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 1-7 of 7 results.

A128122 Numbers m such that 2^m == 6 (mod m).

Original entry on oeis.org

1, 2, 10669, 6611474, 43070220513807782
Offset: 1

Views

Author

Alexander Adamchuk, Feb 15 2007

Keywords

Comments

No other terms below 10^17. - Max Alekseyev, Nov 18 2022
A large term: 862*(2^861-3)/281437921287063162726198552345362315020202285185118249390789 (203 digits). - Max Alekseyev, Sep 24 2016

Examples

			2 == 6 (mod 1), so 1 is a term;
4 == 6 (mod 2), so 2 is a term.
		

Crossrefs

Solutions to 2^m == k (mod m): A000079 (k=0),A187787 (k=1/2), A296369 (k=-1/2), A006521 (k=-1), A296370 (k=3/2), A015919 (k=2), A006517 (k=-2), A050259 (k=3), A015940 (k=-3), A015921 (k=4), A244673 (k=-4), A128121 (k=5), A245318 (k=-5), this sequence (k=6), A245728 (k=-6), A033981 (k=7), A240941 (k=-7), A015922 (k=8), A245319 (k=-8), A051447 (k=9), A240942 (k=-9), A128123 (k=10), A245594 (k=-10), A033982 (k=11), A128124 (k=12), A051446 (k=13), A128125 (k=14), A033983 (k=15), A015924 (k=16), A124974 (k=17), A128126 (k=18), A125000 (k=19), A015925 (k=2^5), A015926 (k=2^6), A015927 (k=2^7), A015929 (k=2^8), A015931 (k=2^9), A015932 (k=2^10), A015935 (k=2^11), A015937 (k=2^12)

Programs

  • Mathematica
    m = 6; Join[Select[Range[m], Divisible[2^# - m, #] &],
    Select[Range[m + 1, 10^6], PowerMod[2, #, #] == m &]] (* Robert Price, Oct 08 2018 *)

Extensions

1 and 2 added by N. J. A. Sloane, Apr 23 2007
a(5) from Max Alekseyev, Nov 18 2022

A296369 Numbers m such that 2^m == -1/2 (mod m).

Original entry on oeis.org

1, 5, 65, 377, 1189, 1469, 25805, 58589, 134945, 137345, 170585, 272609, 285389, 420209, 538733, 592409, 618449, 680705, 778805, 1163065, 1520441, 1700945, 2099201, 2831009, 4020029, 4174169, 4516109, 5059889, 5215769
Offset: 1

Views

Author

Max Alekseyev, Dec 10 2017

Keywords

Comments

Equivalently, 2^(m+1) == -1 (mod m), or m divides 2^(m+1) + 1.
The sequence is infinite, see A055685.

Crossrefs

Solutions to 2^m == k (mod m): A296370 (k=3/2), A187787 (k=1/2), this sequence (k=-1/2), A000079 (k=0), A006521 (k=-1), A015919 (k=2), A006517 (k=-2), A050259 (k=3), A015940 (k=-3), A015921 (k=4), A244673 (k=-4), A128121 (k=5), A245318 (k=-5), A128122 (k=6), A245728 (k=-6), A033981 (k=7), A240941 (k=-7), A015922 (k=8), A245319 (k=-8), A051447 (k=9), A240942 (k=-9), A128123 (k=10), A245594 (k=-10), A033982 (k=11), A128124 (k=12), A051446 (k=13), A128125 (k=14), A033983 (k=15), A015924 (k=16), A124974 (k=17), A128126 (k=18), A125000 (k=19), A015925 (k=2^5), A015926 (k=2^6), A015927 (k=2^7), A015929 (k=2^8), A015931 (k=2^9), A015932 (k=2^10), A015935 (k=2^11), A015937 (k=2^12)

Programs

  • Mathematica
    Select[Range[10^5], Divisible[2^(# + 1) + 1, #] &] (* Robert Price, Oct 11 2018 *)
  • Python
    A296369_list = [n for n in range(1,10**6) if pow(2,n+1,n) == n-1] # Chai Wah Wu, Nov 04 2019

Formula

a(n) = A055685(n) - 1.

Extensions

Incorrect term 4285389 removed by Chai Wah Wu, Nov 04 2019

A251603 Numbers k such that k + 2 divides k^k - 2.

Original entry on oeis.org

3, 4551, 46775, 82503, 106976, 1642796, 4290771, 4492203, 4976427, 21537831, 21549347, 21879936, 51127259, 56786087, 60296571, 80837771, 87761787, 94424463, 96593696, 138644871, 168864999, 221395539, 255881451, 297460451, 305198247, 360306363, 562654203
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Dec 05 2014

Keywords

Comments

Numbers k such that (k^k - 2)/(k + 2) is an integer.
Since k == -2 (mod k+2), also numbers k such that k + 2 divides (-2)^k - 2. - Robert Israel, Jan 04 2015
Numbers k == 0 (mod 4) such that A066602(k/2+1) = 8, and odd numbers k such that k = 3 or A082493(k+2) = 8. - Robert Israel, Apr 08 2015

Examples

			3 is in this sequence because 3 + 2 = 5 divides 3^3 - 2 = 25.
		

Crossrefs

Programs

  • Magma
    [n: n in [0..10000] | Denominator((n^n-2)/(n+2)) eq 1];
    
  • Maple
    isA251603 := proc(n)
        if modp(n &^ n-2,n+2) = 0 then
            true;
        else
            false;
        end if;
    end proc:
    A251603 := proc(n)
        option remember;
        local a;
        if n = 1 then
            3;
        else
            for a from procname(n-1)+1 do
                if isA251603(a) then
                    return a;
                end if;
            end do:
        end if;
    end proc: # R. J. Mathar, Jan 09 2015
  • Mathematica
    Select[Range[10^6], Mod[PowerMod[#, #, # + 2] - 2, # + 2] == 0 &] (* Michael De Vlieger, Dec 20 2014, based on Robert G. Wilson v at A252041 *)
  • PARI
    for(n=1,10^9,if(Mod(n,n+2)^n==+2,print1(n,", "))); \\ Joerg Arndt, Dec 06 2014
    
  • Python
    A251603_list = [n for n in range(1,10**6) if pow(n, n, n+2) == 2] # Chai Wah Wu, Apr 13 2015

Formula

The even terms form A122711, the odd terms are those in A245319 (forming A357125) decreased by 2. - Max Alekseyev, Sep 22 2016

Extensions

a(6)-a(27) from Joerg Arndt, Dec 06 2014

A296370 Numbers m such that 2^m == 3/2 (mod m).

Original entry on oeis.org

1, 111481, 465793, 79036177, 1781269903307, 250369632905747, 708229497085909, 15673900819204067
Offset: 1

Views

Author

Max Alekseyev, Dec 11 2017

Keywords

Comments

Equivalently, 2^(m+1) == 3 (mod m).
Also, numbers m such that 2^(m+1) - 2 is a Fermat pseudoprime base 2, i.e., 2^(m+1) - 2 belongs to A015919 and A006935.
Some larger terms (may be not in order): 2338990834231272653581, 341569682872976768698011746141903924998969680637.

Crossrefs

Solutions to 2^m == k (mod m): this sequence (k=3/2), A187787 (k=1/2), A296369 (k=-1/2), A000079 (k=0), A006521 (k=-1), A015919 (k=2), A006517 (k=-2), A050259 (k=3), A015940 (k=-3), A015921 (k=4), A244673 (k=-4), A128121 (k=5), A245318 (k=-5), A128122 (k=6), A245728 (k=-6), A033981 (k=7), A240941 (k=-7), A015922 (k=8), A245319 (k=-8), A051447 (k=9), A240942 (k=-9), A128123 (k=10), A245594 (k=-10), A033982 (k=11), A128124 (k=12), A051446 (k=13), A128125 (k=14), A033983 (k=15), A015924 (k=16), A124974 (k=17), A128126 (k=18), A125000 (k=19), A015925 (k=2^5), A015926 (k=2^6), A015927 (k=2^7), A015929 (k=2^8), A015931 (k=2^9), A015932 (k=2^10), A015935 (k=2^11), A015937 (k=2^12)

Programs

  • Mathematica
    Select[Range[10^6], Divisible[2^(# + 1) - 3, #] &] (* Robert Price, Oct 11 2018 *)

Formula

a(n) = A296104(n) - 1.

A240942 Numbers k that divide 2^k + 9.

Original entry on oeis.org

1, 11, 121, 323, 117283, 432091, 4132384531, 15516834659, 15941429747, 98953554491, 3272831195051, 7362974489179, 26306805687881, 33869035218491, 280980898827691
Offset: 1

Views

Author

Derek Orr, Aug 04 2014

Keywords

Comments

No other terms below 10^15. Some larger terms: 53496121130110340001650284048539458491, 136243118444105327963550175410279542214992801356720577. - Max Alekseyev, Sep 29 2016

Examples

			2^11 + 9 = 2057 is divisible by 11. Thus 11 is a term of this sequence.
		

Crossrefs

Programs

  • Maple
    select(n -> 9 + 2 &^ n mod n = 0, [$1..10^6]); # Robert Israel, Aug 04 2014
  • PARI
    for(n=1,10^9, if(Mod(2,n)^n==-9, print1(n,", "); ); );

Extensions

a(7)-a(10) from Lars Blomberg, Nov 05 2014
a(11)-a(15) from Max Alekseyev, Sep 29 2016

A357125 Positive integers n such that 2^(n-3) == -1 (mod n).

Original entry on oeis.org

1, 5, 4553, 46777, 82505, 4290773, 4492205, 4976429, 21537833, 21549349, 51127261, 56786089, 60296573, 80837773, 87761789, 94424465, 138644873, 168865001, 221395541, 255881453, 297460453, 305198249, 360306365, 562654205, 635374253, 673867253, 808333573, 1164757553, 1210317349
Offset: 1

Views

Author

Max Alekseyev, Sep 13 2022

Keywords

Comments

Also, odd integers n dividing 2^n + 8.
Some large terms: 5603900696716667005, 446661376165868432471569407934747098747181600670953926245, 1533278864164902082788937853692280620552397221686019535813.

Crossrefs

The odd terms of A245319.

Programs

  • Mathematica
    Select[Range[2155*10^4],PowerMod[2,#-3,#]==#-1&]//Quiet (* The program generates the first 10 terms of the sequence. *) (* Harvey P. Dale, Feb 08 2025 *)

A334634 Numbers m that divide 2^m + 11.

Original entry on oeis.org

1, 13, 16043199041, 91118493923, 28047837698634913
Offset: 1

Views

Author

Max Alekseyev, Sep 10 2020

Keywords

Comments

Equivalently, numbers m such that 2^m == -11 (mod m).
No other terms below 10^17.

Crossrefs

Solutions to 2^n == k (mod n): A296370 (k=3/2), A187787 (k=1/2), A296369 (k=-1/2), A000079 (k=0), A006521 (k=-1), A015919 (k=2), A006517 (k=-2), A050259 (k=3), A015940 (k=-3), A015921 (k=4), A244673 (k=-4), A128121 (k=5), A245318 (k=-5), A128122 (k=6), A245728 (k=-6), A033981 (k=7), A240941 (k=-7), A015922 (k=8), A245319 (k=-8), A051447 (k=9), A240942 (k=-9), A128123 (k=10), A245594 (k=-10), A033982 (k=11), this sequence (k=-11), A128124 (k=12), A051446 (k=13), A128125 (k=14), A033983 (k=15), A015924 (k=16), A124974 (k=17), A128126 (k=18), A125000 (k=19), A015925 (k=2^5), A015926 (k=2^6), A015927 (k=2^7), A015929 (k=2^8), A015931 (k=2^9), A015932 (k=2^10), A015935 (k=2^11), A015937 (k=2^12).

Extensions

a(5) from Sergey Paramonov, Oct 10 2021
Showing 1-7 of 7 results.