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%I #24 Oct 06 2017 08:35:16
%S 1,2,1,5,3,1,16,11,4,1,65,49,19,5,1,326,261,106,29,6,1,1957,1631,685,
%T 193,41,7,1,13700,11743,5056,1457,316,55,8,1,109601,95901,42079,12341,
%U 2721,481,71,9,1,986410,876809,390454,116125,25946,4645,694,89,10,1
%N Square array read by antidiagonals: form the Euler-Seidel matrix for the sequence {k!} and then divide column k by k!.
%C The Euler-Seidel matrix for the sequence {k!} is array A076571 read as a square, whose k-th column entries have a common factor of k!. Removing these common factors gives the current table.
%C This table is closely connected to the constant 1/e. The row, column and diagonal entries of this table occur in series acceleration formulas for 1/e.
%C For a similar table based on the differences of the sequence {k!} and related to the constant e, see A086764. For other arrays similarly related to constants see A143410 (for sqrt(e)), A143411 (for 1/sqrt(e)), A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)).
%H Robert Israel, <a href="/A143409/b143409.txt">Table of n, a(n) for n = 0..10010</a> (antidiagonals 0 to 140, flattened)
%H D. Dumont, <a href="http://www.mat.univie.ac.at/~slc/opapers/s05dumont.html">Matrices d'Euler-Seidel</a>, Sem. Loth. Comb. B05c (1981) 59-78.
%H Eric Weisstein's World of Mathematics <a href="http://mathworld.wolfram.com/Poisson-CharlierPolynomial.html">Poisson-Charlier polynomial</a>
%F T(n,k) = (1/k!)*Sum_{j = 0..n} binomial(n,j)*(k+j)!.
%F T(n,k) = ((n+k)!/k!)*Num_Pade(n,k), where Num_Pade(n,k) denotes the numerator of the Padé approximation for the function exp(x) of degree (n,k) evaluated at x = 1.
%F Recurrence relations:
%F T(n,k) = T(n-1,k) + (k+1)*T(n-1,k+1);
%F T(n,k) = (n+k)*T(n-1,k) + T(n-1,k-1).
%F E.g.f. for column k: exp(y)/(1-y)^(k+1).
%F E.g.f. for array: exp(y)/(1-x-y) = (1 + x + x^2 + ...) + (2 + 3*x + 4*x^2 + ...)*y + (5 + 11*x + 19*x^2 + ...)*y^2/2! + ... .
%F Row n lists the values of the Poisson-Charlier polynomial x^(n) + C(n,1)*x^(n-1) + C(n,2)*x^(n-2) + ... + C(n,n) for x = 1,2,3,..., where x^(m) denotes the rising factorial x*(x+1)*...*(x+m-1).
%F Main diagonal is A001517.
%F Series formulas for 1/e:
%F Row n: 1/e = n!*[1/T(n,0) - 1/(1!*T(n,0)*T(n,1)) + 1/(2!*T(n,1)*T(n,2)) - 1/(3!*T(n,2)*T(n,3)) + ...].
%F Column k: k!/e = A000166(k) + (-1)^(k+1)*[0!/(T(0,k)*T(1,k)) + 1!/(T(1,k)*T(2,k)) + 2!/(T(2,k)*T(3,k)) + ...].
%F Main diagonal: 1/e = 1 - 2*Sum_{n>=0} (-1)^n/(T(n,n)*T(n+1,n+1)) = 1 - 2*[1/(1*3) - 1/(3*19) + 1/(19*193) - ...].
%F Second subdiagonal: 1/e = 2*(1^2/(1*5) - 2^2/(5*49) + 3^2/(49*685) - ...).
%F Compare with A143413.
%F From _Peter Luschny_, Oct 05 2017: (Start)
%F T(n, k) = hypergeom([k+1, k-n], [], -1).
%F When seen as a triangular array then the row sums are A273596 and the alternating row sums are A003470. (End)
%e The Euler-Seidel matrix for the sequence {k!} begins
%e ==============================================
%e n\k|.....0.....1.....2.....3.....4.....5.....6
%e ==============================================
%e 0..|.....1.....1.....2.....6....24...120...720
%e 1..|.....2.....3.....8....30...144...840
%e 2..|.....5....11....38...174...984
%e 3..|....16....49...212..1158
%e 4..|....65...261..1370
%e 5..|...326..1631
%e 6..|..1957
%e ...
%e Dividing the k-th column by k! gives
%e ==============================================
%e n\k|.....0.....1.....2.....3.....4.....5.....6
%e ==============================================
%e 0..|.....1.....1.....1.....1.....1.....1.....1
%e 1..|.....2.....3.....4.....5.....6.....7
%e 2..|.....5....11....19....29....41
%e 3..|....16....49...106...193
%e 4..|....65...261...685
%e 5..|...326..1631
%e 6..|..1957
%e ...
%e Examples of series formula for 1/e:
%e Row 2: 1/e = 2*(1/5 - 1/(1!*5*11) + 1/(2!*11*19) - 1/(3!*19*29) + ...).
%e Column 4: 24/e = 9 - (0!/(1*6) + 1!/(6*41) + 2!/(41*316) + ...).
%e ...
%e Displayed as a triangle:
%e 0 | 1
%e 1 | 2, 1
%e 2 | 5, 3, 1
%e 3 | 16, 11, 4, 1
%e 4 | 65, 49, 19, 5, 1
%e 5 | 326, 261, 106, 29, 6, 1
%e 6 | 1957, 1631, 685, 193, 41, 7, 1
%e 7 | 13700, 11743, 5056, 1457, 316, 55, 8, 1
%p T := (n, k) -> 1/k!*add(binomial(n,j)*(k+j)!, j = 0..n):
%p for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;
%p # Alternate:
%p T:= proc(n,k) option remember;
%p if n = 0 then return 1 fi;
%p (n+k)*procname(n-1,k) + procname(n-1,k-1);
%p end proc:
%p seq(seq(T(s-n,n),n=0..s),s=0..10); # _Robert Israel_, Jul 07 2017
%p # Or:
%p A143409 := (n,k) -> hypergeom([k+1, k-n], [], -1):
%p seq(seq(simplify(A143409(n,k)),k=0..n),n=0..9); # _Peter Luschny_, Oct 05 2017
%t T[n_, k_] := HypergeometricPFQ[{k+1,k-n}, {}, -1];
%t Table[T[n,k], {n,0,9}, {k,0,n}] // Flatten (* _Peter Luschny_, Oct 05 2017 *)
%Y Cf. A008288, A076571, A086764, A108625, A143007, A143410, A143411, A143413, A001517 (main diagonal), A028387 (row 2), A000522 (column 0), A001339 (column 1), A082030 (column 2), A095000 (column 3), A095177 (column 4).
%Y Cf. A273596, A003470.
%K easy,nonn,tabl
%O 0,2
%A _Peter Bala_, Aug 14 2008
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