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.

Previous Showing 11-19 of 19 results.

A199683 a(n) = 3*10^n + 1.

Original entry on oeis.org

4, 31, 301, 3001, 30001, 300001, 3000001, 30000001, 300000001, 3000000001, 30000000001, 300000000001, 3000000000001, 30000000000001, 300000000000001, 3000000000000001, 30000000000000001, 300000000000000001, 3000000000000000001, 30000000000000000001, 300000000000000000001
Offset: 0

Views

Author

Vincenzo Librandi, Nov 09 2011

Keywords

Crossrefs

Primes: A056807, A259866.

Programs

  • Magma
    [3*10^n+1: n in [0..30]];
  • Mathematica
    3*10^Range[0,20]+1 (* or *) LinearRecurrence[{11,-10},{4,31},20] (* Harvey P. Dale, Dec 12 2016 *)

Formula

a(n) = 10*a(n-1) - 9.
a(n) = 11*a(n-1) - 10*a(n-2).
G.f.: (4-13*x)/((1-x)*(1-10*x)).
E.g.f.: exp(x)*(1 + 3*exp(9*x)). - Elmo R. Oliveira, Jun 10 2025

A259866 Primes of the form 3*10^k + 1.

Original entry on oeis.org

31, 3001, 30000001, 30000000001, 30000000000000000000000000001, 3000000000000000000000000000000000001, 30000000000000000000000000000000000000000000000000000000000000000001
Offset: 1

Views

Author

N. J. A. Sloane, Jul 08 2015

Keywords

Crossrefs

Programs

  • Mathematica
    Select[3*10^Range[70]+1,PrimeQ] (* Harvey P. Dale, Jul 05 2020 *)

Formula

a(n) = 3*10^A056807(n)+1. - R. J. Mathar, Jul 15 2015

A232637 Odious numbers of order 2: a(n) = A000069(A000069(n)).

Original entry on oeis.org

1, 2, 7, 13, 14, 21, 25, 26, 31, 37, 41, 42, 49, 50, 55, 61, 62, 69, 73, 74, 81, 82, 87, 93, 97, 98, 103, 109, 110, 117, 121, 122, 127, 133, 137, 138, 145, 146, 151, 157, 161, 162, 167, 173, 174, 181, 185, 186, 193, 194, 199, 205, 206, 213, 217, 218, 223, 229, 233, 234, 241, 242, 247, 253
Offset: 1

Views

Author

Gerasimov Sergey, Nov 27 2013

Keywords

Comments

Odious numbers with odious subscripts.
From Antti Karttunen, Nov 29 2013: (Start)
Starting from 4 and iterating A000069(4), A000069(A000069(4)), A000069(A000069(A000069(4))), etc. gives A004119 from its second term onward: 4, 7, 13, 25, 49, 97, 193, ..., which is thus a subsequence of this sequence from the term 7 onward.
Proof: All of the terms A004119(n) are odious although A004119(n)-1 is evil, and the formula for A000069(n) reduces to a(n) = 2n - 1 when n-1 is evil, and iterating that formula starting from 4 gives A004119 from 7 onward (cf. Philippe Deléham's formula there dated Feb 20 2004).
(End)
These numbers are never multiples of 4. Probably there are infinitely many multiples of m in this sequence for any m not divisible by 4. Equivalently, A233419(n) > 0 for all n. - Charles R Greathouse IV, Dec 05 2013

Examples

			The first odious number, A000069(1) = 1, and A000069(1) = 1, so a(1) = 1.
The second odious number, A000069(2) = 2, and A000069(2) = 2, so a(2) = 2.
Those were the only fixed points of A000069, and after that, we have:
The third odious number, A000069(3) = 4, and A000069(4) = 7, thus a(3) = 7.
The fourth odious number, A000069(4) = 7, and A000069(7) = 13, thus a(4) = 13.
		

Crossrefs

A004119 from term 7 onward is a subsequence.
Subsequence of A042968.

Programs

Formula

a(n) = A000069(A000069(n)).
4n-6 <= a(n) <= 4n-3, see PARI script. - Charles R Greathouse IV, Dec 05 2013

A361899 a(n) = 3*(6858365065530*(2^45 - 1)*n + 153479820268467961)^2.

Original entry on oeis.org

70668165688923686196507258250492563, 174687593550891106640307045856561008882907291372256643, 698750373759134872171732581703201135992894186495330123, 1572188340624731296664944773228844067526467943619713003
Offset: 0

Views

Author

Arkadiusz Wesolowski, Mar 28 2023

Keywords

Comments

The "1/k" heuristic predicts that primes of the form k*2^m + 1 with k odd and m > 0 have almost a 1/k chance of being Fermat divisors (Dubner and Keller). This sequence yields a correction to the "1/k" heuristic, because it generates special values of k.
If:
1) k is of the form 3*a^2, where a is an odd positive integer not divisible by 3,
2) k is not a Sierpiński number,
3) for all odd positive integers m the numbers k*2^m + 1 are composite,
then the probability that a Fermat number is divisible by a prime of the form k*2^m + 1 equal to 0.
Every term meets the first and third condition. For any n, at least one of the primes from A361898 (except 3) divides every integer in the sequence a(n)*2^m + 1 with m odd.
What is the smallest odd integer k such that every prime of the form k*2^m + 1 (m > 0) does not divide any Fermat number?

References

  • H. Suyama, A note on the factors of Fermat numbers II, Abstracts of Papers Presented to the Amer. Math. Soc., Vol. 5 (1984), p. 132.

Crossrefs

Programs

  • Magma
    [3*(6858365065530*(2^45-1)*n+153479820268467961)^2: n in [0..3]];
    
  • Mathematica
    Table[3 (6858365065530 (2^45 - 1) n + 153479820268467961)^2, {n, 0, 3}]
  • PARI
    a(n)=3*(6858365065530*(2^45-1)*n+153479820268467961)^2

Formula

G.f.: (70668165688923686196507258250492563 + 174687593550891106428302548789789950293385516620778954*x + 174687593106461552462815941200289167933694087130037883*x^2)/(1 - x)^3.
a(n) = 3*(2*(Product_{i=1..13} A361898(i))*n + 153479820268467961)^2.
a(n) = 3*((29062/1192737)*(2^48 - 1)*(2^45 - 1)*n + 153479820268467961)^2.

A277084 Pisot sequence L(4,14).

Original entry on oeis.org

4, 14, 49, 172, 604, 2122, 7456, 26198, 92052, 323444, 1136489, 3993295, 14031289, 49301911, 173232725, 608689936, 2138761243, 7514991434, 26405516950, 92781386582, 326007088306, 1145495077635, 4024940008834, 14142480741305, 49692606865991, 174605518105877
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 29 2016

Keywords

Comments

There is no simple g.f. - Ilya Gutkovskiy, May 23 2019

Crossrefs

Cf. A008776 for definitions of Pisot sequences.
Cf. A010904 (Pisot sequence E(4,14)), A251221 (seems to be Pisot sequence P(4,14)).

Programs

  • Mathematica
    RecurrenceTable[{a[0] == 4, a[1] == 14, a[n] == Ceiling[a[n - 1]^2/a[n - 2]]}, a, {n, 25}]

Formula

a(n) = ceiling(a(n-1)^2/a(n-2)), a(0) = 4, a(1) = 14.

A248416 Rectangular array by antidiagonals: for n >= 0, row n gives the positions in the Thue-Morse sequence A010059 at which the first 2^n terms occur.

Original entry on oeis.org

1, 4, 1, 6, 4, 1, 7, 7, 7, 1, 10, 11, 13, 13, 1, 11, 13, 21, 25, 25, 1, 13, 16, 25, 41, 49, 49, 1, 16, 19, 31, 49, 81, 97, 97, 1, 18, 21, 37, 61, 97, 161, 193, 193, 1, 19, 25, 41, 73, 121, 193, 321, 385, 385, 1, 21, 28, 49, 81, 145, 241, 385, 641, 769, 769, 1, 24, 31, 55, 97, 161, 289, 481, 769, 1281, 1537, 1537, 1
Offset: 1

Views

Author

Clark Kimberling, Oct 06 2014

Keywords

Comments

Each row contains contains its successor as a proper subsequence.
Note that this supposes that the Thue-Morse sequence A010059 has offset 1, whereas the true offset is 0. So really the entries should all be reduced by 1. - N. J. A. Sloane, Jul 01 2016
Apparently T(n,3) = A004119(n+1) for n>0. Apparently T(n,4) = A083575(n) for n>0. - R. J. Mathar, Nov 06 2018

Examples

			Northwest corner, n>=0, k>=1:
   1    4    6    7   10   11   13   16   18   19
   1    4    7   11   13   16   19   21   25   28
   1    7   13   21   25   31   37   41   49   55
   1   13   25   41   49   61   73   81   97  109
   1   25   49   81   97  121  145  161  193  217
   1   49   97  161  193  241  289  321  385  433
   1   97  193  321  385  481  577  641  769  865
The Thue-Morse sequence A010059 begins with 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, from which we see that the first 4 terms (=1,0,0,1) occur at positions 1, 7, 13, ..., as indicated for row n=2.
		

Crossrefs

Cf. A010059 (Thue-Morse), A026147 (row 0), A091855 (row 1?), A157971 (row 2?),
Column 1 is essentially A004119 (or A181565).

Programs

  • Maple
    A010060 := proc(n)
        local i;
        add(i, i=convert(n, base, 2)) mod 2 ;
    end proc:
    A010059 := proc(n)
        1-A010060(n) ;
    end proc:
    A248416Off0 := proc(n,k)
        option remember ;
        local strtN,binpat,src,thue ;
        if k = 1 then
            strtN := 0 ;
        else
            strtN := 1+procname(n,k-1) ;
        end if;
        binpat := [seq(A010059(i),i=0..n-1)] ;
        for src from strtN do
            thue := [seq(A010059(i),i=src..src+nops(binpat)-1)] ;
            if binpat=thue then
                return src ;
            end if;
        end do:
    end proc:
    A248416 := proc(n,k)
        1+A248416Off0(2^n,k) ;
    end proc:
    for d from 1 to 11 do
        for k from d to 1 by -1 do
            printf("%d,",A248416(d-k,k)) ;
    end do: # R. J. Mathar, Nov 06 2018
  • Mathematica
    z = 3000; u = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {1, 0}}] &, {0}, 20]; Length[u]
    t[p_, q_] := t[p, q] = Table[u[[k]], {k, p, q}];
    r[n_] := Select[Range[z], t[#, # + 2^(n - 1)] == t[1, 1 + 2^(n - 1)] &]
    TableForm[Table[r[n], {n, 0, 10}]]

Extensions

Definitions and examples clarified. - R. J. Mathar, Nov 06 2018

A259865 Exponent of 2 modulo the prime A039687(n).

Original entry on oeis.org

3, 12, 48, 96, 384, 6144, 393216, 805306368, 103079215104, 549755813888, 110680464442257309696, 1176956575385002643219210516851437453019191645837006471168
Offset: 1

Views

Author

N. J. A. Sloane, Jul 08 2015

Keywords

Comments

Here we corrected a(10) which was 1649267441664 in Golomb's article. - R. J. Mathar, Jul 14 2015

Crossrefs

Cf. A039687.

Programs

A259867 Exponent of 2 modulo the prime A056807(n).

Original entry on oeis.org

5, 1500, 234375, 300000000, 7500000000000000000000000000, 750000000000000000000000000000000000, 1500000000000000000000000000000000000000000000000000000000000000000
Offset: 1

Views

Author

N. J. A. Sloane, Jul 08 2015

Keywords

Crossrefs

Cf. A056807.

Programs

A376102 Array read by ascending antidiagonals: A(n,k) = k*2^(n+1) + 1.

Original entry on oeis.org

1, 1, 3, 1, 5, 5, 1, 9, 9, 7, 1, 17, 17, 13, 9, 1, 33, 33, 25, 17, 11, 1, 65, 65, 49, 33, 21, 13, 1, 129, 129, 97, 65, 41, 25, 15, 1, 257, 257, 193, 129, 81, 49, 29, 17, 1, 513, 513, 385, 257, 161, 97, 57, 33, 19, 1, 1025, 1025, 769, 513, 321, 193, 113, 65, 37, 21
Offset: 0

Views

Author

Stefano Spezia, Sep 14 2024

Keywords

Comments

In 1747, Euler showed that any factor of a Fermat number A000215(n) is of the form k*2^(n+1) + 1. See Wells at p. 148.

Examples

			The array begins as:
  1,   3,   5,   7,   9,  11,  13, ...
  1,   5,   9,  13,  17,  21,  25, ...
  1,   9,  17,  25,  33,  41,  49, ...
  1,  17,  33,  49,  65,  81,  97, ...
  1,  33,  65,  97, 129, 161, 193, ...
  1,  65, 129, 193, 257, 321, 385, ...
  1, 129, 257, 385, 513, 641, 769, ...
  ...
		

References

  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See pp. 70-71, 237-242.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 136.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987.

Crossrefs

Cf. A000012 (k=0), A000051, A000337, A004119, A005408 (n=0), A016813 (n=1), A017077 (n=2), A158057 (n=3).

Programs

  • Mathematica
    A[n_,k_]:=k*2^(n+1)+1; Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten

Formula

G.f.: (1 - 2*x + y)/((1 - x)*(1 - 2*x)*(1 - y)^2).
E.g.f.: exp(x+y)*(1 + 2*exp(x)*y).
Sum_{0<=k<=n} A(n-k,k) = A000295(n+2).
A(n,1) = A000051(n+1).
A(n,3) = A004119(n+2).
A(n,n) = A000337(n+1).
Previous Showing 11-19 of 19 results.