login
Sum at 45 degrees to horizontal in triangle of A081498.
2

%I #23 Jan 17 2022 14:32:19

%S 1,2,4,6,8,11,12,16,15,20,16,22,14,21,8,16,-3,6,-20,-10,-44,-33,-76,

%T -64,-117,-104,-168,-154,-230,-215,-304,-288,-391,-374,-492,-474,-608,

%U -589,-740,-720,-889,-868,-1056,-1034,-1242,-1219,-1448,-1424,-1675,-1650,-1924,-1898,-2196,-2169,-2492,-2464,-2813

%N Sum at 45 degrees to horizontal in triangle of A081498.

%C The leading diagonal is given by A080956(n) = ((n+1)(2-n)/2).

%H Colin Barker, <a href="/A081499/b081499.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3,-3,3,1,-1).

%F a(n) = (n+floor(n/2)+1)*(n-floor(n/2))/2-binomial(ceiling(n/2)+1, ceiling(n/2)-2). - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004

%F G.f.: x*(1 + x - x^2 - x^3 - x^4) / ((1 - x)^4*(1 + x)^3). - _Colin Barker_, Dec 18 2012

%F From _Colin Barker_, Nov 12 2017: (Start)

%F a(n) = (1/96)*(-2*n^3 + 36*n^2 + 32*n) for n even.

%F a(n) = (1/96)*(-2*n^3 + 30*n^2 + 50*n + 18) for n odd.

%F a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7) for n>7.

%F (End)

%e a(7) = 7+5+2+(-2) = 12.

%p seq((n+floor(n/2)+1)*(n-floor(n/2))/2-binomial(ceil(n/2)+1,ceil(n/2)-2),n=1..60); # C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004

%t LinearRecurrence[{1,3,-3,-3,3,1,-1},{1,2,4,6,8,11,12},60] (* _Harvey P. Dale_, Jan 17 2022 *)

%o (PARI) Vec(x*(1 + x - x^2 - x^3 - x^4) / ((1 - x)^4*(1 + x)^3) + O(x^60)) \\ _Colin Barker_, Nov 12 2017

%Y Cf. A080956, A081498.

%K sign,easy

%O 1,2

%A _Amarnath Murthy_, Mar 25 2003

%E More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004