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.

1, 144, 1296, 3600, 176400, 156816, 2985984, 921600, 2702736, 11696400, 18974736, 21566736, 17740944, 5992704, 125888400, 7290000, 8363664, 12027024, 63680400, 210830400, 13838400, 72590400, 15116544, 15397776, 67568400, 128595600, 80784144, 93315600, 33039504

Offset: 1

Author

Michel Lagneau, Aug 28 2015

Keywords

• nonn

Comments

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.

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

Examples

			A056899(2)- A056899(1) = 3-2 = 1^2;
A056899(5)- A056899(4) = 227-83 = 144 = 12^2;
A056899(14)- A056899(13) = 12323-11027 = 1296 = 36^2.
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..1100

Crossrefs

Cf. A056899, A216330, A261659.

Programs

Maple

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:

Mathematica

Select[ Differences[ Select[ Range[0, 1000000], PrimeQ[#^2 + 2] &]^2], IntegerQ@ Sqrt@# &] (* or *)
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 *)