A127690 a(1)=3; for n>1, a(n) is such that a(1)^2+...+a(n)^2 = (1+a(n))^2.

3, 4, 12, 84, 3612, 6526884, 21300113901612, 226847426110843688722000884, 25729877366557343481074291996721923093306518970391612, 331013294649039928396936390888878360035026305412754995683702777533071737279144813617823976263475290370884

Offset

1,1

Links

Table of n, a(n) for n=1..10.

Sierpinski W., Elementary theory of numbers, Monografie Matematyczne 42 (1964), Chapter II, p. 63.

Formula

For n>2, a(n) = (a(1)^2 + a(2)^2 + ... + a(n-1)^2 - 1)/2 = ((a(n-1) + 1)^2 - 1)/2. - Max Alekseyev, Nov 23 2012

a(n) = A053630(n-1)-1 for n>=2. - R. J. Mathar, Apr 23 2007

Example

a(2)=4 because (3^2+4^2=5^2) and (4+1=5), a(3)=12 because (3^2+4^2+12^2=13^2) and (12+1=13) a(5)= 3612 because (3^2+4^2+12^2+84^2+3612^2=3613^2) and (3612+1=3613) etc.

Mathematica

a = {3}; For[k = 1 + a[[Length[a]]], Length[a] < 5, While[ ! ((IntegerQ[Sqrt[(k)^2 + Sum[(a[[t]])^2, {t, 1, Length[a]}]]]) && (Sqrt[(k)^2 + Sum[(a[[t]])^2, {t, 1, Length[a]}]] == k + 1)), k++ ]; AppendTo[a, k]]; a
a = {3}; For[k = 1 + a[[Length[a]]], Length[a] < 12, s2 = Plus @@ (a^2); t = Reduce[{y^2 + s2 == (y + 1)^2}, y, Integers]; t = t /. {Equal -> Rule}; k = y /. t; AppendTo[a, k]]; a (* Daniel Huber *)

Crossrefs

Cf. A018930, A127689, A127691.

Apart from the initial term, the sequence is the same as A053631.

Sequence in context: A127689 A307077 A330068 * A092417 A071543 A242648

Adjacent sequences: A127687 A127688 A127689 * A127691 A127692 A127693

Keyword

nonn

Author

Artur Jasinski, Jan 23 2007, Jan 29 2007

Status

approved