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A349841
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Triangle T(n,k) built by placing all ones on the left edge, [1,0,0,0,0] repeated on the right edge, and filling the body using the Pascal recurrence T(n,k) = T(n-1,k) + T(n-1,k-1).
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4
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1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 4, 1, 1, 1, 5, 10, 10, 5, 2, 0, 1, 6, 15, 20, 15, 7, 2, 0, 1, 7, 21, 35, 35, 22, 9, 2, 0, 1, 8, 28, 56, 70, 57, 31, 11, 2, 0, 1, 9, 36, 84, 126, 127, 88, 42, 13, 2, 1
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OFFSET
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0,8
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COMMENTS
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This is the m=5 member in the sequence of triangles A007318, A059259, A118923, A349839, A349841 which have all ones on the left side, ones separated by m-1 zeros on the other side, and whose interiors obey Pascal's recurrence.
T(n,k) is the (n,n-k)-th entry of the (1/(1-x^5),x/(1-x)) Riordan array.
Sums of antidiagonals give A349843.
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LINKS
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FORMULA
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G.f.: (1-x*y)/((1-(x*y)^5)(1-x-x*y)) in the sense that T(n,k) is the coefficient of x^n*y^k in the series expansion of the g.f.
T(n,0) = 1.
T(n,n) = delta(n mod 5,0).
T(n,1) = n-1 for n>0.
T(n,2) = (n-1)*(n-2)/2 for n>1.
T(n,3) = (n-1)*(n-2)*(n-3)/6 for n>2.
T(n,4) = (n-1)*(n-2)*(n-3)*(n-4)/24 for n>3.
T(n,5) = C(n-1,5) + 1 for n>4.
T(n,6) = C(n-1,6) + n - 6 for n>5.
For 0 <= k < n, T(n,k) = (n-k)*Sum_{j=0..floor(k/5)} binomial(n-5*j,n-k)/(n-5*j).
The g.f. of the n-th subdiagonal is 1/((1-x^5)(1-x)^n).
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EXAMPLE
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Triangle begins:
1;
1, 0;
1, 1, 0;
1, 2, 1, 0;
1, 3, 3, 1, 0;
1, 4, 6, 4, 1, 1;
1, 5, 10, 10, 5, 2, 0;
1, 6, 15, 20, 15, 7, 2, 0;
1, 7, 21, 35, 35, 22, 9, 2, 0;
1, 8, 28, 56, 70, 57, 31, 11, 2, 0;
1, 9, 36, 84, 126, 127, 88, 42, 13, 2, 1;
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MATHEMATICA
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Flatten[Table[CoefficientList[Series[(1 - x*y)/((1 - (x*y)^5)(1 - x - x*y)), {x, 0, 20}, {y, 0, 10}], {x, y}][[n+1, k+1]], {n, 0, 10}, {k, 0, n}]]
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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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