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Row 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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%I #17 Sep 08 2022 08:45:31

%S 0,-1,1,0,4,3,9,8,16,15,25,24,36,35,49,48,64,63,81,80,100,99,121,120,

%T 144,143,169,168,196,195,225,224,256,255,289,288,324,323,361,360,400,

%U 399,441,440,484,483,529,528,576,575,625,624,676,675,729,728,784,783,841

%N Row 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 A000290 and A067998 (starting at second term).

%C First differences are -1, 2, -1, 4, -1, 6, -1, 8, -1, 10, ...: a(n+1) - a(n) = (-1)^(n+1)*A124625(n+2).

%C Main diagonal of T is in A001057, antidiagonal sums are in A131804.

%H Bruno Berselli, <a href="/A131805/b131805.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).

%F a(0) = 0; a(n) = a(n-1) - (n mod 2) + n*(1 - (n mod 2)) for n > 0.

%F G.f.: x*(-1+2*x+x^2)/((1-x)^3*(1+x)^2).

%F a(n) = -A131118(2n) = (2n(n-1)+(2n+3)(-1)^n-3)/8. - _Bruno Berselli_, Mar 27 2012

%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]: k in [1..j] ]: j in [1..m] ];

%o (Magma) m:=29; &cat[ [ n^2, n^2-1 ]: n in [0..m] ];

%o (PARI) {m=58; for(n=0, m, r=n%2; print1(((n-r)/2)^2-r, ","))}

%o (Maxima) makelist((2*n*(n-1)+(2*n+3)*(-1)^n-3)/8,n,0,58); /* _Bruno Berselli_, Mar 27 2012 */

%Y Cf. A000290 (n^2), A067998 (n^2-2*n), A124625, A001057, A131804.

%Y Cf. A131118.

%K sign,easy

%O 0,5

%A _Klaus Brockhaus_, Jul 18 2007