OFFSET
1,1
COMMENTS
For the index of the first of the corresponding four consecutive triangular numbers, see A202391.
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).
FORMULA
a(n) = 7*a(n-1)-7*a(n-2)+a(n-3) for n>3.
G.f.: -x*(x-3)^2 / ((x-1)*(x^2-6*x+1)).
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
EXAMPLE
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.
PROG
(PARI) Vec(-x*(x-3)^2/((x-1)*(x^2-6*x+1)) + O(x^30))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Sep 24 2015
STATUS
approved