A169686 a(n) = sqrt(T(k-1)*T(k)*T(k+1)) as k runs through the terms of A072221 and T(i)=i*(i+1)/2.

0, 30, 5850, 1157730, 229221540, 45384688830, 8985939059790, 1779170548525890, 352266782665431240, 69747043797185672190, 13809562405059974172930, 2734223609158076980818690, 541362465050894178032921580, 107187033856467889149087366750, 21222491341115591157198758976630

Offset: 1

Author

N. J. A. Sloane, Apr 13 2010, based on an email from Neven Juric, Mar 19 2010

Keywords

• nonn

Comments

It is known (see Beiler, p. 198) that the product of three consecutive triangular numbers, T(k-1)T(k)T(k+1), is a square if (and only if?) 2k+1 = 3a for a in A001541. The corresponding values of k are in A072221.

Examples

			sqrt (T(3)T(4)T(5)) = 30
sqrt (T(24)T(25)T(26)) = 5850
sqrt (T(147)T(148)T(149)) = 1157730
sqrt (T(864)T(865)T(866)) = 229221540
		

References

• A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 1966.

Formula

Empirical g.f.: 30*x^2*(x^2-9*x+1) / ((x^2-198*x+1)*(x^2-6*x+1)). - Colin Barker, Jul 26 2013