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

1,2

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

Links

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

Example

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.

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 *)

Crossrefs

Cf. A056899, A216330, A261659.

Sequence in context: A133062 A211473 A203826 * A203819 A222515 A023096

Adjacent sequences: A261652 A261653 A261654 * A261656 A261657 A261658

Keyword

nonn

Author

Michel Lagneau, Aug 28 2015

Status

approved