A198389 Square root of second term of a triple of squares in arithmetic progression.

5, 10, 13, 15, 17, 20, 25, 26, 25, 29, 30, 34, 37, 35, 41, 39, 40, 50, 45, 52, 51, 50, 61, 53, 55, 65, 58, 60, 65, 65, 65, 68, 75, 74, 85, 70, 82, 78, 73, 75, 80, 85, 85, 85, 89, 91, 101, 87, 100, 90, 113, 95, 104, 97, 102, 100, 111, 122, 106, 105, 123, 109

Offset: 1

Author

Reinhard Zumkeller, Oct 24 2011

Keywords

• nonn

Comments

Apart from its initial 1, A001653 is a subsequence: for all n>1 exists an m such that A198388(m)=1 and a(m)=A001653(n). [observed by Zak Seidov, Reinhard Zumkeller, Oct 25 2011]

There is a connection to hypotenuses of Pythagorean triangles. See a comment for the primitive case on A198441 which applies here mutatis mutandis. - Wolfdieter Lang, May 23 2013

Examples

			Connection to Pythagorean triangle hypotenuses: a(20) = 10 because (in the notation of the Zumkeller link) (u,v,w) = 2*(1,5,7) and the Pythagorean triangle is 2*(x=(7-1)/2,y=(1+7)/2,5) = 2*(3,4,5) with hypotenuse 2*5 = 10. - _Wolfdieter Lang_, May 23 2013
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..10000
• Reinhard Zumkeller, Table of initial values
• Keith Conrad, Arithmetic progressions of three squares

Crossrefs

Cf. A198385, A198409, A198440.

Programs

Haskell

a198389 n = a198389_list !! (n-1)
a198389_list = map (\(,x,) -> x) ts where
   ts = [(u,v,w) | w <- [1..], v <- [1..w-1], u <- [1..v-1],
                   w^2 - v^2 == v^2 - u^2]

Mathematica

wmax = 1000;
triples[w_] := Reap[Module[{u, v}, For[u = 1, u < w, u++, If[IntegerQ[v = Sqrt[(u^2 + w^2)/2]], Sow[{u, v, w}]]]]][[2]];
Flatten[DeleteCases[triples /@ Range[wmax], {}], 2][[All, 2]] (* Jean-François Alcover, Oct 20 2021 *)

Formula

A198385(n) = a(n)^2.

A198440(n) = a(A198409(n)).