
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_(n1)[c_1,..,c_(n1)].
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(12c.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./(12c.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)


EXAMPLE

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]+1440[2][4]+640[3]^2+120[2]^3+2304[1][5]
+ 960[1][2][3]+720[1]^2[4]+180[1]^2[2]^2+160[1]^3[3]+30[1]^4[2]+[1]^6
P_7 = 46080[7]+16128[2][5]+13440[3][4]+3360[2]^2[3]+26880[1][6]+10080[1][2][4]+4480[1][3]^2+840[1][2]^3+8064[1]^2[5]+3360[1]^2[2][3]+1680[1]^3[4]+420[1]^3[2]^2+280[1]^4[3]+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 2nth partition polynomial of A036039 with its oddindexed 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 kth zero of the Riemann zeta function and k ranges over all the zeros above the real axis. E.g., (see the Math 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
