This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A129816 #15 Feb 24 2021 04:23:59
%S A129816 2,5,6,7,8,11,12,13,15,17,18,20,22,24,27,28,31,32,33,37,39,40,41,42,
%T A129816 45,48,50,51,52,54,55,56,57,59,60,61,63,69,70,71,72,73,74,76,79,80,84,
%U A129816 87,88,89,90,91,93,96,97,98,99,101,102,104,105,107,108,111,112,114,116,120
%N A129816 Conjectured numbers n such that there do not exist two consecutive primes whose product + n is a square.
%C A129816 For twin primes and k=1, p(n)*p(n+1)+k is always a square. This follows from the fact that for any number x, x(x+2) + 1 = x^2+2x+1 = (x+1)^2. Since twin primes differ by 2, the product of twin primes + 1 is a square (A075369), and 1 is not in the sequence.
%C A129816 Note that the product of the special case of the first 2 consecutive primes 2 and 3 will produce infinitely many squares. 6+3 = 9, 6+10 = 16. 6+k = y^2 or k=y^2 - 6 for y > 4. This leaves us the cases for p(n) > 2 to prove the instances of k such that p(n)*p(n+1) + k != y^2.
%C A129816 Case k=2: Let x = p(n) and x+2m = p(n+1) since the next prime is a multiple of 2 away from the current prime. Now assume x^2+2mx + 2 = y^2.
%C A129816 Completing the square and rearranging terms, we have x^2 + 2mx + m^2 = y^2 -2 + m^2 or (x+m)^2 = y^2 - 2 + m^2 = z^2. Then y^2-z^2 = 2 - m^2. So m=1 is the only possibility.
%C A129816 This gives y^2-z^2 = 1 or y-z= and y+z=1, impossible.
%C A129816 This contradicts the assumption x^2+2mx+2 = y^2 so there are no consecutive primes such that p(n)*p(n+1)+k = y^2.
%C A129816 Case 5: Using the arguments for Case 2, c. so m = 1,2 are the only ppossibilities and y^2-z^2 = 4 or y^2-z^2 = 1 have no integer solutions.
%C A129816 Case 7: y^-z^2 = 7 - m^2. m = 1,2. y^2-z^2 = 6 has no integer solutions. For y^2-z^2 = 3 we have y-z = 1 y+z = 3 y = 2, z=1. Then x^2-2xm+7 = y^2 becomes x^2-2x+3 = 0 which has no integer solution.
%C A129816 Let us consider a working case for k = 14. y^-z^2 = 14 - m^2. m = 1,2,3. For m=1 y-z = 1 y+z = 13 y = 7 Then substituting m,y into x^2 + 2mx + 14 = y^2 we get x^2+2x + 14 = 49. Completing the square we get (x+1)^2 = 49-14+1 = 36 and x=5. So 5*7+14 = 49. I do not see a general proof for all cases that p(n)*p(n+1) + k != y^2.
%C A129816 Complement of A129783. - _Omar E. Pol_, Dec 26 2008
%o A129816 (PARI) primesq2(n) = {local(x); for(x=1,n, if(primesq(10000,x)==0,print1(x",") ) ) } primesq(n,m) = { local(c,k,x,p1,p2,j); c=0; for(k=m,m, for(x=1,n, p1=prime(x); p2=(prime(x+1)); y=p1*p2+k; if(issquare(y), c++; \ print1(k","); break; ) ) ); c; }
%Y A129816 Cf. A129783. - _Omar E. Pol_, Dec 26 2008
%K A129816 easy,nonn,uned
%O A129816 1,1
%A A129816 _Cino Hilliard_, May 20 2007
%E A129816 There is probably no proof that this sequence is correct. - _N. J. A. Sloane_, May 24 2007