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%I #8 Mar 05 2016 11:13:50
%S 9,57,337,1969,11481,66921,390049,2273377,13250217,77227929,450117361,
%T 2623476241,15290740089,89120964297,519435045697,3027489309889,
%U 17645500813641,102845515571961,599427592618129,3493720040136817,20362892648202777,118683635849079849
%N The index of the first of two consecutive positive triangular numbers (A000217) the sum of which is equal to the sum of four consecutive positive triangular numbers.
%C For the index of the first of the corresponding four consecutive triangular numbers, see A202391.
%H Colin Barker, <a href="/A262490/b262490.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).
%F a(n) = 7*a(n-1)-7*a(n-2)+a(n-3) for n>3.
%F G.f.: -x*(x-3)^2 / ((x-1)*(x^2-6*x+1)).
%F a(n) = -1+(1-1/sqrt(2))*(3-2*sqrt(2))^n+(1+1/sqrt(2))*(3+2*sqrt(2))^n. - _Colin Barker_, Mar 05 2016
%e 9 is in the sequence because T(9)+T(10) = 45+55 = 100 = 15+21+28+36 = T(5)+T(6)+T(7)+T(8), where T(k) is the k-th triangular number.
%o (PARI) Vec(-x*(x-3)^2/((x-1)*(x^2-6*x+1)) + O(x^30))
%Y Cf. A000217, A202391, A262489, A262491, A262492.
%K nonn,easy
%O 1,1
%A _Colin Barker_, Sep 24 2015