

A129829


Conjectured first occurrence of numbers n with the property that there exist two consecutive primes p and q such that pq + n is a cube.


0



2, 9, 12, 20, 21, 25, 29, 32, 48, 49, 58, 62, 65, 73, 75, 81, 90, 101, 110, 119, 122, 139, 156, 176, 181, 184, 186, 189, 200
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OFFSET

1,1


COMMENTS

It remains to prove that for certain n, pq+n != y^3 for all consecutive primes p and q. This list was computed for p and q with prime indices up to 100000. Also note that this is not a complete list but rather the first occurrence of a solution. For example, n = 101 has 2 solutions p=29,q=31 for 29*31+101=1000 = 10^3 and p=197,p=199 for 197*199+101 = 38304 = 34^3. n square tend to have more solutions.


LINKS

Table of n, a(n) for n=1..29.


EXAMPLE

p=2,q=3,k=2. 2*3+2 = 8 a cube.


PROG

(PARI) primecube(n, m) = { local(c, k, x, p1, p2, j); c=0; for(k=1, m, for(x=1, n, p1=prime(x); p2=(prime(x+1)); y=p1*p2+k; if(iscube(y), c++; print1(k", "); break; ) ) ); c; } iscube(n) = { local(r); r = n^(1/3); if(floor(r+.5)^3== n, 1, 0) }


CROSSREFS

Cf. A129783.
Sequence in context: A048768 A070226 A273669 * A053900 A318681 A183207
Adjacent sequences: A129826 A129827 A129828 * A129830 A129831 A129832


KEYWORD

easy,nonn


AUTHOR

Cino Hilliard, May 20 2007


STATUS

approved



