

A129816


Conjectured numbers n such that there do not exist two consecutive primes whose product + n is a square.


1



2, 5, 6, 7, 8, 11, 12, 13, 15, 17, 18, 20, 22, 24, 27, 28, 31, 32, 33, 37, 39, 40, 41, 42, 45, 48, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 63, 69, 70, 71, 72, 73, 74, 76, 79, 80, 84, 87, 88, 89, 90, 91, 93, 96, 97, 98, 99, 101, 102, 104, 105, 107, 108, 111, 112, 114, 116, 120
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OFFSET

1,1


COMMENTS

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.
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.
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.
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^2z^2 = 2  m^2. So m=1 is the only possibility.
This gives y^2z^2 = 1 or yz= and y+z=1, impossible.
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.
Case 5: Using the arguments for Case 2, c. so m = 1,2 are the only ppossibilities and y^2z^2 = 4 or y^2z^2 = 1 have no integer solutions.
Case 7: y^z^2 = 7  m^2. m = 1,2. y^2z^2 = 6 has no integer solutions. For y^2z^2 = 3 we have yz = 1 y+z = 3 y = 2, z=1. Then x^22xm+7 = y^2 becomes x^22x+3 = 0 which has no integer solution.
Let us consider a working case for k = 14. y^z^2 = 14  m^2. m = 1,2,3. For m=1 yz = 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 = 4914+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.
Complement of A129783.  Omar E. Pol, Dec 26 2008


LINKS

Table of n, a(n) for n=1..68.


PROG

(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; }


CROSSREFS

Cf. A129783.  Omar E. Pol, Dec 26 2008
Sequence in context: A028726 A047269 A039027 * A137708 A122806 A122546
Adjacent sequences: A129813 A129814 A129815 * A129817 A129818 A129819


KEYWORD

easy,nonn,uned


AUTHOR

Cino Hilliard, May 20 2007


EXTENSIONS

There is probably no proof that this sequence is correct.  N. J. A. Sloane, May 24 2007


STATUS

approved



