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 1-10 of 14 results. Next

A054992 Number of prime factors of 2^n + 1 (counted with multiplicity).

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 2, 1, 4, 3, 2, 2, 2, 3, 4, 1, 2, 4, 2, 2, 4, 3, 2, 3, 4, 4, 6, 2, 3, 6, 2, 2, 5, 4, 5, 4, 3, 4, 4, 2, 3, 6, 2, 3, 7, 5, 3, 3, 3, 7, 6, 3, 3, 6, 6, 3, 5, 3, 4, 4, 2, 5, 7, 2, 6, 6, 3, 4, 5, 7, 3, 5, 3, 5, 7, 4, 6, 10, 2, 3, 10, 5, 6, 5, 4, 5, 5, 4, 4, 11, 6, 2, 5, 4, 5, 3, 5, 6, 9, 6, 2, 9, 3
Offset: 1

Views

Author

Arne Ring (arne.ring(AT)epost.de), May 30 2000

Keywords

Comments

The length of row n in A001269.

Examples

			a(3) = 2 because 2^3 + 1 = 9 = 3*3.
		

Crossrefs

bigomega(b^n+1): A057934 (b=10), A057935 (b=9), A057936 (b=8), A057937 (b=7), A057938 (b=6), A057939 (b=5), A057940 (b=4), A057941 (b=3), this sequence (b=2).
Cf. A046051 (number of prime factors of 2^n-1).
Cf. A086257 (number of primitive prime factors).

Programs

Formula

a(n) = A046051(2n) - A046051(n). - T. D. Noe, Jun 18 2003
a(n) = A001222(A000051(n)). - Amiram Eldar, Oct 04 2019

Extensions

Extended by Patrick De Geest, Oct 01 2000
Terms to a(500) in b-file from T. D. Noe, Nov 10 2007
Deleted duplicate (and broken) Wagstaff link. - N. J. A. Sloane, Jan 18 2019
a(500)-a(1062) in b-file from Amiram Eldar, Oct 04 2019
a(1063)-a(1128) in b-file from Max Alekseyev, Jul 15 2023, Mar 15 2025

A002587 Largest prime factor of 2^n + 1.

Original entry on oeis.org

2, 3, 5, 3, 17, 11, 13, 43, 257, 19, 41, 683, 241, 2731, 113, 331, 65537, 43691, 109, 174763, 61681, 5419, 2113, 2796203, 673, 4051, 1613, 87211, 15790321, 3033169, 1321, 715827883, 6700417, 20857, 26317, 86171, 38737, 25781083, 525313
Offset: 0

Views

Author

Keywords

Comments

a(n) != 1 (mod n) for n = 3, 51, 141, 309, 321, 348, ... - Giovanni Resta & Thomas Ordowski, Jan 05 2014
a(n) != 1 (mod n) iff a(m) = a(n) for some m < n. Then n = 3m for m = 1, 17, 47, 103, 107, 116, ... - Thomas Ordowski, Jan 08 2014

References

  • J. Brillhart et al., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.
  • M. Kraitchik, Recherches sur la Théorie des Nombres. Gauthiers-Villars, Paris, Vol. 1, 1924, Vol. 2, 1929, see Vol. 2, p. 85.
  • E. Lucas, Théorie des fonctions numériques simplement périodiques, Amer. J. Math., 1 (1878), 184-239, 289-321.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. similar sequences listed in A274903.

Programs

Formula

Charles proves that a(n) >> n^(4/3) infinitely often under the abc conjecture, and reports that Andrew Granville has improved this to a(n) >> n^2. - Charles R Greathouse IV, Apr 29 2013
a(n) = A006530(A000051(n)). - Vincenzo Librandi, Jul 12 2016

Extensions

More terms from James Sellers, Jul 06 2000
Offset 0, a(0) = 2 from Vincenzo Librandi, Jul 12 2016

A066135 a(n) = least number m > 1 such that sigma_n(m) = k*m for some k.

Original entry on oeis.org

6, 10, 6, 34, 6, 10, 6, 84, 6, 10, 6, 34, 6, 10, 6, 84, 6, 10, 6, 34, 6, 10, 6, 194, 6, 10, 6, 34, 6, 10, 6, 84, 6, 10, 6, 34, 6, 10, 6, 84, 6, 10, 6, 34, 6, 10, 6, 228, 6, 10, 6, 34, 6, 10, 6, 84, 6, 10, 6, 34, 6, 10, 6, 84, 6, 10, 6, 34, 6, 10, 6, 194, 6, 10, 6
Offset: 1

Views

Author

Labos Elemer, Dec 06 2001

Keywords

Comments

a(n) <= 2p, where p = A002586(n) is the smallest prime factor of (1 + 2^n). (Proof. Since sigma_n(2p) = (1 + 2^n)(1 + p^n) and p is odd, 2p divides sigma_n(2p).) - Jonathan Sondow, Nov 23 2012

Crossrefs

Cf. A218860, A218861 (unique values and where they first occur).

Programs

  • Mathematica
    Table[m = 2; While[Mod[DivisorSigma[n, m], m] > 0, m++]; m, {n, 100}] (* T. D. Noe, Nov 23 2012 *)

Formula

Sum{d^n} = ka(n), d runs over the divisors of a(n), where k is an integer and a(n) is the smallest suitable number.

Extensions

Definition and formulas corrected by Jonathan Sondow, Nov 23 2012

A366719 Smallest prime dividing 12^n + 1.

Original entry on oeis.org

2, 13, 5, 7, 89, 13, 5, 13, 17, 7, 5, 13, 89, 13, 5, 7, 153953, 13, 5, 13, 41, 7, 5, 13, 17, 13, 5, 7, 89, 13, 5, 13, 769, 7, 5, 13, 89, 13, 5, 7, 17, 13, 5, 13, 89, 7, 5, 13, 7489, 13, 5, 7, 89, 13, 5, 13, 17, 7, 5, 13, 41, 13, 5, 7, 36097, 13, 5, 13, 89, 7
Offset: 0

Views

Author

Sean A. Irvine, Oct 17 2023

Keywords

Crossrefs

Programs

Formula

a(n) = A020639(A178248(n)). - Paul F. Marrero Romero, Oct 25 2023

A366609 Smallest prime dividing 4^n + 1.

Original entry on oeis.org

2, 5, 17, 5, 257, 5, 17, 5, 65537, 5, 17, 5, 97, 5, 17, 5, 641, 5, 17, 5, 257, 5, 17, 5, 193, 5, 17, 5, 257, 5, 17, 5, 274177, 5, 17, 5, 97, 5, 17, 5, 65537, 5, 17, 5, 257, 5, 17, 5, 641, 5, 17, 5, 257, 5, 17, 5, 449, 5, 17, 5, 97, 5, 17, 5, 59649589127497217
Offset: 0

Views

Author

Sean A. Irvine, Oct 14 2023

Keywords

Crossrefs

A366670 Smallest prime dividing 6^n + 1.

Original entry on oeis.org

2, 7, 37, 7, 1297, 7, 13, 7, 17, 7, 37, 7, 1297, 7, 37, 7, 353, 7, 13, 7, 41, 7, 37, 7, 17, 7, 37, 7, 281, 7, 13, 7, 2753, 7, 37, 7, 577, 7, 37, 7, 17, 7, 13, 7, 89, 7, 37, 7, 193, 7, 37, 7, 1297, 7, 13, 7, 17, 7, 37, 7, 41, 7, 37, 7, 4926056449, 7, 13, 7, 137
Offset: 0

Views

Author

Sean A. Irvine, Oct 15 2023

Keywords

Crossrefs

Programs

Formula

a(n) = A020639(A062394(n)). - Paul F. Marrero Romero, Oct 17 2023

A066284 a(n) = A066135(4*n).

Original entry on oeis.org

34, 84, 34, 84, 34, 194, 34, 84, 34, 84, 34, 228, 34, 84, 34, 84, 34, 194, 34, 84, 34, 84, 34, 228, 34, 84, 34, 84, 34, 194, 34, 84, 34, 84, 34, 386, 34, 84, 34, 84, 34, 194, 34, 84, 34, 84, 34, 228, 34, 84, 34, 84, 34, 194, 34, 84, 34, 84, 34, 228, 34, 84, 34, 84, 34, 194
Offset: 1

Views

Author

Labos Elemer, Dec 11 2001

Keywords

Comments

a(n) <= 2p, where p = A002586(4n) is the least prime factor of (1 + 16^n). (See the Comment in A066135.) - Jonathan Sondow, Nov 23 2012

Examples

			First 3 terms correspond to entries of other sequences as follows: a(1)=A046763(2), a(2)=A055712(2), a(3)=A055716(2).
From _Michael De Vlieger_, Jul 17 2017: (Start)
First position of values, with observations pertaining to values for 1 <= n <= 3000:
    Value   Position   Observations:
    --------------------------------
       34     1        All odd.
       84     2        In A047235.
      194     6        In A017593.
      228    12
      386    36
     1282    72
     1538   144
     3084   288
   147468   576
     1956   864
  1046532  1152
    24578  2304
     3252  2880
(End)
		

Crossrefs

Programs

  • Mathematica
    Table[m = 2; While[Mod[DivisorSigma[4 n, m], m] > 0, m++]; m, {n, 66}] (* Michael De Vlieger, Jul 17 2017 *)
  • PARI
    a(n) = {n *= 4; my(m = 2); while (sigma(m, n) % m, m++); m;} \\ Michel Marcus, Oct 02 2016

Formula

a(n) = Min{x : sigma_4n(x) mod x = 0, x > 1}

A366671 Smallest prime dividing 8^n + 1.

Original entry on oeis.org

2, 3, 5, 3, 17, 3, 5, 3, 97, 3, 5, 3, 17, 3, 5, 3, 193, 3, 5, 3, 17, 3, 5, 3, 97, 3, 5, 3, 17, 3, 5, 3, 641, 3, 5, 3, 17, 3, 5, 3, 97, 3, 5, 3, 17, 3, 5, 3, 193, 3, 5, 3, 17, 3, 5, 3, 97, 3, 5, 3, 17, 3, 5, 3, 769, 3, 5, 3, 17, 3, 5, 3, 97, 3, 5, 3, 17, 3, 5
Offset: 0

Views

Author

Sean A. Irvine, Oct 15 2023

Keywords

Comments

a(n) = 3 if n is odd. a(n) = 5 if n == 2 (mod 4). - Robert Israel, Nov 20 2023

Crossrefs

Programs

  • Maple
    P1000:= mul(ithprime(i),i= 4..1000):
    f:= proc(n) local t;
      if n::odd then return 3 elif n mod 4 = 2 then return 5 fi;
      t:= igcd(8^n+1,P1000);
      if t <> 1 then min(numtheory:-factorset(t)) else min(numtheory:-factorset(8^n+1)) fi
    end proc:
    map(f, [$0..100]); # Robert Israel, Nov 20 2023
  • Mathematica
    Table[FactorInteger[8^n + 1][[1,1]], {n, 0, 78}] (* Paul F. Marrero Romero, Oct 20 2023 *)
  • Python
    from sympy import primefactors
    def A366671(n): return min(primefactors((1<<3*n)+1)) # Chai Wah Wu, Oct 16 2023

Formula

a(n) = A020639(A062395(n)). - Paul F. Marrero Romero, Oct 20 2023
a(n) = A002586(3*n) for n >= 1. - Robert Israel, Nov 20 2023

A374237 Irregular triangle read by rows where row n lists, in increasing order, the divisors of 2^n + 1.

Original entry on oeis.org

1, 2, 1, 3, 1, 5, 1, 3, 9, 1, 17, 1, 3, 11, 33, 1, 5, 13, 65, 1, 3, 43, 129, 1, 257, 1, 3, 9, 19, 27, 57, 171, 513, 1, 5, 25, 41, 205, 1025, 1, 3, 683, 2049, 1, 17, 241, 4097, 1, 3, 2731, 8193, 1, 5, 29, 113, 145, 565, 3277, 16385, 1, 3, 9, 11, 33, 99, 331, 993, 2979, 3641, 10923, 32769
Offset: 0

Views

Author

Paolo Xausa, Jul 02 2024

Keywords

Examples

			Triangle begins:
   [0]  1,   2;
   [1]  1,   3;
   [2]  1,   5;
   [3]  1,   3,  9;
   [4]  1,  17;
   [5]  1,   3, 11,  33;
   [6]  1,   5, 13,  65;
   [7]  1,   3, 43, 129;
   [8]  1, 257;
   [9]  1,   3,  9,  19,  27,   57, 171, 513;
  [10]  1,   5, 25,  41, 205, 1025;
  ...
		

Crossrefs

Subsequence of A027750.
Cf. A000051, A002586 (2nd column), A046798 (row lengths), A069060 (row products), A069061 (row sums).
Cf. A361438 (analogous for 2^n - 1).

Programs

  • Maple
    T:= n-> sort([numtheory[divisors](2^n+1)[]])[]:
    seq(T(n), n=0..15);  # Alois P. Heinz, Oct 20 2024
  • Mathematica
    Divisors[2^Range[0, 20] + 1]

A292834 Numbers m, not powers of 2, such that the least prime factor of 2^m + 1 is congruent to 1 (mod m).

Original entry on oeis.org

24, 48, 112, 160, 192, 272, 448, 496, 656, 688, 832, 896, 1088, 1152, 1168, 1328, 1360, 1408, 1472, 1520, 1664, 1744, 1920, 1984, 2176, 2304, 2432, 2560, 2688, 2752, 2816, 2944, 2960, 3056, 3072, 3200, 3328, 3520, 3664, 3712, 3776, 4672, 4864, 4928, 5120, 5376, 5552, 5888, 6144
Offset: 1

Views

Author

Thomas Ordowski, Sep 24 2017

Keywords

Comments

Problem: are there infinitely many such numbers?
Theorem: there are no numbers m in the sequence such that, for each prime factor p of 2^m + 1, p == 1 (mod m).
Proof: if all prime factors p of 2^m + 1 are p == 1 (mod m), then 2^m + 1 == 1 (mod m), thus 2^m == 0 (mod m), so m = 2^k.
From Theorem in A002586, all terms are == 0 (mod 8). - Robert G. Wilson v, Jan 02 2018

Crossrefs

Programs

  • Mathematica
    Select[Range[200], And[! IntegerQ @ Log2 @ #, Mod[FactorInteger[2^# + 1][[1, 1]], #] == 1] &] (* Michael De Vlieger, Sep 24 2017 *)
    fQ[n_] := If[ OddQ@ n || IntegerQ@ Log2@ n || PrimeQ[2^n +1], False, Block[{p = 3}, While[PowerMod[2, n, p] +1 != p, p = NextPrime@ p]; Mod[p, n] == 1]] (* Robert G. Wilson v, Jan 01 2018 *)
  • PARI
    isok(n) = my(e = valuation(n, 2)); (2^e != n) && ((vecmin(factor(2^n+1)[,1]) % n) == 1); \\ Michel Marcus, Nov 13 2017

Extensions

a(9)-a(15) from Robert G. Wilson v, Jan 01 2018
a(16)-a(49) from Robert G. Wilson v, Jan 02 2018
Showing 1-10 of 14 results. Next