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Square root of second term of a triple of squares in arithmetic progression.
5

%I #32 Oct 20 2021 09:21:02

%S 5,10,13,15,17,20,25,26,25,29,30,34,37,35,41,39,40,50,45,52,51,50,61,

%T 53,55,65,58,60,65,65,65,68,75,74,85,70,82,78,73,75,80,85,85,85,89,91,

%U 101,87,100,90,113,95,104,97,102,100,111,122,106,105,123,109

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

%C 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]

%C 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

%H Ray Chandler, <a href="/A198389/b198389.txt">Table of n, a(n) for n = 1..10000</a>

%H Reinhard Zumkeller, <a href="/A198384/a198384_2.txt">Table of initial values</a>

%H Keith Conrad, <a href="http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/3squarearithprog.pdf">Arithmetic progressions of three squares</a>

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

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

%e 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

%t wmax = 1000;

%t 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]];

%t Flatten[DeleteCases[triples /@ Range[wmax], {}], 2][[All, 2]] (* _Jean-François Alcover_, Oct 20 2021 *)

%o (Haskell)

%o a198389 n = a198389_list !! (n-1)

%o a198389_list = map (\(_,x,_) -> x) ts where

%o ts = [(u,v,w) | w <- [1..], v <- [1..w-1], u <- [1..v-1],

%o w^2 - v^2 == v^2 - u^2]

%Y Cf. A198385, A198409, A198440.

%K nonn

%O 1,1

%A _Reinhard Zumkeller_, Oct 24 2011