A143409 Square array read by antidiagonals: form the Euler-Seidel matrix for the sequence {k!} and then divide column k by k!.

1, 2, 1, 5, 3, 1, 16, 11, 4, 1, 65, 49, 19, 5, 1, 326, 261, 106, 29, 6, 1, 1957, 1631, 685, 193, 41, 7, 1, 13700, 11743, 5056, 1457, 316, 55, 8, 1, 109601, 95901, 42079, 12341, 2721, 481, 71, 9, 1, 986410, 876809, 390454, 116125, 25946, 4645, 694, 89, 10, 1

Offset: 0

Author

Peter Bala, Aug 14 2008

Keywords

• easy
• nonn
• tabl

Comments

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.

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.

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)).

Examples

			The Euler-Seidel matrix for the sequence {k!} begins
==============================================
n\k|.....0.....1.....2.....3.....4.....5.....6
==============================================
0..|.....1.....1.....2.....6....24...120...720
1..|.....2.....3.....8....30...144...840
2..|.....5....11....38...174...984
3..|....16....49...212..1158
4..|....65...261..1370
5..|...326..1631
6..|..1957
...
Dividing the k-th column by k! gives
==============================================
n\k|.....0.....1.....2.....3.....4.....5.....6
==============================================
0..|.....1.....1.....1.....1.....1.....1.....1
1..|.....2.....3.....4.....5.....6.....7
2..|.....5....11....19....29....41
3..|....16....49...106...193
4..|....65...261...685
5..|...326..1631
6..|..1957
...
Examples of series formula for 1/e:
Row 2: 1/e = 2*(1/5 - 1/(1!*5*11) + 1/(2!*11*19) - 1/(3!*19*29) + ...).
Column 4: 24/e = 9 - (0!/(1*6) + 1!/(6*41) + 2!/(41*316) + ...).
...
Displayed as a triangle:
0 |     1
1 |     2,     1
2 |     5,     3,    1
3 |    16,    11,    4,    1
4 |    65,    49,   19,    5,   1
5 |   326,   261,  106,   29,   6,  1
6 |  1957,  1631,  685,  193,  41,  7, 1
7 | 13700, 11743, 5056, 1457, 316, 55, 8, 1
		

Links

• Robert Israel, Table of n, a(n) for n = 0..10010 (antidiagonals 0 to 140, flattened)
• D. Dumont, Matrices d'Euler-Seidel, Sem. Loth. Comb. B05c (1981) 59-78.
• Eric Weisstein's World of Mathematics Poisson-Charlier polynomial

Crossrefs

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).

Cf. A273596, A003470.

Programs

Maple

T := (n, k) -> 1/k!*add(binomial(n,j)*(k+j)!, j = 0..n):
for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;
# Alternate:
T:= proc(n,k) option remember;
  if n = 0 then return 1 fi;
  (n+k)*procname(n-1,k) + procname(n-1,k-1);
end proc:
seq(seq(T(s-n,n),n=0..s),s=0..10); # Robert Israel, Jul 07 2017
# Or:
A143409 := (n,k) -> hypergeom([k+1, k-n], [], -1):
seq(seq(simplify(A143409(n,k)),k=0..n),n=0..9); # Peter Luschny, Oct 05 2017

Mathematica

T[n_, k_] := HypergeometricPFQ[{k+1,k-n}, {}, -1];
Table[T[n,k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)

Formula

T(n,k) = (1/k!)*Sum_{j = 0..n} binomial(n,j)*(k+j)!.

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.

Recurrence relations:

T(n,k) = T(n-1,k) + (k+1)*T(n-1,k+1);

T(n,k) = (n+k)*T(n-1,k) + T(n-1,k-1).

E.g.f. for column k: exp(y)/(1-y)^(k+1).

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! + ... .

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).

Main diagonal is A001517.

Series formulas for 1/e:

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)) + ...].

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)) + ...].

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) - ...].

Second subdiagonal: 1/e = 2*(1^2/(1*5) - 2^2/(5*49) + 3^2/(49*685) - ...).

Compare with A143413.

From Peter Luschny, Oct 05 2017: (Start)

T(n, k) = hypergeom([k+1, k-n], [], -1).

When seen as a triangular array then the row sums are A273596 and the alternating row sums are A003470. (End)