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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. 4

%I

%S 1,1,1,5,3,1,29,17,5,1,233,131,37,7,1,2329,1281,353,65,9,1,27949,

%T 15139,4105,743,101,11,1,391285,209617,56189,10049,1349,145,13,1,

%U 6260561,3325923,883885,156679,20841,2219,197,15,1,112690097,59475329,15700313

%N 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 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 Eric Weisstein's World of Mathematics <a href="http://mathworld.wolfram.com/Poisson-CharlierPolynomial.html">Poisson-Charlier polynomial</a>

%F T(n,k) = (-1)^n/k!*sum {j = 0..n} (-2)^j*C(n,j)*(k+j)!.

%F Relation with Poisson-Charlier polynomials c_n(x,a): T(n,k) = c_n(-(k+1),-1/2).

%F 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 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 + ...) + (1 + 3*x + 5*x^2 + ...)*y + (5 + 17*x + 37*x^2 + ...)*y^2/2! + ... .

%F Series acceleration formulas for sqrt(e):

%F 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 Column k: sqrt(e) = (1+(1/2)/1!+(1/2)^2/2!+...+(1/2)^k/k!) + 1/(2^k*k!) * sum {n = 0..inf}((-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 Main diagonal: sqrt(e) = 1 + 2*[1/(1*3) - 1/(3*37) + 1/(37*743) - ...]. See A143412.

%F T(n, k) = (-1)^n*(-1/2)^(k + 1)*KummerU(k + 1, k + n + 2, -1/2). - _Peter Luschny_, Jan 02 2020

%e Table of differences of {2^k*k!}

%e =====================================================

%e Column................0.....1.....2.....3.....4.....5

%e =====================================================

%e Sequence 2^k*k! ......1.....2.....8....48...384..3840

%e First differences.....1.....6....40...336..3456

%e Second differences....5....34...296..3120

%e Third differences....29...262..2824

%e Fourth differences..233..2562

%e ...

%e Remove the common factor 2^k*k! from k-th column entries:

%e ====================================

%e n\k|...0......1......2......3......4

%e ====================================

%e 0..|...1......1......1......1......1

%e 1..|...1......3......5......7......9

%e 2..|...5.....17.....37.....65....101

%e 3..|..29....131....353....743...1349

%e 4..|.233...1281...4105..10049..20841

%e ...

%p T := (n, k) -> (-1)^n/k!*add((-2)^j*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;

%Y Cf. A008288, A076571, A086764, A108625, A143007, A143409, A143411, A143412.

%K easy,nonn,tabl

%O 0,4

%A _Peter Bala_, Aug 19 2008

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Last modified February 29 02:18 EST 2020. Contains 332353 sequences. (Running on oeis4.)