OFFSET
1,2
COMMENTS
a(n+1) is the second element of the n-th set of three consecutive triangular numbers whose product is a perfect square. - Arkadiusz Wesolowski, Apr 27 2012
The triangular numbers of this sequence satisfy the Diophantine equation T(k) = k*(k+1)/2 = m^2 + 1, which is equivalent to (2k+1)^2 - 2*(2m)^2 = 9. Now, with x=2k+1 and y=2m, the Pell-Fermat equation x^2 - 2*y^2 = 9 appears. The solutions x and y of this equation are respectively in A106329 and A075848. The indices k=(x-1)/2 of the triangular numbers of this sequence are in A072221, while the indices m=y/2 of the corresponding square numbers are in A106328. - Bernard Schott, Mar 09 2019
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..500
K. B. Subramaniam, Almost Square Triangular Numbers, The Fibonacci Quarterly, Vol. 37, No. 3 (1999), pp. 194-197.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).
FORMULA
From R. J. Mathar, Sep 22 2009: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
G.f.: x*(1-25*x+10*x^2)/((1-x)*(x^2-34*x+1)). (End)
a(n) = (14 + 9*(17+12*sqrt(2))^(1-n) - 9*(-17+12*sqrt(2))*(17+12*sqrt(2))^n) / 32. - Colin Barker, Mar 23 2019
a(n) = 9 * A001110(n) + 1 (Subramaniam, 1999). - Amiram Eldar, Jan 13 2022
EXAMPLE
10 is in this sequence because it is a triangular number A000217(4) and is equal to a square plus 1: 10 = 3^2 + 1.
MATHEMATICA
LinearRecurrence[{35, -35, 1}, {1, 10, 325}, 50] (* G. C. Greubel, Sep 09 2017 *)
PROG
(Haskell)
a164055 n = a164055_list !! (n-1)
a164055_list = 1 : 10 : 325 : zipWith (+) a164055_list
(map (* 35) $ tail $ zipWith (-) (tail a164055_list) a164055_list)
-- Reinhard Zumkeller, Apr 29 2012
(PARI) my(x='x+O('x^50)); Vec(x*(1-25*x+10*x^2)/((1-x)*(x^2-34*x+1))) \\ G. C. Greubel, Sep 09 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Tanya Khovanova & Alexey Radul, Aug 08 2009
EXTENSIONS
Comment molded into formula by R. J. Mathar, Sep 22 2009
STATUS
approved