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.

Showing 1-3 of 3 results.

A229853 a(n) = 384*n + 1.

Original entry on oeis.org

1, 385, 769, 1153, 1537, 1921, 2305, 2689, 3073, 3457, 3841, 4225, 4609, 4993, 5377, 5761, 6145, 6529, 6913, 7297, 7681, 8065, 8449, 8833, 9217, 9601, 9985, 10369, 10753, 11137, 11521, 11905, 12289, 12673, 13057, 13441, 13825, 14209, 14593, 14977, 15361, 15745
Offset: 0

Views

Author

Arkadiusz Wesolowski, Oct 01 2013

Keywords

Comments

Every composite Fermat number has a divisor of the form 384*n + 1, n > 0.

Crossrefs

Programs

  • Magma
    [384*n+1 : n in [0..40]];
    
  • Maple
    seq(384*n+1, n=0..40);
  • Mathematica
    Table[384*n + 1, {n, 0, 40}]
  • PARI
    for(n=0, 40, print1(384*n+1, ", "));

Formula

G.f.: (1 + 383*x)/(1 - x)^2.
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: exp(x)*(1 + 384*x).
a(n) = 2*a(n-1) - a(n-2). (End)

A229855 a(n) = 384*n + 257.

Original entry on oeis.org

257, 641, 1025, 1409, 1793, 2177, 2561, 2945, 3329, 3713, 4097, 4481, 4865, 5249, 5633, 6017, 6401, 6785, 7169, 7553, 7937, 8321, 8705, 9089, 9473, 9857, 10241, 10625, 11009, 11393, 11777, 12161, 12545, 12929, 13313, 13697, 14081, 14465, 14849, 15233, 15617, 16001
Offset: 0

Views

Author

Arkadiusz Wesolowski, Oct 01 2013

Keywords

Comments

Every composite Fermat number has at least two divisors of the form 384*n + 257, n > 0.

Crossrefs

Programs

  • Magma
    [384*n+257 : n in [0..40]];
    
  • Maple
    seq(384*n+257, n=0..40);
  • Mathematica
    Table[384*n + 257, {n, 0, 40}]
  • PARI
    for(n=0, 40, print1(384*n+257, ", "));

Formula

G.f.: (257 + 127*x)/(1 - x)^2.
a(n) = 128*A016789(n) + 1.
From Elmo R. Oliveira, Dec 08 2024: (Start)
E.g.f.: exp(x)*(257 + 384*x).
a(n) = 2*a(n-1) - a(n-2) for n > 1. (End)

A229850 Number of prime factors congruent to 1 mod 3 that divide the Fermat number 2^(2^n) + 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 3, 2
Offset: 0

Views

Author

Arkadiusz Wesolowski, Oct 01 2013

Keywords

Comments

a(n) < A046052(n) because all Fermat numbers greater than 3 are equal to 2 (mod 3).
a(n) = 1 if A046052(n) = 2.
If A046052(n) = 3, then a(n) = 0 or 2.
a(n) <= A228846(n) - n - 1 for n = 0 to 11.

References

  • M. Krizek, F. Luca, L. Somer, 17 Lectures on Fermat Numbers: From Number Theory to Geometry, CMS Books in Mathematics, vol. 9, Springer-Verlag, New York, 2001, pp. 61-63, 65-66.

Crossrefs

Showing 1-3 of 3 results.