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A193605
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Triangle: (row n) = partial sums of partial sums of row n of Pascal's triangle.
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3
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1, 1, 3, 1, 4, 8, 1, 5, 12, 20, 1, 6, 17, 32, 48, 1, 7, 23, 49, 80, 112, 1, 8, 30, 72, 129, 192, 256, 1, 9, 38, 102, 201, 321, 448, 576, 1, 10, 47, 140, 303, 522, 769, 1024, 1280, 1, 11, 57, 187, 443, 825, 1291, 1793, 2304, 2816, 1, 12, 68, 244, 630, 1268, 2116, 3084, 4097, 5120, 6144
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OFFSET
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0,3
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COMMENTS
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The n-th row is contains the partial sums of the n-th row of the array interpretation of A052509. - R. J. Mathar, Apr 22 2013
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LINKS
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FORMULA
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Writing the general term as T(n,k), for 0<=k<=n:
G.f.: -(1-x*y)^2/(4*x^3*y^3+(4*x^3-8*x^2)*y^2+(5*x-4*x^2)*y+x-1). - Vladimir Kruchinin, Aug 19 2019
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EXAMPLE
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1
1....3
1....4....8
1....5....12....20
1....6....17....32....48
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MAPLE
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if k = 0 then
1;
else
procname(n, k-1)+binomial(n, k) ;
end if;
end proc:
if k = 0 then
1;
else
end if;
gf := ((x*y-1)/(1-2*x*y))^2/(1-x*y-x): ser := series(gf, x, 12):
p := n -> coeff(ser, x, n): row := n -> seq(coeff(p(n), y, k), k=0..n):
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MATHEMATICA
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u[n_, k_] := Sum[Binomial[n, h], {h, 0, k}]
p[n_, k_] := Sum[u[n, h], {h, 0, k}]
Table[p[n, k], {n, 0, 12}, {k, 0, n}]
Flatten[%] (* A193605 as a sequence *)
TableForm[Table[p[n, k], {n, 0, 12}, {k, 0, n}]] (* A193605 as a triangle *)
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PROG
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(Maxima)
T(n, k):=sum(((i+3)*2^(i-2))*binomial(n-i, k-i), i, 1, min(n, k))+binomial(n, k);
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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