

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.


2



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
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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) = 32 = 1^2;
A056899(5) A056899(4) = 22783 = 144 = 12^2;
A056899(14) A056899(13) = 1232311027 = 1296 = 36^2.


MAPLE

q:=2:for n from 1 to 10^7 do:p:=n^2+2:if isprime(p) then x:=pq: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



