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A091491
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Triangle, read by rows, where the n-th diagonal is generated from the n-th row by the sum of the products of the n-th row terms with binomial coefficients.
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9
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1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 9, 8, 4, 1, 1, 23, 22, 13, 5, 1, 1, 65, 64, 41, 19, 6, 1, 1, 197, 196, 131, 67, 26, 7, 1, 1, 626, 625, 428, 232, 101, 34, 8, 1, 1, 2056, 2055, 1429, 804, 376, 144, 43, 9, 1, 1, 6918, 6917, 4861, 2806, 1377, 573, 197, 53, 10, 1
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OFFSET
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0,5
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COMMENTS
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Apart from the first column, any term is the partial sum of terms of the row above, when summing from the right. - Ralf Stephan, Apr 27 2004
Matrix inverse equals triangle A104402.
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LINKS
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FORMULA
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T(n, k) = Sum_{j=0..n-k} T(n-k, j)*C(k+j-1, k-1).
G.f.: 2/(2-y*(1-sqrt(1-4*x)))/(1-x).
T(n, k) = T(n-1, k-1) + T(n, k+1) for n>0, with T(n, 0)=1.
Recurrence: for k>0, T(n, k) = Sum_{j=k..n} T(n-1, j). - Ralf Stephan, Apr 27 2004
T(n,k) = k * Sum_{i=0..n-k} binomial(2*(n-i)-k-1, n-i-1)/(n-i) for k>0; T(n,0)=1. - Vladimir Kruchinin, Feb 07 2011
The n-th row of the triangle is the top row of M^n, where M is the following infinite square production matrix in which a column of (1,0,0,0,...) is prepended to an infinite lower triangular matrix of all 1's and the rest zeros:
1, 1, 0, 0, 0, 0, ...
0, 1, 1, 0, 0, 0, ...
0, 1, 1, 1, 0, 0, ...
0, 1, 1, 1, 1, 0, ...
0, 1, 1, 1, 1, 1, ...
(End)
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EXAMPLE
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T(7,3) = T(4,0)*C(2,2) + T(4,1)*C(3,2) + T(4,2)*C(5,2) + T(4,3)*C(6,2) = (1)*1 + (4)*3 + (3)*6 + (1)*10 = 41.
Rows begin:
1;
1, 1;
1, 2, 1;
1, 4, 3, 1;
1, 9, 8, 4, 1;
1, 23, 22, 13, 5, 1;
1, 65, 64, 41, 19, 6, 1;
1, 197, 196, 131, 67, 26, 7, 1;
1, 626, 625, 428, 232, 101, 34, 8, 1;
1, 2056, 2055, 1429, 804, 376, 144, 43, 9, 1;
1, 6918, 6917, 4861, 2806, 1377, 573, 197, 53, 10, 1;
1, 23714, 23713, 16795, 9878, 5017, 2211, 834, 261, 64, 11, 1;
1, 82500, 82499, 58785, 35072, 18277, 8399, 3382, 1171, 337, 76, 12, 1;
...
As to the production matrix M, top row of M^3 = [1, 4, 3, 1, 0, 0, 0, ...].
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MATHEMATICA
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nmax = 11; t[n_, k_] := k*(2n-k-1)!*HypergeometricPFQ[{1, k-n, -n}, {k/2-n+1/2, k/2-n+1}, 1/4]/(n!*(n-k)!); t[_, 0] = 1; Flatten[ Table[ t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Nov 14 2011, after Vladimir Kruchinin *)
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PROG
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(PARI) T(n, k)=if(k>n || n<0 || k<0, 0, if(k==0 || k==n, 1, sum(j=0, n-k, T(n-k, j)*binomial(k+j-1, k-1)); ); )
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
(PARI) T(n, k)=local(X=x+x*O(x^n), Y=y+y*O(y^k)); polcoeff(polcoeff(2/(2-Y*(1-sqrt(1-4*X)))/(1-X), n, x), k, y)
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
(PARI) T(n, k)=if(n<k || k<0, 0, if(n==k || k==0, 1, T(n-1, k-1)+T(n, k+1)))
for(n=0, 10, for(k=0, n, print1(T(n, k), ", ")); print(""))
(Haskell)
a091491 n k = a091491_tabl !! n !! k
a091491_row n = a091491_tabl !! n
a091491_tabl = iterate (\row -> 1 : scanr1 (+) row) [1]
(Magma)
A091491:= func< n, k | k eq 0 select 1 else k*(&+[Binomial(2*(n-j)-k-1, n-j-1)/(n-j): j in [0..n-k]]) >;
(Sage)
def A091491(n, k): return 1 if (k==0) else k*sum(binomial(2*(n-j)-k-1, n-j-1)/(n-j) for j in (0..n-k))
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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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