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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!.
5

%I #26 Jul 09 2024 20:10:47

%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]; (* antidiagonals *)

%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) >; // antidiagonal 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) # antidiagonal 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