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%I #13 Dec 08 2013 12:42:44
%S 3,4,12,84,3612,6526884,21300113901612,226847426110843688722000884,
%T 25729877366557343481074291996721923093306518970391612,
%U 331013294649039928396936390888878360035026305412754995683702777533071737279144813617823976263475290370884
%N a(1)=3; for n>1, a(n) is such that a(1)^2+...+a(n)^2 = (1+a(n))^2.
%H Sierpinski W., <a href="http://pldml.icm.edu.pl/pldml/details/contents.action?id=bwmeta1.element.dl-catalog-556369c7-b6cc-4a5b-be36-bfc8e0ca7cfa">Elementary theory of numbers</a>, Monografie Matematyczne 42 (1964), Chapter II, p. 63.
%F 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
%F a(n) = A053630(n-1)-1 for n>=2. - _R. J. Mathar_, Apr 23 2007
%e 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.
%t 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
%t 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 *)
%Y Cf. A018930, A127689, A127691.
%Y Apart from the initial term, the sequence is the same as A053631.
%K nonn
%O 1,1
%A _Artur Jasinski_, Jan 23 2007, Jan 29 2007
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