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%I #14 Sep 08 2022 08:45:30
%S 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,
%T 114,127,133,147,154,169,176,192,200,217,225,243,252,271,280,300,310,
%U 331,341,363,374,397,408,432,444,469,481,507,520,547,560,588,602,631
%N 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.
%C Interleaving of A077043 and A006578.
%C First differences are in A124072.
%C 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, ... .
%C Row sums of triangle A168316 = (1, 1, 3, 4, 7, 8, 12, ...). - _Gary W. Adamson_, Nov 22 2009
%F 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.
%F G.f.: x^2*(1+x^2+x^3)/((1-x)^3*(1+x)^2*(1+x^2)).
%F 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
%e First seven rows of T are
%e [ 0 ]
%e [ 0, 1 ]
%e [ 0, 1, 2 ]
%e [ 0, 1, 3, 2 ]
%e [ 0, 1, 4, 2, 3 ]
%e [ 0, 1, 5, 2, 4, 3 ]
%e [ 0, 1, 6, 2, 5, 3, 4 ].
%o (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
%o (PARI) {vector(59, n, (n-2+n%2)*(n+n%2)/8+floor((n-2-n%2)^2/16))} // _Klaus Brockhaus_, Jul 16 2007
%Y Cf. A077043, A006578, A124072.
%Y Cf. A168316. - _Gary W. Adamson_, Nov 22 2009
%K nonn
%O 0,5
%A _Paul Curtz_, May 20 2007
%E Edited and extended by _Klaus Brockhaus_, Jul 16 2007
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