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A261655 Squares equal to the difference between two successive primes of the form k^2+2 in the order in which they appear in A056899.

%I A261655 #33 Sep 22 2015 04:59:54
%S A261655 1,144,1296,3600,176400,156816,2985984,921600,2702736,11696400,
%T A261655 18974736,21566736,17740944,5992704,125888400,7290000,8363664,
%U A261655 12027024,63680400,210830400,13838400,72590400,15116544,15397776,67568400,128595600,80784144,93315600,33039504
%N A261655 Squares equal to the difference between two successive primes of the form k^2+2 in the order in which they appear in A056899.
%C A261655 Note that all terms after the first are divisible by 144: for n>1 the sequence b(n) = sqrt(a(n)/144) is 1, 3, 5, 35, 33, 144, 80, 137, 285, 363, 387, 351, 204, 935, 225, 241, 289, ..., see A261659.
%C A261655 The proof that all terms are == 0 (mod 144) is simple once you realize that the primes == 11 (mod 72), see comment in A056899. - _Robert G. Wilson v_, Sep 03 2015
%H A261655 Robert G. Wilson v, <a href="/A261655/b261655.txt">Table of n, a(n) for n = 1..1100</a>
%e A261655 A056899(2)- A056899(1) = 3-2 = 1^2;
%e A261655 A056899(5)- A056899(4) = 227-83 = 144 = 12^2;
%e A261655 A056899(14)- A056899(13) = 12323-11027 = 1296 = 36^2.
%p A261655 q:=2:for n from 1 to 10^7 do:p:=n^2+2:if isprime(p) then x:=p-q:q:=p: z:=sqrt(x):if z=floor(z) then printf(`%d, `, x):else fi:fi:od:
%t A261655 Select[ Differences[ Select[ Range[0, 1000000], PrimeQ[#^2 + 2] &]^2], IntegerQ@ Sqrt@# &] (* or *)
%t A261655 k = 1; p = 3; lst = {1}; While[k < 10000001, q = (6k +3)^2 + 2; If[ PrimeQ@ q, If[ IntegerQ@ Sqrt[q - p], AppendTo[lst, q - p]]; p = q]; k++] (* _Robert G. Wilson v_, Sep 03 2015 *)
%Y A261655 Cf. A056899, A216330, A261659.
%K A261655 nonn
%O A261655 1,2
%A A261655 _Michel Lagneau_, Aug 28 2015