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 A231846 Polynomials for total Pontryagin classes. Refinement of double Pochhammer triangle. 4
 1, 1, 2, 1, 8, 6, 1, 48, 32, 12, 12, 1, 384, 240, 160, 80, 60, 20, 1, 3840, 1440, 640, 120, 2304, 960, 720, 180, 160, 30, 1, 46080, 16128, 13440, 3360, 26880, 10080, 4480, 840, 8064, 3360, 1680, 420, 280, 42, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 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. LINKS M. Coffey, Relations and positivity results for derivatives of the Riemann xi function, J. Comput. Appl. Math. 166(2) (2004), 525-534. T. 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 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) 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 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. A000165, A001147, A036039, A055140, A119275. Cf. A263916. Sequence in context: A142336 A193735 A114193 * A039683 A318389 A108085 Adjacent sequences: A231843 A231844 A231845 * A231847 A231848 A231849 KEYWORD nonn,tabf,more AUTHOR Tom Copeland, Nov 14 2013 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 STATUS approved

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Last modified November 27 09:16 EST 2022. Contains 358367 sequences. (Running on oeis4.)