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A202241
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Array F(n,m) read by antidiagonals: F(0,m)=1, F(n,0) = A130713(n), and column m+1 is recursively defined as the partial sums of column m.
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2
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1, 2, 1, 1, 3, 1, 0, 4, 4, 1, 0, 4, 8, 5, 1, 0, 4, 12, 13, 6, 1, 0, 4, 16, 25, 19, 7, 1, 0, 4, 20, 41, 44, 26, 8, 1, 0, 4, 24, 61, 85, 70, 34, 9, 1, 0, 4, 28, 85, 146, 155, 104, 43, 10, 1, 0, 4, 32, 113, 231, 301, 259, 147, 53, 11, 1, 0, 4, 36, 145, 344, 532, 560, 406, 200, 64, 12, 1
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OFFSET
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0,2
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COMMENTS
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The array F(n,m), beginning with row n=0, is:
1, 1, 1, 1, 1, 1, 1,
2, 3, 4, 5, 6, 7, 8,
1, 4, 8, 13, 19, 26, 34,
0, 4, 12, 25, 44, 70, 104,
0, 4, 16, 41, 85, 155, 259,
0, 4, 20, 61, 146, 301, 560,
0, 4, 24, 85, 231, 532, 1092.
Columns after A130713, A113311, A008574 have signatures (3,-3,1), (4,-6,4,-1), (5,-10,10,-5,1), (6,-15,20,-15,6,-1) (from A135278(n+3)).
Inserting columns of zeros and pushing the columns down, plus alternating sign switches defines the following triangle T(n,2m) = (-1)^(m/2)*F(n-2m,m):
1,
2 0,
1 0 -1,
0 0 -3 0,
0 0 -4 0 1,
0 0 -4 0 4 0,
0 0 -4 0 8 0 -1
The row sums in the triangle are (-1)^n*A099838(n).
1
1 0
1 0 0
1 0 -2 0
1 0 -4 0 1
1 0 -6 0 5 0
1 0 -8 0 13 0 -2
1 0 -10 0 25 0 -12 0
1 0 -12 0 41 0 -38 0 4
1 0 -14 0 61 0 -88 0 28 0
1 0 -16 0 85 0 -170 0 104 0 -8
5th column: A001844; 7th column: -A035597=-2*A005900(n+1); 9th column: 4*A006325(n+2); 11th column: -8*(1,8,34,104) (from columns 4,5,6,7 of F(n,m)).
As a triangular array, this is the Riordan array ((1+x)^2, x/(1-x)). - Philippe Deléham, Feb 21 2012
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LINKS
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FORMULA
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F(1,m) = m+2.
Sum_{m=0..d} F(d-m,m) = A116453(d-3), d >= 3 (antidiagonal sums).
As a triangular array T(n,k), 0 <= k <= n, satisfies: T(n,k) = T(n-1,k) + T(n-1,k-1) with T(0,0) = 1, T(1,0) = 2, T(2,0) = 1, T(3,0) = 0. - Philippe Deléham, Feb 21 2012
Unsigned diagonals of A267633 (beginning with its main diagonal) appear to be the reverse rows of this entry's triangle beginning with the fourth row. - Tom Copeland, Jan 26 2016
T(n,k) = C(n, n-k) + C(n-1, n-k-1) - C(n-2, n-k-2) - C(n-3, n-k-3), where C(n, k) = n!/(k!*(n-k)!) if 0 <= k <= n, otherwise 0. - Peter Bala, Mar 20 2018
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EXAMPLE
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Triangle T(n,k) begins:
1
2, 1
1, 3, 1
0, 4, 4, 1
0, 4, 8, 5, 1
0, 4, 12, 13, 6, 1
0, 4, 16, 25, 19, 7, 1
0, 4, 20, 41, 44, 26, 8, 1
0, 4, 24, 61, 85, 70, 34, 9, 1
0, 4, 28, 85, 146, 155, 104, 43, 10, 1
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MAPLE
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if n <= 2 and n >=0 then
op(n+1, [1, 2, 1]) ;
else
0;
end if;
end proc:
option remember;
if n < 0 then
0 ;
elif m = 0 then
else
procname(n, m-1)+procname(n-1, m) ;
end if;
end proc:
for d from 0 to 12 do
for m from 0 to d do
end do:
C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if end proc:
for n from 0 to 10 do
seq(C(n, n-k) + C(n-1, n-k-1) - C(n-2, n-k-2) - C(n-3, n-k-3), k = 0..n);
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MATHEMATICA
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rows = 12;
T[0] = PadRight[{1, 2, 1}, rows];
T[n_ /; n<rows] := Accumulate[T[n-1]];
A = Array[T, rows, 0] // Transpose;
F[n_ /; n<rows, m_ /; m<rows] := A[[n+1, m+1]];
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PROG
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(Sage)
def Trow(n): return [binomial(n, n-k) + binomial(n-1, n-k-1) - binomial(n-2, n-k-2) - binomial(n-3, n-k-3) for k in (0..n)]
(GAP) Flat(List([0..12], n->List([0..n], k->Binomial(n, n-k)+Binomial(n-1, n-k-1)-Binomial(n-2, n-k-2)-Binomial(n-3, n-k-3)))); # Muniru A Asiru, Mar 22 2018
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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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