log[1+(x+x^2)/t] = sum{n>=1, F(n,1/t,1/t,0,..) x^n/n}, so
2-(2xt+1)xt/(t+xt+(xt)^2) = 2 + sum{n>=1, t^n F(n,1//t,1/t,0,..) x^n} is an o.g.f. for the row polynomials of this entry.
(End)

Examples

			x_1^5+x_2^5 = -p^5 + 5p^3q - 5pq^2, so row 5 reads -1, 5, -5.
		

References

  • M. Herkenhoff Konersmann, Sprokkel XXXI: x_1^n+x_2^n, Nieuw Tijdschr. Wisk, 42 (1954-55) 180.

Crossrefs

Formula

T(n, k) = (-1)^(n+k)*A034807(n, k).
O.g.f.: 2-(2xt+1)xt/(t+xt+(xt)^2) = (2+x)/(1+x+x^2/t). - Tom Copeland, Nov 07 2015

A152060 Triangle read by rows, characteristic polynomials of Cartan ring matrices.

Original entry on oeis.org

1, 1, -2, 1, -4, 3, 1, -6, 9, -4, 1, -8, 20, -16, 4, 1, -10, 35, -50, 25, -4, 1, -12, 54, -112, 105, -36, 4, 1, -14, 77, -210, 294, -196, 49, -4, 1, -16, 104, -352, 660, -672, 336, -64, 4, 1, -18, 135, -546, 1287, -1782, 1386, -540, 81, -4, 1, -20, 170, -800, 2275, -4004, 4290, -2640, 825, -100, 4
Offset: 0

Views

Author

Keywords

Examples

			Triangle begins:
1;
1, -2;
1, -4, 3;
1, -6, 9, -4;
1, -8, 20, -16, 4;
1, -10, 35, -50, 25, -4;
1, -12, 54, -112, 105, -36, 4;
1, -14, 77, -210, 294, -196, 49, -4;
1, -16, 104, -352, 660, -672, 336, -64, 4;
1, -18, 135, -546, 1287, -1782, 1386, -540, 81, -4;
1, -20, 170, -800, 2275, -4004, 4290, -2640, 825, -100, 4;
...
Example: x^5 -10x^4 + 35x^3 -50x^2 + 25x - 4 = (x - 4) * (x^2 - 3x + 1)^2 is the characteristic polynomial of the matrix
[ 2,-1, 0, 0, 1]
[-1, 2,-1, 0, 0]
[ 0,-1, 2,-1, 0]
[ 0, 0,-1, 2,-1]
[ 1, 0, 0,-1, 2].
		

References

  • William G. Harter, University of Arkansas; personal communication

Crossrefs

Programs

  • Mathematica
    M[n_] := SparseArray[{Band[{1, 1}] -> 2, Band[{1, 2}] -> -1, Band[{2, 1}] -> -1, {1, n} -> 1, {n, 1} -> 1}, {n, n}];
    row[0] = {1}; row[1] = {1, -2};
    row[n_] := (-1)^n CharacteristicPolynomial[M[n], x] // CoefficientList[#, x]& // Reverse;
    Table[row[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Aug 08 2018 *)

Formula

Triangle read by rows, n-th row = characteristic polynomial of n X n Cartan ring matrix, defined as a Cartan matrix with 1's in the upper right and lower left corners, i.e., positions (1,n) and (n,1).
The coefficients of characteristic polynomials of matrices C_n, defined by
C_n=
(2 -1 0 ... 0 1)
(-1 2 -1 0 ... 0)
(0 -1 2 -1 0 ... 0)
...
(0 ... 0 -1 2 -1)
(1 0 ... 0 -1 2),
give the same triangle T(n,k), for n>0, k=0,...,n, with T(0,0)=1. - L. Edson Jeffery, Mar 27 2011
It appears that for n >= 3 the n-th row polynomial equals 2*T(2*n,sqrt(x)/2) + 2*(-1)^n, where T(n,x) denotes the Chebyshev polynomial of the first kind (A008310). Checked for n = 3 through n = 12. - Peter Bala, May 04 2014
Apparently, omitting the diagonal here, this triangular array is signed, reversed A156308 (cf. A127677, A217476, A263916). For relations among the characteristic polynomials of Cartan matrices of the Coxeter root groups, Chebyshev polynomials, cyclotomic polynomials, and the polynomials of this entry, see Damianou (p. 12, 20, and 21) and Damianou and Evripidou (p. 7). - Tom Copeland, Nov 07 2015

Extensions

Edited by L. Edson Jeffery, Mar 26 2011
Some terms corrected from Peter Bala, May 04 2014

A263646 Coefficients for an expansion of the Schwarzian derivative of a power series.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 1, 1, 5, 1, 1, 1, 6, 1, 1, 1, 7, 1, 1, 1, 1, 8, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 14, 1, 1, 1, 1, 1, 1, 1, 15, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Tom Copeland, Oct 31 2015

Keywords

Comments

Coefficients for an expansion of the Schwarzian derivative of a power series f(x), with f'(0) = 1, expressed in terms of an expansion of the natural logarithm of the derivative of the function G(x) = log(D f(x)) = Sum_{n >= 1} -F(n) x^n/n.

Examples

			Partitions by powers of x^n:
n=0: -(F2 + F1^2/2)
n=1: -(2 F3 + F1 F2)
n=2: -(3 F4 + F1 F3 + F2^2/2) =  -[3 F4 + (F1 F3 + F2 F2 + F3 F1) / 2]
n=3: -(4 F5 + F1 F4 + F2 F3)  =  -[4 F5 + (F1 F4 + F2 F3 + F3 F2 + F4 F1) / 2]
n=4: -(5 F6 + F1 F5 + F2 F4 + F3^2/2)
--------------------
Example series:
f(x)= (1/2) / (1-x)^2 = 1/2 + x + (3/2) x^2 + 2x^3 + (5/2)x^4 + ... .
log(f'(x)) = log(1 + 3x + 6x^2 + 10x^3 + ...) = 3x + 3 x^2/2 + 3 x^3/3  + ... .
Then F(n) = -3 for n>=1, and the Schwarzian derivative series is
S{f(x)} = - [(-3 + 3^2/2) + (-2*3 + 3^2) x + (-3*3 + 3^2 + 3^2/2) x^2 + ...] = -3/2 - 3x - (9/2)x^2 - 6x^3 - (15/2)x^4 - ... .
--------------------
The Schwarzian vanishes if and only if acting on a Moebius, or linear fractional, transformation. This corresponds to F(n) = (-1)^(n+1) 2 * d^n, where d is an arbitrary constant.
--------------------
Example polynomial:
f(x) = (x-x1)(x-x2)/-(x1+x2)
log(f'(x)) = log[1 - 2x/(x1+x2)] = Sum_{n>= 1} -(2/(x1+x2))^n x^n/n.
Then F(n) = (2/(x1+x2))^n, and the Schwarzian derivative series is
S{f(x)} = (Sum_{n>= 0} -6(n+1) 2^n (x/(x1+x2))^n) / (x1+x2)^2 = -6 / (x1+x2-2*x)^2 (cf. A001787 and A085750).
		

Crossrefs

Programs

  • Python
    print(sum(([n]+[1]*((n+1)//2) for n in range(1, 18)), [])) # Andrey Zabolotskiy, Mar 07 2024

Formula

Schwarzian{f(x)} = S{f(x)} = (D^3 f(x)) / (D f(x)) - (3/2) [(D^2 f(x)) / D f(x)]^2 = D [(D^2 f(x)) / D f(x)] - (1/2) [(D^2 f(x)) / D f(x)]^2 = D^2 log[D f(x)] - (1/2) [D log[D f(x)]]^2.
Then, with G(x) = log[D f(x)], S{f(x)} = D^2 G(x) - (1/2) [D G(x)]^2.
With f'(0) = 1, G(x) = log[D f(x)] = sum[n >= 1, -F(n) * x^n/n], and F(n) as Fn,
S{f(x)} = -[(F2 + F1^2/2) + (2 F3 + F1 F2) x + (3 F4 + F1 F3 + F2^2/2) x^2 + (4 F5 + F1 F4 + F2 F3) x^3 + (5 F6 + F1 F5 + F2 F4 + F3^2/2) x^4 + (6 F7 + F1 F6 + F2 F5 + F3 F4) x^5 + (7 F8 + F1 F7 + F2 F6 + F3 F5 + F4^2/2) x^6 + ...] .
This entry's a(m) are the numerators of the coefficients of the binary partitions in the brackets. For the singular partition of the integer n, the coefficient is (n-1); for the symmetric partition, 1/2; and for the rest, 1.
More symmetrically, x^2 S{f(x)}= - sum{n>=2, x^n [(n-1)F(n) + (1/2) sum(k=1 to n-1, F(n-k) F(k))]}.
With f(x)= c(0) + x + c(2) x^2 + ... , F(n) are given by the Faber polynomials of A263916: F(n) = Faber(n,2c(2),3c(3),..,(n+1)c(n+1)).

Extensions

More terms from Tom Copeland, Oct 01 2016

A330415 Coefficient of h(y) in Sum_{k > 0, i > 0} x_i^k = p(1) + p(2) + p(3) + ..., where h is the basis of homogeneous symmetric functions, p is the basis of power-sum symmetric functions, and y is the integer partition with Heinz number n.

Original entry on oeis.org

0, 1, 2, -1, 3, -3, 4, 1, -2, -4, 5, 4, 6, -5, -5, -1, 7, 5, 8, 5, -6, -6, 9, -5, -3, -7, 2, 6, 10, 12, 11, 1, -7, -8, -7, -9, 12, -9, -8, -6, 13, 14, 14, 7, 7, -10, 15, 6, -4, 7, -9, 8, 16, -7, -8, -7, -10, -11, 17, -21, 18, -12, 8, -1, -9, 16, 19, 9, -11, 16
Offset: 1

Views

Author

Gus Wiseman, Dec 14 2019

Keywords

Comments

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).
Up to sign, a(n) is the number of acyclic spanning subgraphs of an undirected n-cycle whose component sizes are the prime indices of n.

Crossrefs

The unsigned version (except with a(1) = 1) is A319225.
The transition from p to e by Heinz numbers is A321752.
The transition from p to h by Heinz numbers is A321754.
Different orderings with and without signs and first terms are A115131, A210258, A263916, A319226, A330417.

Programs

  • Mathematica
    Table[If[n==1,0,(-1)^(PrimeOmega[n]-1)*Total[Cases[FactorInteger[n],{p_,k_}:>k*PrimePi[p]]]*(PrimeOmega[n]-1)!/(Times@@Factorial/@FactorInteger[n][[All,2]])],{n,30}]

Formula

a(n) = (-1)^(Omega(n) - 1) * A056239(n) * (Omega(n) - 1)! / Product c_i! where c_i is the multiplicity of prime(i) in the prime factorization of n.

A330417 Coefficient of e(y) in Sum_{k > 0, i > 0} x_i^k = p(1) + p(2) + p(3) + ..., where e is the basis of elementary symmetric functions, p is the basis of power-sum symmetric functions, and y is the integer partition with Heinz number n.

Original entry on oeis.org

0, 1, -2, 1, 3, -3, -4, 1, 2, 4, 5, -4, -6, -5, -5, 1, 7, 5, -8, 5, 6, 6, 9, -5, 3, -7, -2, -6, -10, -12, 11, 1, -7, 8, -7, 9, -12, -9, 8, 6, 13, 14, -14, 7, 7, 10, 15, -6, 4, 7, -9, -8, -16, -7, 8, -7, 10, -11, 17, -21, -18, 12, -8, 1, -9, -16, 19, 9, -11
Offset: 1

Views

Author

Gus Wiseman, Dec 14 2019

Keywords

Comments

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).
Up to sign, a(n) is the number of acyclic spanning subgraphs of an undirected n-cycle whose component sizes are the prime indices of n.

Crossrefs

The unsigned version (except with a(1) = 1) is A319225.
The transition from p to e by Heinz numbers is A321752.
The transition from p to h by Heinz numbers is A321754.
Different orderings with and without signs and first terms are A115131, A210258, A263916, A319226, A330415.

Programs

  • Mathematica
    Table[If[n==1,0,With[{tot=Total[Cases[FactorInteger[n],{p_,k_}:>k*PrimePi[p]]]},(-1)^(tot-PrimeOmega[n])*tot*(PrimeOmega[n]-1)!/(Times@@Factorial/@FactorInteger[n][[All,2]])]],{n,30}]

Formula

a(n) = (-1)^(A056239(n) - Omega(n)) * A056239(n) * (Omega(n) - 1)! / Product c_i! where c_i is the multiplicity of prime(i) in the prime factorization of n.

A355201 Normalized Schur self-convolution expansion coefficients K_{n+1}^n / n giving the coefficients of the Laurent series (compositionally) inverse to f(z) = c_0 z + c_1 + c_2 / z + c_3 / z^2 + ... . Irregular triangle for partition polynomials, with row lengths A000041(n) - 1 except for the first two, which are both of length 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 3, 3, 1, 1, 6, 4, 2, 12, 6, 2, 4, 4, 1, 1, 10, 5, 10, 30, 10, 10, 10, 20, 10, 5, 5, 5, 1, 1, 15, 6, 30, 60, 15, 5, 60, 30, 60, 20, 15, 15, 30, 30, 15, 3, 6, 6, 6, 1, 1, 21, 7, 70, 105, 21, 35, 210, 70, 140, 35, 35, 105, 105, 105, 105, 35, 7, 42, 21, 21, 42, 42, 21, 7, 7, 7, 7, 1
Offset: 0

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Author

Tom Copeland, Jun 23 2022

Keywords

Comments

For the formal Laurent series f(z) = a_0 z + a_1 + a_2 / z + a_3 / z^2 + ..., the formal compositional inverse is g(z) = b_0 z + b_1 + b_2 / z + b_3 / z^2 + ..., whose coefficients are partition polynomials whose numerical factors are those of this irregular triangle T(n,k). For the Schur coefficients defined in the formula section, -b_n = K_{n}^{n-1} / (n-1) for n > 1.
Analytic proofs of the relationship between the partition polynomials of the compositional inverse pair of Laurent series and Schur's self-convolution expansion coefficients are given in Schur (beware of a sign error) and in Airault and Ren.
Explicit examples (with a_0=1) of K_{n}^{n-1} up through n=5 are in Airault and Bouali on p. 182.
Various formulas for the b_n in terms of the associahedra (A133437), noncrossing (A134264), reciprocal (A263633), and Faber partition (A263916) polynomials are given in Copeland as well as a derivation of the explicit multi-factorial expression in the formula section and a combinatorial model.

Examples

			Triangle begins:
1) 1
2) 1
3) 1
4) 1,  1
5) 1,  1,  2,  1
6) 1,  3,  3,  3,  3,  1
7) 1,  6,  4,  2, 12,  6,  2,  4,  4,  1
8) 1, 10,  5, 10, 30, 10, 10, 10, 20, 10,  5,  5,  5,  1
  ...
The first few partition polynomials, with the monomials in the order of the partitions on p. 831 of Abramowitz and Stegun, are
b0 =    1 / a0
b1 = - a1 / a0
b2 = - a2
b3 = -(a1 a2 + a0 a3)
b4 = -(a1^2 a2 + a0 a2^2 + 2 a0 a1 a3 + a0^2 a4)
b5 = -(a1^3 a2+ 3 a0 a1 a2^2 + 3 a0 a1^2 a_3 + 3 a0^2 a2 a3 + 3 a0^2 a1 a4
      + a0^3 a_5)
b6 = -(a1^4 a2 + 6 a0 a1^2 a2^2 + 4 a0 a1^3 a3 + 2 a0^2 a2^3 + 12 a0^2 a1 a2 a3
      + 6 a0^2 a1^2 a4  + 2 a0^3 a3^2 + 4a0^3 a2 a4 + 4 a0^3 a1 a5 + a0^4 a6)
b7 = -(a1^5 a2 + 10 a_0 a1^3 a2^2 + 5 a0 a1^4 a3 + 10 a0^2 a1 a2^3
      + 30 a0^2 a1^2 a2 a3 + 10 a0^2 a1^3 a4 + 10 a0^3 a2^2 a3 + 10 a0^3 a1 a3^2
      + 20 a0^3 a1 a2 a4 + 10 a0^3 a1^2 a5 + 5 a0^4 a3 a4 + 5 a0^4 a2 a5
      + 5 a0^4 a1 a6 + a0^5 a7)
_____________________
		

Crossrefs

Programs

  • Mathematica
    row[0] = row[1] = {1};
    row[n_] := With[{s = Expand[Coefficient[Sum[c[k] x^k, {k, 0, n}]^(n-1), x, n] / (n-1)]}, Table[Coefficient[s, Product[c[t], {t, p}]], {p, Reverse[Sort[Sort /@ IntegerPartitions[n, {n-1}, Range[0, n]]]]}]];
    Table[row[n], {n, 0, 8}] // Flatten (* Andrey Zabolotskiy, Feb 05 2023 *)

Formula

The index notations b(n), b_n, and bn are used interchangeably in this entry for indeterminates.
For n > 1, b_n(a_0,a_1,...,a_n) is a sum of monomials of the form a0^{e0} a1^{e1} a2^{e2} ... an^{en} with e1 * 1 + e2 * 2 + ... + en * n = n. When a_0 is not set to unity, e0 + e1 + ... + en = n - 1. (a1^n is not present.)
From a combinatorial argument in Copeland, the unsigned numerical coefficient of the monomial is given by (n-2)! / [(n - 1 - (e1 + e2 + ... + en))! e1! e2! ... en!].
The partition polynomials are generated by a subset of the Schur self-convolution expansion coefficients as -b_n = K_{n}^{n-1} / (n-1) =(D_{x=0}^n / n!) (a_0 + a_1 x + a_2 x^2 + ... + a_n x^n)^{n-1} / (n-1).
Row sums are the Catalan numbers A000108, ignoring the overall sign, for b_1 onwards.
Reduces to the Narayana triangle A001263 with a_0 = t and all the other indeterminates unity, ignoring the overall sign, for b_2 onwards.
Reduces to A091869 (reversed A091187) with a_1 = t and all the other indeterminates unity, ignoring the overall sign, for b_2 onwards.
b_n(c_1,...,c_n) = - Sum_{k=0}^{n-1} b_k(c_1,...,c_k) N_{n-k}(c_1,...,c_{n-k}) with c_0 = 1 and N_k(c_1,...,c_k) the noncrossing partition polynomials of A134264.
[b] = [R][N], representing the substitution of the noncrossing partition polynomials of A134264 for the indeterminates of the signed reciprocal polynomials of A263633 defined by R_n = 1 / (1 + c_1 x + c_2 x^2 + ...).
Conversely, [R][b] = [N] since the substitution transformation denoted by [R] is an involution, i.e., [R]^2 = [I], the identity substitution.
[b] = [R][A][R], a substitutional conjugation of the set of associahedra partition polynomials of A133147, or A111785, with re-indexing and (1') = 1, e.g., A_0 = 1, A_1 = -c_1, and A_2 = 2 c_1^2 - c_2.
Conversely, [A] = [R][b][R].

Extensions

Rows 8-9 added by Andrey Zabolotskiy, Feb 05 2023
Previous Showing 21-27 of 27 results.