A124100 Sum_(x^i*y^j*z^k) with i + j + k = m and (x, y, z) = the primitive Pythagorean triple (8, 15, 17).

1, 40, 1089, 25160, 531521, 10625640, 204744769, 3844391560, 70827391041, 1286290883240, 23101397290049, 411249127989960, 7269184506192961, 127745926316548840, 2234231991096868929, 38920247688751940360

Offset: 0

Author

Giorgio Balzarotti and Paolo P. Lava, Nov 26 2006

Keywords

• nonn

Examples

			a(2) = 1089 because x^2 + y^2 + z^2 + x*y + x*z + y*z = 8^2 + 15^2 + 17^2 + 8*15 + 8*17 + 15*17 = 1089 and x^2 + y^2 = z^2.
		

References

• G. Balzarotti and P. P. Lava, Le sequenze di numeri interi, Hoepli, 2008, p. 196.

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..811
• Index entries for linear recurrences with constant coefficients, signature (40, -511, 2040).

Crossrefs

Cf. A019682, A020000, A020340-A020342, A020344-A020346, A021664, A021684, A021844, A025942, A077515.

Programs

Maple

seq(sum(8^(m-n)*sum(15^p*17^(n-p),p=0..n),n=0..m),m=0..N);

Mathematica

LinearRecurrence[{40,-511,2040},{1,40,1089},30] (* Harvey P. Dale, May 25 2025 *)

Formula

a(m) = (x^(m+2)*(z-y) + y^(m+2)*(x-z) + z^(m+2)*(y-x))/((x-y)*(y-z)*(z-x)).

From Chai Wah Wu, Sep 24 2016: (Start)

a(n) = 40*a(n-1) - 511*a(n-2) + 2040*a(n-3) for n > 2.

G.f.: 1/((1 - 8*x)*(1 - 15*x)*(1 - 17*x)). (End)

a(n) = 2^(3*n+6)/63 - 15^(n+2)/14 + 17^(n+2)/18. - Vaclav Kotesovec, Sep 25 2016