cp's OEIS Frontend

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.

A235365 Smallest odd prime factor of 3^n + 1, for n > 1.

Original entry on oeis.org

5, 7, 41, 61, 5, 547, 17, 7, 5, 67, 41, 398581, 5, 7, 21523361, 103, 5, 2851, 41, 7, 5, 23535794707, 17, 61, 5, 7, 41, 523, 5, 6883, 926510094425921, 7, 5, 61, 41, 18427, 5, 7, 17, 33703, 5, 82064241848634269407, 41, 7, 5, 16921, 76801, 547, 5, 7, 41, 78719947, 5, 61, 17, 7, 5, 3187, 41
Offset: 2

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Author

Jonathan Sondow, Jan 19 2014

Keywords

Comments

Levi Ben Gerson (1288-1344) proved that 3^n + 1 = 2^m has no solution in integers if n > 1, by showing that 3^n + l has an odd prime factor. His proof uses remainders after division of powers of 3 by 8 and powers of 2 by 8; see the Lenstra and Peterson links. For an elegant short proof, see the Franklin link.

Examples

			3^2 + 1 = 10 = 2*5, so a(2) = 5.
		

References

  • L. E. Dickson, History of the Theory of Numbers, Vol. II, Chelsea, NY 1992; see p. 731.

Crossrefs

See A235366 for 3^n - 1.
Cf. also A003586 (products 2^m * 3^n), A006899, A061987, A108906.

Programs

  • Magma
    [PrimeDivisors(3^n +1)[2]: n in [2..60] ] ; // Vincenzo Librandi, Mar 16 2019
  • Mathematica
    Table[FactorInteger[3^n + 1][[2, 1]], {n, 2, 50}]

Formula

a(2+4n) = 5 as 3^(2+4n) + 1 = (3^2)*(3^4)^n + 1 = 9*81^n + 1 = 9*(80+1)^n + 1 == 9 + 1 == 0 (mod 5).
a(3+6n) = 7 as 3^(3+6n) + 1 = (3^3)*(3^6)^n + 1 = 27*729^n + 1 = 27*(728+1)^n + 1 == 27 + 1 == 0 (mod 7), but 27 * 729^n + 1 == 2*(-1)^n + 1 !== 0 (mod 5).

Extensions

Terms to a(132) in b-file from Vincenzo Librandi, Mar 16 2019
a(133)-a(658) in b-file from Amiram Eldar, Feb 05 2020
a(659)-a(768) in b-file from Max Alekseyev, Apr 27 2022