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A193845
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Mirror of the triangle A193844.
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4
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1, 3, 1, 7, 5, 1, 15, 17, 7, 1, 31, 49, 31, 9, 1, 63, 129, 111, 49, 11, 1, 127, 321, 351, 209, 71, 13, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1
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OFFSET
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0,2
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COMMENTS
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This triangle is obtained by reversing the rows of the triangle A193844.
Subtriangle of the triangle in A112857.
Riordan array (1/((1-2*x)*(1-x), x/(1-2*x)). (End)
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-2)^0 + A_1*(x-2)^1 + A_2*(x-2)^2 + ... + A_n*(x-2)^n. This sequence gives A_0, ... A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014
The n-th row lists the coefficients of the polynomial sum_{k=0..n} (X+2)^k, in order of increasing powers. - M. F. Hasler, Oct 15 2014
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LINKS
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FORMULA
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T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - T(n-2,k-1), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 17 2014
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EXAMPLE
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First six rows:
1
3....1
7....5....1
15...17...7....1
31...49...31...9...1
63...129..111..49..11..1
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MATHEMATICA
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z = 10;
p[n_, x_] := (x + 1)^n;
q[n_, x_] := (x + 1)^n
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]] (* A193844 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]] (* A193845 *)
Table[2^k*Binomial[n + 1, k]*Hypergeometric2F1[1, -k, -k + n + 2, 1/2], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Nov 09 2021 *)
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PROG
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(PARI) for(n=0, 20, for(k=0, n, print1(1/k!*sum(i=0, n, (2^(i-k)*prod(j=0, k-1, i-j))), ", "))) \\ Derek Orr, Oct 14 2014
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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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