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%I #4 Mar 30 2012 17:35:16
%S 1,25,1105,12025,21025,66625,252601,292825,751825,1026745,1671865,
%T 1892185,4210945,4534945,8529625,8958625,10251025,16040401,24019801,
%U 28404025,29138425,29604625,47859265,51396865,53438905,62747425
%N Numbers that can be written as the sum of two squares in three ways, using three consecutive squares.
%C Numbers can be parameterized as follows: choose an odd number a and k a divisor of (a^2+1)/2, k <= sqrt((a^2+1)/2) (otherwise you will get a duplicate value with a negative number being squared) and let d = a/k * (a^2+3ak+2k^2+1) / k. Then the three consecutive squares are (d-1)/2, (d+1)/2 and (d+3)/2. The squares added to them are (d/a + a) / 2, (d/a - a) / 2 and ((d+2)/(a+2k) - (a+2k))/2. All members of this sequence are congruent to 1 or 25 (mod 120); all are the product exclusively of primes congruent to 1 (mod 4). No number can written as the sum of two squares using four consecutive squares.
%e 1105 = 31^2 + 12^2 = 32^2 + 9^2 = 33^2 + 4^2, using the consecutive values 31,32,33. 1 is included for the triple -1,0,1.
%Y Cf. A002144, A001481.
%K nonn
%O 1,2
%A _Franklin T. Adams-Watters_, Oct 04 2006