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%I #18 Sep 03 2019 02:31:44
%S 0,3,8,25,54,153,322,899,1884,5247,10988,30589,64050,178293,373318,
%T 1039175,2175864,6056763,12681872,35301409,73915374,205751697,
%U 430810378,1199208779,2510946900,6989500983,14634871028,40737797125
%N Numbers k such that the sum of the k-th triangular number and (k+2)-nd triangular number is a triangular number.
%C T(a(n)) + T(a(n)+2) = A069017(n+1) where T(k) = k*(k+1)/2.
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,6,-6,-1,1).
%F Let b(n) = A001109(n). Then we have a pair of recursion formulas:
%F a(2n+2) = 2*a(2n+1) - a(2n) + 2*b(n+1);
%F a(2n+3) = 2*a(2n+2) - a(2n+1) + 2*b(n+2).
%F G.f.: x*(3 + 5*x - x^2 - x^3)/((1-x)*(1 - 6*x^2 + x^4)).
%F a(n) = -3 + (1/8)*(-1^n)((7 + 5*sqrt(2))*(-1 - sqrt(2))^n + (7 - 5*sqrt(2))*(-1 + sqrt(2))^n - (1 + sqrt(2))^n - (1 - sqrt(2))^n).
%Y Cf. A000217, A001109, A069017.
%K easy,nonn
%O 1,2
%A Bruce Corrigan (scentman(AT)myfamily.com), Oct 29 2002
%E Edited by _Jon E. Schoenfield_, Sep 02 2019
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