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A129819
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Antidiagonal sums of triangular array T: T(j,k) = (k+1)/2 for odd k, T(j,k) = 0 for k = 0, T(j,k) = j+1-k/2 for even k > 0; 0 <= k <= j.
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7
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0, 0, 1, 1, 3, 4, 7, 8, 12, 14, 19, 21, 27, 30, 37, 40, 48, 52, 61, 65, 75, 80, 91, 96, 108, 114, 127, 133, 147, 154, 169, 176, 192, 200, 217, 225, 243, 252, 271, 280, 300, 310, 331, 341, 363, 374, 397, 408, 432, 444, 469, 481, 507, 520, 547, 560, 588, 602, 631
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OFFSET
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0,5
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COMMENTS
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If the values of the second, fourth, sixth, ... column are replaced by the corresponding negative values, the antidiagonal sums of the resulting triangular array are 0, 0, 1, 1, -1, -2, -1, -2, -6, -8, -7, -9, ... .
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LINKS
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FORMULA
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a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-4) - a(n-5) - a(n-6) + a(n-7) for n > 6, with a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 1, a(4) = 3, a(5) = 4, a(6) = 7.
G.f.: x^2*(1+x^2+x^3)/((1-x)^3*(1+x)^2*(1+x^2)).
a(n) = (3/16)*(n+2)*(n+1) - (5/8)*(n+1) + 7/32 + (3/32)*(-1)^n + (1/16)*(n+1)*(-1)^n - (1/8)*cos(n*Pi/2) + (1/8)*sin(n*Pi/2). - Richard Choulet, Nov 27 2008
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EXAMPLE
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First seven rows of T are
[ 0 ]
[ 0, 1 ]
[ 0, 1, 2 ]
[ 0, 1, 3, 2 ]
[ 0, 1, 4, 2, 3 ]
[ 0, 1, 5, 2, 4, 3 ]
[ 0, 1, 6, 2, 5, 3, 4 ].
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PROG
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(Magma) m:=59; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do for k:=2 to j do if k mod 2 eq 0 then M[j, k]:= k div 2; else M[j, k]:=j-(k div 2); end if; end for; end for; [ &+[ M[j-k+1, k]: k in [1..(j+1) div 2] ]: j in [1..m] ]; // Klaus Brockhaus, Jul 16 2007
(PARI) {vector(59, n, (n-2+n%2)*(n+n%2)/8+floor((n-2-n%2)^2/16))} // Klaus Brockhaus, Jul 16 2007
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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