A127689 a(1)=3; for n>1, a(n) is least number such that a(n) > a(n-1) and a(1)^2+...+a(n)^2 is a square.

3, 4, 12, 84, 132, 12324, 15960, 26280, 27300, 66660, 115188, 9777193284, 23465263884, 48701491080, 40900397690640, 680008604512020, 127049882801497788, 247290967245178188, 335580091290976716, 1045885075937364972, 1607091702050097396, 3419793695168900508, 5138020847719969956, 10059508412964112740

Offset

1,1

Comments

Without the a(n) > a(n-1) constraint, the sequence would be A018930.

Links

Table of n, a(n) for n=1..24.

Example

a(2)=4 because 3^2+4^2=5^2, a(3)=12 because 3^2+4^2+12^2=13^2 etc.

Mathematica

a = {3}; For[k = 1 + a[[Length[a]]], Length[a] < 11, While[ ! IntegerQ[Sqrt[(k)^2 + Sum[(a[[t]])^2, {t, 1, Length[a]}]]], k++ ]; AppendTo[a, k]]; a

Prog

(PARI) q=3; s=9; for(n=1, 30, b=0; fordiv(s, d, if(d*d>=s, break); t=(s\d-d)/2; if(t>q, b=t); ); q=b; print1(q, ", "); s+=q^2); \\ Max Alekseyev, Nov 23 2012

Crossrefs

Cf. A018930, A127690, A127691.

Sequence in context: A122903 A059792 A018930 * A307077 A330068 A127690

Adjacent sequences: A127686 A127687 A127688 * A127690 A127691 A127692

Keyword

nonn

Author

Artur Jasinski, Jan 23 2007

Extensions

More terms from Max Alekseyev, Nov 23 2012

Status

approved