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%I #23 Oct 05 2023 02:22:50
%S 1,3,1,13,5,1,79,33,7,1,633,277,61,9,1,6331,2849,643,97,11,1,75973,
%T 34821,7993,1225,141,13,1,1063623,493825,114751,17793,2071,193,15,1,
%U 17017969,7977173,1870837,292681,34361,3229,253,17,1
%N Square array, read by antidiagonals: form the Euler-Seidel matrix for the sequence {2^k*k!} and then divide column k by 2^k*k!.
%C This table is closely connected to the constant 1/sqrt(e). The row, column and diagonal entries of this table occur in series acceleration formulas for 1/sqrt(e). For a similar table based on the differences of the sequence {2^k*k!} and related to the constant sqrt(e), see A143410. For other arrays similarly related to constants see A086764 (for e), A143409 (for 1/e), A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)).
%H G. C. Greubel, <a href="/A143411/b143411.txt">Antidiagonals n = 0..50, flattened</a>
%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} 2^j*binomial(n,j)*(k+j)!.
%F Relation with Poisson-Charlier polynomials c_n(x,a):
%F T(n,k) = (-1)^n*c_n(-(k+1),1/2).
%F Recurrence relations:
%F T(n,k) = 2*n*T(n-1,k) + T(n,k-1);
%F T(n,k) = 2*(n+k)*T(n-1,k) + T(n-1,k-1);
%F T(n,k) = 2*(k+1)*T(n-1,k+1) + T(n-1,k).
%F Recurrence for row n entries: 2*k*T(n,k) = (2*n+2*k-1)*T(n,k-1) + T(n,k-2).
%F E.g.f. for column k: exp(y)/(1 - 2*y)^(k+1).
%F E.g.f. for array: exp(y)/(1 - x - 2*y) = (1 + x + x^2 + ...) + (3 + 5*x + 7*x^2 + ...)*y + (13 + 37*x + 61*x^2 + ...)*y^2/2! + ... .
%F Series acceleration formulas for 1/sqrt(e):
%F Row n: 1/sqrt(e) = 2^n*n!*(1/T(n,0) - 1/(2*1!*T(n,0)*T(n,1)) + 1/(2^2*2!*T(n,1)*T(n,2)) - 1/(2^3*3!*T(n,2)*T(n,3)) + ...). For example, row 3 gives 1/sqrt(e) = 48*(1/79 - 1/(2*79*277) + 1/(8*277*643) - 1/(48*643*1225) + ...).
%F Column k: 1/sqrt(e) = (1 - (1/2)/1! + (1/2)^2/2! - ... + (-1/2)^k/k!) + (-1)^(k+1)/(2^k*k!)*( Sum_{n = 0..inf} 2^n*n!/(T(n,k)*T(n+1,k)) ). For example, column 3 gives 1/sqrt(e) = 29/48 + 1/48*( 1/(1*9) + 2/(9*97) + 8/(97*1225) + 48/(1225*17793) + ... ).
%F Main diagonal: 1/sqrt(e) = 1 - 2*( 1/(1*5) - 1/(5*61) + 1/(61*1225) - ... ). See A065919.
%e The Euler-Seidel matrix for the sequence {2^k*k!} begins
%e ========================================
%e n\k| 0 1 2 3 4 5
%e ========================================
%e 0 | 1 2 8 48 384 3840
%e 1 | 3 10 56 432 4224
%e 2 | 13 66 488 4656
%e 3 | 79 554 5144
%e 4 | 633 5698
%e 5 | 6331
%e ...
%e .
%e Dividing the k-th column by 2^k*k! gives
%e ========================================
%e n\k| 0 1 2 3 4 5
%e ========================================
%e 0 | 1 1 1 1 1 1
%e 1 | 3 5 7 9 11
%e 2 | 13 33 61 97
%e 3 | 79 277 643
%e 4 | 633 2849
%e 5 | 6331
%e ...
%p with combinat: T := (n, k) -> 1/k!*add(2^j*binomial(n,j)*(k+j)!, j = 0..n): for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;
%t A[n_, k_]:= (1/k!)*Sum[Binomial[n,j]*(k+j)!*2^j, {j,0,n}]; (* array *)
%t A143411[n_, k_]:= A[n-k,k]; (* anti-diagonals *)
%t Table[A143411[n,k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Oct 05 2023 *)
%o (Magma)
%o A:= func< n,k | (&+[Binomial(n,j)*Factorial(k+j)*2^j/Factorial(k): j in [0..n]]) >; // Array
%o A143411:= func< n,k | A(n-k,k) >; // anti-diagonal triangle
%o [A143411(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Oct 05 2023
%o (SageMath)
%o def A(n,k): return sum(binomial(n,j)*factorial(j+k)*2^j/factorial(k) for j in range(n+1)) # array
%o def A143411(n,k): return A(n-k,k) # anti-diagonal triangle
%o flatten([[A143411(n,k) for k in range(n+1)] for n in range(13)]) # _G. C. Greubel_, Oct 05 2023
%Y Cf. A008288, A065919 (main diagonal), A076571, A086764, A108625, A143007, A143409, A143410.
%K easy,nonn,tabl
%O 0,2
%A _Peter Bala_, Aug 19 2008
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