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%I #68 Feb 19 2024 12:11:00
%S 1,1,2,1,8,6,1,48,32,12,12,1,384,240,160,80,60,20,1,3840,2304,1440,
%T 640,720,960,120,160,180,30,1,46080,26880,16128,13440,8064,10080,4480,
%U 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
%N Polynomials for total Pontryagin classes. Refinement of double Pochhammer triangle.
%C 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.
%H M. Coffey, <a href="https://doi.org/10.1016/j.cam.2003.09.003">Relations and positivity results for derivatives of the Riemann xi function</a>, J. Comput. Appl. Math. 166(2) (2004), 525-534.
%H Tom Copeland, <a href="https://tcjpn.wordpress.com/2020/10/08/appells-and-roses-newton-leibniz-euler-riemann-and-symmetric-polynomials/">Appells and Roses: Newton, Leibniz, Euler, Riemann and Symmetric Polynomials</a>, 2020.
%H M. Fecko, <a href="http://sophia.dtp.fmph.uniba.sk/~fecko/referaty/regensburg_2011.pdf">Selected topological concepts used in physics</a> pp. 37-41 and 56-58.
%H Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/3693429/sums-of-reciprocals-of-powers-of-the-imaginary-part-of-the-nontrivial-zeros-of-t">Sums of the reciprocals of powers of the imaginary part of the nontrivial Riemann zeros</a>, a MSE question posed by T. Copeland and answered by G. Helms, 2020.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Pontryagin_class">Pontryagin class</a>
%H Y. Zhang, <a href="http://www.lepp.cornell.edu/~yz98/notes/A brief introduction to characteristic classes from the differentiable viewpoint.pdf">A brief introduction to characteristic classes from the differentiable viewpoint</a> p. 27.
%F From _Tom Copeland_, Oct 11 2016: (Start)
%F A generating function for the polynomials PB_n[b_2,b_4,..,b_(2n)] of this array is
%F 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.
%F 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)].
%F 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.
%F 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)].
%F Another generator: G(L,0,c_2,c_3,..) (c_1)^n = PC_n(c_1,c_2,..,c_n).
%F 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.
%F Letting c_1 = x and all other c_n = 1 gives the row polynomials of A055140.
%F (End)
%e In terms of the trace of a curvature form Tr(F^n)={n} or indeterminates c_n=[n]:
%e P_0 = 1,
%e P_1 = Tr(F^2) = {2}
%e = c_1 = [1],
%e P_2 = 2Tr(F^4)+Tr(F^2)^2 = 2{4}+{2}^2
%e = 2c_2+ (c_1)^2 = 2[2]+[1]^2,
%e P_3 = 8Tr(F^6)+6Tr(F^2)Tr(F^4)+Tr(F^2)^3= 8{6}+6{2}{4}+{2}^3
%e = 8c_3+6c_1 c_2+(c_1)^3 = 8[3]+6[1][2]+[1]^3,
%e P_4 = 48{8}+32{2}{6}+12{4}^2+12{2}^2{4}+{2}^4
%e = 48[4]+32[1][3]+12[2]^2+12[1]^2[2]+[1]^4,
%e P_5 = 384{10}+240{2}{8}+160{4}{6}+80{2}^2{6}
%e + 60{2}{4}^2+20{2}^3{4}+{2}^5
%e = 384[5]+240[1][4]+160[2][3]+80[1]^2[3]
%e + 60[1][2]^2+20[1]^3[2]+[1]^5
%e P_6 = 3840[6]+2304[1][5]+1440[2][4]+640[3]^2+720[1]^2[4]
%e +960[1][2][3]+120[2]^3+160[1]^3[3]+180[1]^2[2]^2+30[1]^4[2]+[1]^6
%e P_7 = 46080[7]+26880[1][6]+16128[2][5]+13440[3][4]+8064[1]^2[5]
%e +10080[1][2][4]+4480[1][3]^2+3360[2]^2[3]+1680[1]^3[4]
%e +3360[1]^2[2][3]+840[1][2]^3+280[1]^4[3]+420[1]^3[2]^2+42[1]^5[2]+[1]^7
%e ....
%e 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.
%e 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
%t 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]]}]];
%t rows[8] // Flatten (* _Andrey Zabolotskiy_, Feb 19 2024 *)
%Y Cf. A000165, A001147, A036039, A055140, A119275.
%Y Cf. A263916.
%Y The terms are indexed by partitions in the Abramowitz and Stegun order, A036036.
%K nonn,tabf
%O 0,3
%A _Tom Copeland_, Nov 14 2013
%E Polynomials P_6 and P_7 added by _Tom Copeland_, Oct 11 2016
%E Correction to P_3 in Example by _Tom Copeland_, May 27 2020
%E Terms in rows 6-7 reordered, row 8 added by _Andrey Zabolotskiy_, Feb 19 2024
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