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%I #24 Feb 12 2021 16:23:10
%S 1,1,0,2,1,0,6,8,1,0,24,58,22,1,0,120,444,328,52,1,0,720,3708,4400,
%T 1452,114,1,0,5040,33984,58140,32120,5610,240,1,0,40320,341136,785304,
%U 644020,195800,19950,494,1,0,362880,3733920,11026296,12440064,5765500,1062500
%N Triangle related to the o.g.f.s. of the right-hand columns of A130534 (E(x,m=1,n)).
%C The asymptotic expansions of the higher-order exponential integral E(x,m=1,n) lead to triangle A130524, see A163931 for information on E(x,m,n). The o.g.f.s. of the right-hand columns of triangle A130534 have a nice structure: gf(p) = W1(z,p)/(1-z)^(2*p-1) with p = 1 for the first right-hand column, p = 2 for the second right-hand column, etc. The coefficients of the W1(z,p) polynomials lead to the triangle given above, n >= 1 and 1 <= m <= n. Our triangle is the same as A112007 with an extra right-hand column, see also the second Eulerian triangle A008517. The row sums of our triangle lead to A001147.
%C We observe that the row sums of the triangles A163936 (m=1), A163937 (m=2), A163938 (m=3) and A163939 (m=4) for z=1 lead to A001147, A001147 (minus a(0)), A001879 and A000457 which are the first four left-hand columns of the triangle of the Bessel coefficients A001497 or, if one wishes, the right-hand columns of A001498. We checked this phenomenon for a few more values of m and found that this pattern persists: m = 5 leads to A001880, m=6 to A001881, m=7 to A038121 and m=8 to A130563 which are the next left- (right-) hand columns of A001497 (A001498). An interesting phenomenon.
%C If one assumes the triangle not (1,1) based but (0,0) based, one has T(n, k) = E2(n, n-k), where E2(n, k) are the second-order Eulerian numbers A340556. - _Peter Luschny_, Feb 12 2021
%H G. C. Greubel, <a href="/A163936/b163936.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F a(n, m) = Sum_{k=0..(m-1)} (-1)^(n+k+1)*binomial(2*n-1,k)*Stirling1(m+n-k-1,m-k), for 1 <= m <= n.
%F Assuming offset = 0 the T(n, k) are the coefficients of recursively defined polynomials. T(n, k) = [x^k] x^n*E2poly(n, 1/x), where E2poly(n, x) = x*(x - 1)^(2*n)*d_{x}((1 - x)^(1 - 2*n)*E2poly(n - 1, x))) for n >= 1 and E2poly(0, x) = 1. - _Peter Luschny_, Feb 12 2021
%e Triangle starts:
%e [ 1] 1;
%e [ 2] 1, 0;
%e [ 3] 2, 1, 0;
%e [ 4] 6, 8, 1, 0;
%e [ 5] 24, 58, 22, 1, 0;
%e [ 6] 120, 444, 328, 52, 1, 0;
%e [ 7] 720, 3708, 4400, 1452, 114, 1, 0;
%e [ 8] 5040, 33984, 58140, 32120, 5610, 240, 1, 0;
%e [ 9] 40320, 341136, 785304, 644020, 195800, 19950, 494, 1, 0;
%e The first few W1(z,p) polynomials are
%e W1(z,p=1) = 1/(1-z);
%e W1(z,p=2) = (1 + 0*z)/(1-z)^3;
%e W1(z,p=3) = (2 + 1*z + 0*z^2)/(1-z)^5;
%e W1(z,p=4) = (6 + 8*z + 1*z^2 + 0*z^3)/(1-z)^7.
%p with(combinat): a := proc(n, m): add((-1)^(n+k+1)*binomial(2*n-1, k)*stirling1(m+n-k-1, m-k), k=0..m-1) end: seq(seq(a(n, m), m=1..n), n=1..9); # _Johannes W. Meijer_, revised Nov 27 2012
%t Table[Sum[(-1)^(n + k + 1)*Binomial[2*n - 1, k]*StirlingS1[m + n - k - 1, m - k], {k, 0, m - 1}], {n, 1, 10}, {m, 1, n}] // Flatten (* _G. C. Greubel_, Aug 13 2017 *)
%o (PARI) for(n=1,10, for(m=1,n, print1(sum(k=0,m-1,(-1)^(n+k+1)* binomial(2*n-1,k)*stirling(m+n-k-1,m-k, 1)), ", "))) \\ _G. C. Greubel_, Aug 13 2017
%o (PARI) \\ assuming offset = 0:
%o E2poly(n,x) = if(n == 0, 1, x*(x-1)^(2*n)*deriv((1-x)^(1-2*n)*E2poly(n-1,x)));
%o { for(n = 0, 9, print(Vec(E2poly(n,x)))) } \\ _Peter Luschny_, Feb 12 2021
%Y Row sums equal A001147.
%Y A000142, A002538, A002539, A112008, A112485 are the first few left hand columns.
%Y A000007, A000012, A005803(n+2), A004301, A006260 are the first few right hand columns.
%Y Cf. A163931 (E(x,m,n)), A048994 (Stirling1) and A008517 (Euler).
%Y Cf. A112007, A163937 (E(x,m=2,n)), A163938 (E(x,m=3,n)) and A163939 (E(x,m=4,n)).
%Y Cf. A001497 (Bessel), A001498 (Bessel), A001147 (m=1), A001147 (m=2), A001879 (m=3) and A000457 (m=4), A001880 (m=5), A001881 (m=6) and A038121 (m=7).
%Y Cf. A340556.
%K easy,nonn,tabl
%O 1,4
%A _Johannes W. Meijer_, Aug 13 2009
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