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Triangle read by rows: T(n, k) = (k+1)*A132393(n+1, k+1), for 0 <= k <= n.
31

%I #86 Feb 25 2024 15:03:36

%S 1,1,2,2,6,3,6,22,18,4,24,100,105,40,5,120,548,675,340,75,6,720,3528,

%T 4872,2940,875,126,7,5040,26136,39396,27076,9800,1932,196,8,40320,

%U 219168,354372,269136,112245,27216,3822,288,9

%N Triangle read by rows: T(n, k) = (k+1)*A132393(n+1, k+1), for 0 <= k <= n.

%C Previous name was: Number triangle f(n, k) from n-th differences of the sequence {1/m^2}_{m >= 1}, for n >= 0; the n-th difference sequence is {(-1)^n*n!*P(n, m)/D(n, m)^2}_{m >= 1} where P(n, x) is the row polynomial P(n, x) = Sum_{k=0..n} f(n,k)*x^k and D(n, x) = x*(x+1)*...*(x+n).

%C From _Johannes W. Meijer_, Oct 07 2009: (Start)

%C The higher-order exponential integrals E(x,m,n) are defined in A163931 and the general formula of the asymptotic expansion of E(x,m,n) can be found in A163932.

%C We used the general formula and the asymptotic expansion of E(x,m=1,n), see A130534, to determine that E(x,m=2,n) ~ (exp(-x)/x^2)*(1 - (1+2*n)/x + (2 + 6*n + 3*n^2)/x^2 - (6 + 22*n + 18*n^2 + 4*n^3)/x^3 + ...) which can be verified with the EA(x,2,n) formula, see A163932. The coefficients in the denominators of this expansion lead to the sequence given above.

%C The asymptotic expansion of E(x,m=2,n) leads for n from one to ten to known sequences, see the cross-references. With these sequences one can form the triangles A165674 (left hand columns) and A093905 (right hand columns).

%C (End)

%C For connections to an operator relation between log(x) and x^n(d/dx)^n, see A238363. - _Tom Copeland_, Feb 28 2014

%C From _Wolfdieter Lang_, Nov 25 2018: (Start)

%C The signed triangle t(n, k) := (-1)^{n-k}*f(n, k) gives (n+1)*N(-1;n,x) = Sum_{k=0..n} t(n, k)*x^k, where N(-1;n,x) are the Narumi polynomials with parameter a = -1 (see the Weisstein link).

%C The members of the n-th difference sequence of the sequence {1/m^2}_{m>=1} mentioned above satisfies the recurrence delta(n, m) = delta(n-1, m+1) - delta(n-1, m), for n >= 1, m >= 1, with input delta(0, m) = 1/m^2. The solution is delta(n, m) = (n+1)!*N(-1;n,-m)/risefac(m, n+1)^2, with Narumi polynomials N(-1;n,x) and the rising factorials risefac(x, n+1) = D(n, x) = x*(x+1)*...*(x+n).

%C The above mentioned row polynomials P satisfy P(n, x) = (-1)^n*(n + 1)*N(-1;n,-x), for n >= 0. The recurrence is P(n, x) = (-x^2*P(n-1, x+1) + (n+x)^2*P(n-1, x))/n, for n >= 1, and P(0, x) = 1. (End)

%C The triangle is the exponential Riordan square (cf. A321620) of -log(1-x) with an additional main diagonal of zeros. - _Peter Luschny_, Jan 03 2019

%H G. C. Greubel, <a href="/A028421/b028421.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%H Eric Weisstein's World of Mathematics, <a href="http://www.mathworld.wolfram.com/NarumiPolynomial.html">Narumi Polynomial</a> [here for a = -1].

%F E.g.f.: d/dt(-log(1-t)/(1-t)^x). - _Vladeta Jovovic_, Oct 12 2003

%F The e.g.f. with offset 1: y = x + (1 + 2*t)*x^2/2! + (2 + 6*t + 3*t^2)*x^3/3! + ... has series reversion with respect to x equal to y - (1 + 2*t)*y^2/2! + (1 + 3*t)^2*y^3/3! - (1 + 4*t)^3*y^4/4! + .... This is an e.g.f. for a signed version of A139526. - _Peter Bala_, Jul 18 2013

%F Recurrence: T(n, k) = 0 if n < k; if k = 0 then T(0, 0) = 1 and T(n, 0) = n * T(n-1, 0) for n >= 1, otherwise T(n, k) = n*T(n-1, k) + ((k+1)/k)*T(n-1, k-1). From the unsigned Stirling1 recurrence. - _Wolfdieter Lang_, Nov 25 2018

%e The triangle T(n, k) begins:

%e n\k 0 1 2 3 4 5 6 7 8 9 10

%e ------------------------------------------------------------------------------------

%e 0: 1

%e 1: 1 2

%e 2: 2 6 3

%e 3: 6 22 18 4

%e 4: 24 100 105 40 5

%e 5: 120 548 675 340 75 6

%e 6: 720 3528 4872 2940 875 126 7

%e 7: 5040 26136 39396 27076 9800 1932 196 8

%e 8: 40320 219168 354372 269136 112245 27216 3822 288 9

%e 9: 362880 2053152 3518100 2894720 1346625 379638 66150 6960 405 10

%e 10: 3628800 21257280 38260728 33638000 17084650 5412330 1104411 145200 11880 550 11

%e ... - _Wolfdieter Lang_, Nov 23 2018

%p A028421 := proc(n,k) (-1)^(n+k)*(k+1)*Stirling1(n+1,k+1) end:

%p seq(seq(A028421(n,k), k=0..n), n=0..8);

%p # _Johannes W. Meijer_, Oct 07 2009, Revised Sep 09 2012

%p egf := (1 - t)^(-x - 1)*(1 - x*log(1 - t)):

%p ser := series(egf, t, 16): coefft := n -> expand(coeff(ser,t,n)):

%p seq(seq(n!*coeff(coefft(n), x, k), k = 0..n), n = 0..8); # _Peter Luschny_, Jun 12 2022

%t f[n_, k_] = (k + 1) StirlingS1[n + 1, k + 1] // Abs; Flatten[Table[f[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* _Jean-François Alcover_, Jun 01 2011, after formula *)

%o (Sage) # uses[riordan_square from A321620]

%o riordan_square(-ln(1 - x), 10, True) # _Peter Luschny_, Jan 03 2019

%Y Row sums give A000254(n+1), n >= 0.

%Y Cf. A132393 (unsigned Stirling1), A061356, A139526, A321620.

%Y From _Johannes W. Meijer_, Oct 07 2009: (Start)

%Y A000142, A052517, 3*A000399, 5*A000482 are the first four left hand columns; A000027, A002411 are the first two right hand columns.

%Y The asymptotic expansion of E(x,m=2,n) leads to A000254 (n=1), A001705 (n=2), A001711 (n=3), A001716 (n=4), A001721 (n=5), A051524 (n=6), A051545 (n=7), A051560 (n=8), A051562 (n=9), A051564 (n=10), A093905 (triangle) and A165674 (triangle).

%Y Cf. A163931 (E(x,m,n)), A130534 (m=1), A163932 (m=3), A163934 (m=4), A074246 (E(x,m=2,n+1)). (End)

%K tabl,nonn

%O 0,3

%A Peter Wiggen (wiggen(AT)math.psu.edu)

%E Edited by _Wolfdieter Lang_, Nov 23 2018