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A143411
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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!.
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5
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1, 3, 1, 13, 5, 1, 79, 33, 7, 1, 633, 277, 61, 9, 1, 6331, 2849, 643, 97, 11, 1, 75973, 34821, 7993, 1225, 141, 13, 1, 1063623, 493825, 114751, 17793, 2071, 193, 15, 1, 17017969, 7977173, 1870837, 292681, 34361, 3229, 253, 17, 1
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OFFSET
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0,2
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COMMENTS
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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)).
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LINKS
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FORMULA
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T(n,k) = (1/k!)*Sum_{j = 0..n} 2^j*binomial(n,j)*(k+j)!.
Relation with Poisson-Charlier polynomials c_n(x,a):
T(n,k) = (-1)^n*c_n(-(k+1),1/2).
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).
Recurrence for row n entries: 2*k*T(n,k) = (2*n+2*k-1)*T(n,k-1) + T(n,k-2).
E.g.f. for column k: exp(y)/(1 - 2*y)^(k+1).
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! + ... .
Series acceleration formulas for 1/sqrt(e):
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) + ...).
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) + ... ).
Main diagonal: 1/sqrt(e) = 1 - 2*( 1/(1*5) - 1/(5*61) + 1/(61*1225) - ... ). See A065919.
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EXAMPLE
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The Euler-Seidel matrix for the sequence {2^k*k!} begins
========================================
n\k| 0 1 2 3 4 5
========================================
0 | 1 2 8 48 384 3840
1 | 3 10 56 432 4224
2 | 13 66 488 4656
3 | 79 554 5144
4 | 633 5698
5 | 6331
...
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Dividing the k-th column by 2^k*k! gives
========================================
n\k| 0 1 2 3 4 5
========================================
0 | 1 1 1 1 1 1
1 | 3 5 7 9 11
2 | 13 33 61 97
3 | 79 277 643
4 | 633 2849
5 | 6331
...
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MAPLE
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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;
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MATHEMATICA
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A[n_, k_]:= (1/k!)*Sum[Binomial[n, j]*(k+j)!*2^j, {j, 0, n}]; (* array *)
A143411[n_, k_]:= A[n-k, k]; (* anti-diagonals *)
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PROG
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(Magma)
A:= func< n, k | (&+[Binomial(n, j)*Factorial(k+j)*2^j/Factorial(k): j in [0..n]]) >; // Array
A143411:= func< n, k | A(n-k, k) >; // anti-diagonal triangle
(SageMath)
def A(n, k): return sum(binomial(n, j)*factorial(j+k)*2^j/factorial(k) for j in range(n+1)) # array
def A143411(n, k): return A(n-k, k) # anti-diagonal triangle
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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