A176542 Numbers n such that there are only a finite nonzero number of sets of n consecutive triangular numbers that sum to a square.

32, 50, 98, 128, 200, 242, 338, 392, 512, 578, 722, 800, 968, 1058, 1250, 1352, 1568, 1682, 1922, 2048, 2312, 2450, 2738, 2888, 3200, 3362, 3698, 3872, 4232, 4418, 4802, 5000, 5408, 5618, 6050, 6272, 6728, 6962, 7442, 7688, 8192, 8450, 8978, 9248, 9800

Offset: 1

Author

Andrew Weimholt, Apr 20 2010

Keywords

• nonn

Comments

Members of A176541, for which there are only a finite number of solutions.

Integer n is in this sequence if n=2*m^2 and the equation (2*x-m*y)*(2*x+m*y)=A077415(n)/2 has integer solutions with y>=n. - Max Alekseyev, May 10 2010

It seems that a(n) = 2*A001651(n+2)^2. - Colin Barker, Sep 25 2015

Examples

			32 is in this sequence because there is only one set of 32 consecutive triangular numbers that sum to a square (namely, A000217(26) thru A000217(57), which sum to 29584 = 172^2).
3 is NOT in this sequence, because there are infinitely many sets of 3 consecutive triangular numbers that sum to a square (cf. A165517).
4 is NOT in this sequence, because there are infinitely many sets of 4 consecutive triangular numbers that sum to a square (cf. A202391).
5 is NOT in this sequence, because there are NO sets of 5 consecutive triangular numbers that sum to a square.
11 is NOT in this sequence, since there are infinitely many sets of 11 consecutive triangular numbers that sum to a square (cf. A116476).
		

Crossrefs

Cf. A176541, A000217, A000292, A077415.

Formula

Conjectures from Colin Barker, Sep 24 2015: (Start)

a(n) = (9*n^2+24*n+16)/2 for n even.

a(n) = (9*n^2+30*n+25)/2 for n odd.

a(n) = a(n-1)+2*a(n-2)-2*a(n-3)-a(n-4)+a(n-5) for n>5.

G.f.: -2*x*(4*x^4-3*x^3-8*x^2+9*x+16) / ((x-1)^3*(x+1)^2).

(End)

Extensions

Terms a(6) onward from Max Alekseyev, May 10 2010