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A181619 Numbers k such that k^2+1 = 2p,(k+1)^2+1 = 5q, (k+2)^2+1 = 10r where p, q, and r are primes.

%I A181619 #11 Sep 01 2013 11:55:13
%S A181619 11,51,61,101,221,261,571,2271,2821,2871,5071,5651,5761,6561,6951,
%T A181619 9751,10461,10851,11621,11711,14961,15911,16551,17171,17601,18511,
%U A181619 19071,19551,23151,25261,27351,27751
%N A181619 Numbers k such that k^2+1 = 2p,(k+1)^2+1 = 5q, (k+2)^2+1 = 10r where p, q, and r are primes.
%C A181619 a(n) == 1 (mod 10).
%H A181619 Harvey P. Dale, <a href="/A181619/b181619.txt">Table of n, a(n) for n = 1..1000</a>
%e A181619 a(2) = 51 because 51^2+1 = 2*1301, 52^2+1 = 5*541, 53^2+1 = 10*281.
%p A181619 with(numtheory):for n from 1 to 30000 do : x:=n^2+1:y:=(n+1)^2+1:z:=(n+2)^2+1:x1:=factorset(x):y1:=factorset(y):z1:=factorset(z):n1:=bigomega(x):n2:=bigomega(y):n3:=bigomega(z):if
%p A181619   x1[1]=2 and n1=2 and y1[1]=5 and n2 = 2 and z1[1]=2 and z1[2]=5 and n3=3 then
%p A181619   printf(`%d, `, n):else fi:od:
%t A181619 ksQ[k_]:=And@@PrimeQ[{(k^2+1)/2,((k+1)^2+1)/5,((k+2)^2+1)/10}]; Select[ Range[30000],ksQ] (* _Harvey P. Dale_, Sep 01 2013 *)
%o A181619 (PARI) forstep(k=1,1e5,10,if(isprime(k^2\2+1)&isprime((k+1)^2\5+1)&isprime((k+2)^2\10+1),print1(k", ")))
%Y A181619 Cf. A002144, A005574, A002731.
%K A181619 nonn
%O A181619 1,1
%A A181619 _Michel Lagneau_, Jan 31 2011