OFFSET
1,1
COMMENTS
It is well known that T(n)+T(n+1) is always a square. T(n)+T(n+1)+T(n+2) is a square for n in A165517. T(n)+T(n+1)+T(n+2)+T(n+3) is a square for n in A202391. There is no sequence of 5, 6, 7, 8, 9 or 10 consecutive T(i)'s which sum to a square, cf. A176541. The next possible length is 11, see A116476. Then comes this sequence, corresponding to length 13.
Positive integers y in the solutions to 2*x^2-13*y^2-169*y-728 = 0. - Colin Barker, May 04 2015
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,102,-102,0,0,-1,1).
FORMULA
G.f.: x*(3*x^8+7*x^7+6*x^6+26*x^5-260*x^4-357*x^3-46*x^2-26*x-3) / ((x-1)*(x^4-10*x^2-1)*(x^4+10*x^2-1)). - Colin Barker, May 04 2015
MATHEMATICA
Select[Range[10^5], IntegerQ[Sqrt[(#^2+13*#+56)*13/2]]&] (* Ivan N. Ianakiev, May 04 2015 *)
LinearRecurrence[{1, 0, 0, 102, -102, 0, 0, -1, 1}, {3, 29, 75, 432, 998, 3624, 8310, 44717, 102443}, 50] (* Vincenzo Librandi, May 05 2015 *)
PROG
(PARI) for(n=0, 10^8, issquare(binomial(n+14, 3)-binomial(n+1, 3))&&print1(n", "))
(PARI) Vec(x*(3*x^8+7*x^7+6*x^6+26*x^5-260*x^4-357*x^3-46*x^2-26*x-3) / ((x-1)*(x^4-10*x^2-1)*(x^4+10*x^2-1)) + O(x^100)) \\ Colin Barker, May 04 2015
(Magma) I:=[3, 29, 75, 432, 998, 3624, 8310, 44717, 102443]; [n le 9 select I[n] else Self(n-1)+102*Self(n-4)-102*Self(n-5)-Self(n-8)+Self(n-9): n in [1..40]]; // Vincenzo Librandi, May 05 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
M. F. Hasler, May 04 2015
STATUS
approved