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%I #25 Nov 14 2019 18:22:59
%S 11,191,495611,1628171,1893371,7806431,14167691,15085451,16048031,
%T 20124191,23951231,27415691,28558331,30096191,41396351,46022651,
%U 55591931,87871871,88811771,99720191,103347551,130051211,139617851,152670731,181333151,198922811,222427391
%N Prime(n-1) such that prime(n-1)+5 is square and equal to prime(n+1)-1.
%C Conjecture: If the condition holds, prime(n-1) and prime(n) are twin primes of the form 10k+1 and 10k+3. I.e., the last digits of the primes are 1 and 3. The 1 ending is evident in the entries here.
%C The conjecture is true: For p = prime(n-1) > 2 in this sequence we know that {p+1, p+3, p+5} are all even, p+6 = prime(n+1), p+5 = r^2 is a square, and p+4 = r^2-1 = (r-1)*(r+1) is composite. So p+2 = prime(n). Then p mod 10 = 1 because no prime > 5 has rightmost digit 5 and no square has rightmost digit 2. - _Alois P. Heinz_, Sep 07 2016
%C Subsequence of A001359 and of A030430. - _Alois P. Heinz_, Sep 07 2016
%H Alois P. Heinz, <a href="/A158509/b158509.txt">Table of n, a(n) for n = 1..10000</a>
%H S. M. Ruiz, <a href="http://groups.yahoo.com/group/primenumbers/message/19899">Integer then equal</a>.
%H Sebastian Martin Ruiz and others, <a href="/A158460/a158460.txt">Integers then Equals</a>, digest of 7 messages in primenumbers Yahoo group, Mar 14 - Mar 20, 2009.
%F a(n) = prime(A158460(n)-1). - _R. J. Mathar_, Sep 07 2016
%e For n = 6, prime(6-1)=11,11+5=16;prime(6+1)=17,17-1=16. So prime(6-1)=11 is the first entry in the sequence.
%t Transpose[Select[Partition[Prime[Range[7400000]],3,1],Last[#]-First[#] == 6 && IntegerQ[Sqrt[First[#]+5]]&]][[1]] (* _Harvey P. Dale_, Dec 22 2013 *)
%o (PARI) integerequal(m,n,a,b) =
%o {
%o local(x,p1,p2);
%o for(x=m,n,
%o p1=prime(x-1);
%o if(issquare(p1+a),
%o p2=prime(x+1);if((p1+a)==(p2-b),
%o print(x","p1","prime(x))
%o )
%o )
%o )
%o }
%Y Cf. A001359, A030430, A158460.
%K nonn
%O 1,1
%A _Cino Hilliard_, Mar 20 2009
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