A214493 Numbers of the form ((6k+5)^2+9)/2 or 2(3k+4)^2-9.

17, 23, 65, 89, 149, 191, 269, 329, 425, 503, 617, 713, 845, 959, 1109, 1241, 1409, 1559, 1745, 1913, 2117, 2303, 2525, 2729, 2969, 3191, 3449, 3689, 3965, 4223, 4517, 4793, 5105, 5399, 5729, 6041, 6389, 6719, 7085, 7433, 7817, 8183, 8585, 8969, 9389, 9791, 10229, 10649, 11105, 11543, 12017, 12473, 12965, 13439, 13949

Offset: 0

Author

Yasir Karamelghani Gasmallah, Jul 19 2012

Keywords

• nonn
• easy

Comments

For every n=2k the triple (a(2k-1)^2, a(2k)^2 , a(2k+1)^2) is an arithmetic progression, i.e., 2*a(2k)^2 = a(2k-1)^2 + a(2k+1)^2.

In general a triple((x-y)^2,z^2,(x+y)^2) is an arithmetic progression if and only if x^2+y^2=z^2.

The first differences of this sequence is the interleaved sequence 6,42,24,60,42,78.... = 9*n*(39-27*(-1)^n)/2.

Examples

			For n = 7, a(7)=2*a(6)-2*a(4)+a(3)=2*269-2*149+89=329.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

Cf. A178218, A214345.

Programs

Magma

I:=[17, 23, 65, 89]; [n le 4 select I[n] else 2*Self(n-1)-2*Self(n-3)+Self(n-4): n in [1..75]];

Mathematica

LinearRecurrence[{2,0,-2,1},{17,23,65,89},60] (* Harvey P. Dale, Aug 07 2015 *)

Formula

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

G.f.: (17-11*x+19*x^2-7*x^3)/((1+x)*(1-x)^3).

a(n) = (6*n*(3*n+10)+27*(-1)^n+41)/4.

2*a(2n)^2 = a(2n-1)^2 + a(2n+1)^2.