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%I #42 Jun 25 2018 22:59:26
%S 1,3,2,11,12,3,50,70,30,4,274,450,255,60,5,1764,3248,2205,700,105,6,
%T 13068,26264,20307,7840,1610,168,7,109584,236248,201852,89796,22680,
%U 3276,252,8,1026576,2345400,2171040,1077300,316365,56700,6090,360,9
%N Triangle of coefficients, read by rows, where the n-th row forms the polynomial P(n,x) = {Sum_{k=1..n} 1/(k+x)}*{Product_{k=1..n} (k+x)}.
%C The n-th row polynomial, P(n,x), has ordered zeros {z_k < z_(k+1), 0<k<n} that satisfy z_k + z_(n-k) = -(n+1) and integerpart(z_k) = -k. For even rows, polynomial P(2n,x) has zero z_n = -(n+1)/2. Example: at n=6, P(6,x) has zeros z_1 = -1.336553473264694, z_2 = -2.426299641757407, z_3 = -3.5, z_4 = -4.573700358242594, z_5 = -5.663446526735307.
%C The higher-order exponential integrals E(x,m,n) are defined in A163931 and the asymptotic expansion of E(x,m=2,n) can be found in A028421. We determined with the latter that E(x,m=2,n+1) = (exp(-x)/x^2)*(1 - (3+2*n)/x + (11+12*n+3*n^2)/x^2 - (50+70*n+30*n^2+ 4*n^3)/x^3 + .... ). The polynomial coefficients in the numerators lead to the coefficients of the triangle given above. The numerators of the o.g.f.s of the right hand columns of this triangle lead for z = 1 to A001147. - _Johannes W. Meijer_, Oct 16 2009
%H G. C. Greubel, <a href="/A074246/b074246.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F First column is A000254 (Stirling numbers of first kind s(n, 2): a(n+1)=(n+1)*a(n)+n!), while sum of rows is A001705 (generalized Stirling numbers). Also related to Harmonic numbers: P(n, 0)=n!*H(n), H(n)=harmonic number.
%F T(n,k) = (-1)^(n+k)*k*Stirling1(n+1,k+1). - _Johannes W. Meijer_, Oct 16 2009
%F E.g.f.: 1/(1 - z)^(x+1)*log(1/(1 - z)). Cf. A028421. - _Peter Bala_, Jan 06 2015
%e Polynomials begin:
%e P(1,x) = 1,
%e P(2,x) = 3 + 2x,
%e P(3,x) = 11 + 12x + 3x^2,
%e P(4,x) = 50 + 70x + 30x^2 + 4x^3,
%e P(5,x) = 274 + 450x + 255x^2 + 60x^3 + 5x^4,
%e P(6,x) = 1764 + 3248x + 2205x^2 + 700x^3 + 105x^4 + 6x^5,
%e P(7,x) = 13068 + 26264x + 20307x^2 + 7840x^3 + 1610x^4 + 168x^5 + 7x^6,
%e P(8,x) = 109584 + 236248x + 201852x^2 + 89796x^3 + 22680x^4 + 3276x^5 + 252x^6 + 8x^7,
%e P(9,x) = 1026576 + 2345400x + 2171040x^2 + 1077300x^3 + 316365x^4 + 56700x^5 + 6090x^6 + 360x^7 + 9x^8,
%e P(10,x) = 10628640 + 25507152x + 25228500x^2 + 13667720x^3 + 4510275x^4 + 946638x^5 + 127050x^6 + 10560x^7 + 495x^8 + 10x^9, ...
%p with(combinat): A074246 := proc(n,m): (-1)^(n+m)*binomial(m,1)*stirling1(n+1,m+1) end: seq(seq(A074246(n,m),m=1..n),n=1..9); # _Johannes W. Meijer_, Oct 16 2009, Revised Sep 09 2012
%t p[n_, x_] := Sum[1/(k+x), {k, 1, n}] Product[k+x, {k, 1, n}] ; Flatten[Table[ CoefficientList[ p[n, x] // Simplify[#, ComplexityFunction -> Length] &, x], {n, 1, 9}]] (* _Jean-François Alcover_, May 04 2011 *)
%o (PARI) P(n) = Vecrev(sum(k=1, n, prod(k=1, n, (k+x))/(k+x)));
%o for (n=1, 10, print(P(n))) \\ _Michel Marcus_, Jan 22 2017
%Y See references and formulas at A000254, A001705. Cf. A028421.
%Y A027480 is the second right hand column. - _Johannes W. Meijer_, Oct 16 2009
%K easy,nice,nonn,tabl
%O 1,2
%A _Paul D. Hanna_, Sep 19 2002
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