OFFSET
1,2
COMMENTS
From T(k)+T(k+1) = (k*(k+1)+(k+1)*(k+2))/2 = (k+1)^2 any two consecutive triangular numbers sum to a square, the above sequence gives the sums that are also triangular. The units digit cycles through 0, 5, 4, 3, 8, 9, 0, 5, ...
Let P(b,e) be the polynomial 1+4*b+4*b^2+4*e+4*e^2. It appears that sequences A076708 and A076049 are special cases of the sequence of integers b such that P(b,b+n) is a perfect square. A076708 and A076049 for example are respectively the sequences of b's such that P(b,b+1) and P(b,b+2) are perfect squares. In fact it appears to be true that the sequence of integers b such that P(b,b+n) is a perfect square has the property that t(b)+t(b+n) is a triangular number. I have not had time to prove this but I do have evidence produced by Mathematica to support the assertion. - Robert Phillips (bobanne(AT)bellsouth.net), Sep 04 2009; corrected Sep 08 2009
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
B. Polster, M. Ross, Marching in squares, arXiv preprint arXiv:1503.04658, 2015
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).
FORMULA
Recursion: a(n+2) = 6*a(n+1)-a(n)+4, with a(0)=0 and a(1)=5.
G.f.: (5*x^2-x^3)/((1-x)*(1-6*x+x^2)).
Closed form: a(n)= ( sqrt(2)*( (3+2*sqrt(2))^(n+1) - (3-2*sqrt(2))^(n+1) )-8 )/8.
Also, if the entries in A001109 are denoted by b(n) then a(n) = b(n+1)-1.
a(n) = sqrt(A001110(n)) - 1. - Ivan N. Ianakiev, May 03 2014
EXAMPLE
a(1) = (sqrt(2)*((3+2*sqrt(2))^2-(3-2*sqrt(2))^2)-8)/8 = (sqrt(2)*(9+12*sqrt(2)+8-9+12*sqrt(2)-8)-8)/8 = (sqrt(2)*24*sqrt(2)-8)/8 = (48-8)/8 = 40/8 = 5.
T(5) + T(6) = 15 + 21 = 36 = T(8).
MATHEMATICA
Table[((3 + 2 Sqrt[2])^n - (3 - 2 Sqrt[2])^n)/(4 Sqrt[2]) - 1, {n, 1, 20}] (* Zerinvary Lajos, Jul 14 2009 *)
PROG
(PARI) concat(0, Vec(x^2*(x-5)/((x-1)*(x^2-6*x+1)) + O(x^100))) \\ Colin Barker, May 15 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
STATUS
approved