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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.

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A217277 Numbers n such that n^2-9 is divisible by consecutive primes beginning with 2.

Original entry on oeis.org

5, 9, 15, 21, 27, 33, 51, 57, 87, 93, 123, 147, 213, 297, 333, 483, 753, 1053, 1347, 2307, 2643, 3237, 4287, 6003, 10293, 12477, 14403, 18147, 26247, 58803, 74253, 739203, 1166883, 3801333
Offset: 1

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Author

Robin Garcia, Sep 29 2012

Keywords

Comments

Searched to 10^8. If this sequence is finite with 34 terms, then there is only one solution to m! + 9 = n^2: m=6, n=27.
Consider the Diophantine equations m^2 - 9 = k*(n^2 - 9), which lead to the Pell-like equations m^2 - k*n^2 = 9*(-k + 1) (1) where the only primes dividing k are in P = {2,3,5,7,11,13,17,19} and such that for every k there is at least one solution (m,n) that belongs to the sequence.
As an example, if k = 10 = 2*5, the 2 pairs of solutions of m^2 - 10n^2 = -81; (27,9) and (333,1053) belong to the sequence but the other solutions seem to not be P-smooth. If k = 30 = 2*3*5; m^2 - 30n^2 = -261 and (147, 27) is a solution belonging to the sequence, ...
If the infinitely many solutions of the Pell-like equations are never P-smooth, then this sequence is finite and there is a precise answer to the extended Brocard's problem: There are exactly 3 solutions to m!+t^2 = n^2 with t=1; 1 solution with t = 3; ... - Robin Garcia, Oct 01 2012
All terms beyond the first are 3 mod 6, since otherwise n^2 - 9 = 2^k for some k and so n - 3 and n + 3 are both powers of 2. For n > 7 all terms are +-3 mod 30, a consequence of Størmer's theorem. - Charles R Greathouse IV, Oct 01 2012
No further terms up to 10^10. - Charles R Greathouse IV, Oct 01 2012

Examples

			5^2 - 9 = 2^4; 9^2 -9 = 2^3*3^2; 27^2 -9 = 2^4*3^2*5 = 6!; 87^2 - 9 = 2^3*3^3*5*7; 333^2 - 9 = 2^5*3^2*5*7*11.

Crossrefs

Cf. A055932, A194099.

Programs

  • PARI
    is(n)=my(m=n^2-9,t);forprime(p=2,,t=valuation(m,p);if(t,m/=p^t,return(m==1)))
print1(5);forstep(n=9,1e6,6,if(is(n),print1(", "n)))
\\ Charles R Greathouse IV, Oct 01 2012
  • PARI
    is(n)=my(f=factor(n^2-9)[,1]~);f==primes(#f) \\ Charles R Greathouse IV, Oct 01 2012
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