A195035 Multiples of 15 and of 8 interleaved: a(2n-1) = 15n, a(2n) = 8n.

15, 8, 30, 16, 45, 24, 60, 32, 75, 40, 90, 48, 105, 56, 120, 64, 135, 72, 150, 80, 165, 88, 180, 96, 195, 104, 210, 112, 225, 120, 240, 128, 255, 136, 270, 144, 285, 152, 300, 160, 315, 168, 330, 176, 345, 184, 360, 192, 375, 200, 390, 208, 405, 216

Offset: 1

Author

Omar E. Pol, Sep 12 2011

Keywords

• nonn
• easy

Comments

First differences of A195036.

a(n) is also the length of the n-th edge of a square spiral in which the first two edges are the legs of the primitive Pythagorean triple [15, 8, 17]. Zero together with partial sums give A195036; the vertices of the spiral.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..10000
• Ron Knott, Pythagorean triangles and Triples
• Eric Weisstein's World of Mathematics, Pythagorean Triple
• Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Crossrefs

Cf. A004526, A010686, A010706.

Cf. A195019, A195031, A195033, A195036.

Programs

Magma

&cat[[15*n, 8*n]: n in [1..27]];  // Bruno Berselli, Sep 30 2011

Mathematica

Riffle[15*#, 8*#] & [Range[50]] (* Paolo Xausa, Mar 21 2024 *)

PARI

a(n)=(n+1)\2*if(n%2,15,8) \\ Charles R Greathouse IV, Oct 07 2015

Formula

From Bruno Berselli, Sep 30 2011: (Start)

G.f.: x*(15+8*x)/((1-x)^2*(1+x)^2).

a(n) = A010686(n)*A010706(n-1)*A004526(n+1) = (23*n-(7*n+15)*(-1)^n+15)/4.

a(n) = 2*a(n-2) - a(n-4).

a(-n) = -a(A014681(n-1)). (End)