A319588 First of three consecutive triangular numbers that add up to a perfect square.

0, 15, 105, 2016, 11628, 203203, 1152921, 19968040, 113108320, 1957220895, 11084786065, 191793185496, 1086209028828, 18793829460003, 106437529743441, 1841604033412080, 10429792989769440, 180458406785594575, 1022013288177368025, 17683082313822046576

Offset: 1

Author

Harvey P. Dale, Sep 23 2018

Keywords

• nonn

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,108,-108,-982,982,108,-108,-1,1).

Crossrefs

Cf. A000217, A165517.

Programs

Mathematica

Select[Partition[Accumulate[Range[2500000]],3,1],IntegerQ[Sqrt[ Total[ #]]]&] [[All,1]]
(#(#+1))/2&/@LinearRecurrence[{1,10,-10,-1,1},{0,5,14,63,152},20]
LinearRecurrence[{1, 108, -108, -982, 982, 108, -108, -1, 1},{0, 15, 105, 2016, 11628, 203203, 1152921, 19968040, 113108320},20] (* Ray Chandler, Mar 01 2024 *)

PARI

concat(0, Vec(x^2*(15 + 90*x + 291*x^2 - 108*x^3 - 83*x^4 + 2*x^5 + x^6) / ((1 - x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)*(1 - 10*x^2 + x^4)) + O(x^40))) \\ Colin Barker, Sep 24 2018

Formula

a(n) = A000217(A165517(n)). - Alois P. Heinz, Sep 24 2018

From Colin Barker, Sep 24 2018: (Start) Heinz's formula implies a g.f. and a recurrence:

G.f.: x^2*(15 + 90*x + 291*x^2 - 108*x^3 - 83*x^4 + 2*x^5 + x^6) / ((1 - x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)*(1 - 10*x^2 + x^4)).

a(n) = a(n-1) + 108*a(n-2) - 108*a(n-3) - 982*a(n-4) + 982*a(n-5) + 108*a(n-6) - 108*a(n-7) - a(n-8) + a(n-9) for n>9.

(End)