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A143410 Form the difference table of the sequence {2^k*k!}, then divide k-th column entries by 2^k*k!. Array read by ascending antidiagonals, T(n, k) for n, k >= 0.

%I A143410 #25 Feb 16 2025 08:33:08
%S A143410 1,1,1,5,3,1,29,17,5,1,233,131,37,7,1,2329,1281,353,65,9,1,27949,
%T A143410 15139,4105,743,101,11,1,391285,209617,56189,10049,1349,145,13,1,
%U A143410 6260561,3325923,883885,156679,20841,2219,197,15,1,112690097,59475329,15700313
%N A143410 Form the difference table of the sequence {2^k*k!}, then divide k-th column entries by 2^k*k!. Array read by ascending antidiagonals, T(n, k) for n, k >= 0.
%C A143410 This table is closely connected to the constant sqrt(e). The row, column and diagonal entries of this table occur in series acceleration formulas for sqrt(e). For a similar table based on the Euler-Seidel matrix of the sequence {2^k*k!} and related to the constant 1/sqrt(e), see A143411. 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 A143410 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Poisson-CharlierPolynomial.html">Poisson-Charlier polynomial</a>.
%F A143410 T(n,k) = ((-1)^n/k!)*Sum {j = 0..n} (-2)^j*C(n,j)*(k+j)!.
%F A143410 Relation with Poisson-Charlier polynomials c_n(x,a): T(n,k) = c_n(-(k+1),-1/2).
%F A143410 Recurrence relations: T(n,k) = 2*n*T(n-1,k) + T(n,k-1); T(n,k) = 2*(n+k)*T(n-1,k) - T(n-1,k-1); T(n,k) = 2*(k+1)*T(n-1,k+1) - T(n-1,k);
%F A143410 Recurrence for row n entries: 2*k*T(n,k) = (2*n+2*k+1)*T(n,k-1) - T(n,k-2).
%F A143410 E.g.f. for column k: exp(-y)/(1-2*y)^(k+1).
%F A143410 E.g.f. for array: exp(-y)/(1-x-2*y) = (1 + x + x^2 + ...) + (1 + 3*x + 5*x^2 + ...)*y + (5 + 17*x + 37*x^2 + ...)*y^2/2! + ... .
%F A143410 Series acceleration formulas for sqrt(e):
%F A143410 Row n: sqrt(e) = 2^n*n!*(1/T(n,0) + (-1)^n*(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 sqrt(e) = 48*(1/29 - 1/(2*29*131) - 1/(8*131*353) - 1/(48*353*743) - ...).
%F A143410 Column k: sqrt(e) = (1 + (1/2)/1! + (1/2)^2 / 2! + ... + (1/2)^k/k!) + 1/(2^k*k!) * Sum_{n>= 0} ((-2)^n *n!/(T(n,k)*T(n+1,k))). For example, column 3 gives sqrt(e) = 79/48 + (1/48)*(1/(1*7) - 2/(7*65) + 8/(65*743) - 48/(743*10049) + ...).
%F A143410 Main diagonal: sqrt(e) = 1 + 2*(1/(1*3) - 1/(3*37) + 1/(37*743) - ...). See A143412.
%F A143410 T(n, k) = (-1)^n*(-1/2)^(k + 1)*KummerU(k + 1, k + n + 2, -1/2). - _Peter Luschny_, Jan 02 2020
%e A143410 Table of differences of {2^k*k!}
%e A143410   =====================================================
%e A143410   Column                0     1     2     3     4     5
%e A143410   =====================================================
%e A143410   Sequence 2^k*k!       1     2     8    48   384  3840
%e A143410   First differences     1     6    40   336  3456
%e A143410   Second differences    5    34   296  3120
%e A143410   Third differences    29   262  2824
%e A143410   Fourth differences  233  2562
%e A143410   ...
%e A143410 Remove the common factor 2^k*k! from k-th column entries:
%e A143410   ====================================
%e A143410   n\k|   0      1      2      3      4
%e A143410   ====================================
%e A143410   0  |   1      1      1      1      1
%e A143410   1  |   1      3      5      7      9
%e A143410   2  |   5     17     37     65    101
%e A143410   3  |  29    131    353    743   1349
%e A143410   4  | 233   1281   4105  10049  20841
%p A143410 T := (n, k) -> (-1)^n/k!*add((-2)^j*binomial(n,j)*(k+j)!, j = 0..n):
%p A143410 for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;
%Y A143410 Cf. A008288, A076571, A086764, A108625, A143007, A143409, A143411, A143412.
%K A143410 easy,nonn,tabl
%O A143410 0,4
%A A143410 _Peter Bala_, Aug 19 2008