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.

A066601 a(n) = 3^n mod n.

Original entry on oeis.org

0, 1, 0, 1, 3, 3, 3, 1, 0, 9, 3, 9, 3, 9, 12, 1, 3, 9, 3, 1, 6, 9, 3, 9, 18, 9, 0, 25, 3, 9, 3, 1, 27, 9, 12, 9, 3, 9, 27, 1, 3, 15, 3, 37, 18, 9, 3, 33, 31, 49, 27, 29, 3, 27, 12, 9, 27, 9, 3, 21, 3, 9, 27, 1, 48, 3, 3, 13, 27, 39, 3, 9, 3, 9, 57
Offset: 1

Views

Author

Amarnath Murthy, Dec 22 2001

Keywords

Examples

			a(7) = 3 as 3^7 = 2187 = 7*312 + 3.
		

Crossrefs

Cf. k^n mod n: A015910 (k=2), this sequence (k=3), A066602 (k=4), A066603 (k=5), A066604 (k=6), A066438 (k=7), A066439 (k=8), A066440 (k=9), A056969 (k=10), A066441 (k=11), A066442 (k=12), A116609 (k=13).

Programs

  • Maple
    seq(irem(3^n,n),n=1..75); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Table[PowerMod[3, n, n], {n, 75}]
  • PARI
    a(n) = { lift(Mod(3, n)^n) } \\ Harry J. Smith, Mar 09 2010
    
  • Python
    def A066601(n): return pow(3,n,n) # Chai Wah Wu, Aug 24 2023

Extensions

More terms from Robert G. Wilson v, Dec 27 2001

A056969 a(n) = 10^n modulo n.

Original entry on oeis.org

0, 0, 1, 0, 0, 4, 3, 0, 1, 0, 10, 4, 10, 2, 10, 0, 10, 10, 10, 0, 13, 12, 10, 16, 0, 22, 1, 4, 10, 10, 10, 0, 10, 32, 5, 28, 10, 24, 25, 0, 10, 22, 10, 12, 10, 8, 10, 16, 31, 0, 31, 16, 10, 28, 10, 16, 31, 42, 10, 40, 10, 38, 55, 0, 30, 34, 10, 4, 34, 60, 10, 64, 10, 26, 25, 44, 54, 40
Offset: 1

Views

Author

Henry Bottomley, Jul 20 2000

Keywords

Examples

			a(7) = 3 since 10000000 = 7*1428571+3
		

Crossrefs

Cf. k^n mod n: A015910 (k=2), A066601 (k=3), A066602 (k=4), A066603 (k=5), A066604 (k=6), A066438 (k=7), A066439 (k=8), A066440 (k=9), this sequence (k=10), A066441 (k=11), A066442 (k=12), A116609 (k=13).

Programs

  • Maple
    seq(irem(10^n,n),n=1..78); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Table[PowerMod[10, n, n], {n, 80} ]
  • PARI
    a(n) = lift(Mod(10, n)^n); \\ Michel Marcus, Oct 19 2017

Formula

a(n) = 10*A056968(n) mod n = A011557(n) mod n.

A066438 a(n) = 7^n mod n.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 0, 1, 1, 9, 7, 1, 7, 7, 13, 1, 7, 1, 7, 1, 7, 5, 7, 1, 7, 23, 1, 21, 7, 19, 7, 1, 13, 15, 28, 1, 7, 11, 31, 1, 7, 7, 7, 25, 37, 3, 7, 1, 0, 49, 37, 9, 7, 1, 43, 49, 1, 49, 7, 1, 7, 49, 28, 1, 37, 37, 7, 21, 67, 49, 7, 1, 7, 49, 43, 45, 28, 25, 7, 1
Offset: 1

Views

Author

Robert G. Wilson v, Dec 27 2001

Keywords

Crossrefs

Cf. k^n mod n; A015910 (k=2), A066601 (k=3), A066602 (k=4), A066603 (k=5), A066604 (k=6), this sequence (k=7), A066439 (k=8), A066440 (k=9), A056969 (k=10), A066441 (k=11), A066442 (k=12), A116609 (k=13).

Programs

  • Maple
    seq(irem(7^n,n),n=1..80); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Table[PowerMod[7, n, n], {n, 80} ]
  • PARI
    a(n) = { lift(Mod(7, n)^n) } \\ Harry J. Smith, Feb 14 2010

A066603 a(n) = 5^n mod n.

Original entry on oeis.org

0, 1, 2, 1, 0, 1, 5, 1, 8, 5, 5, 1, 5, 11, 5, 1, 5, 1, 5, 5, 20, 3, 5, 1, 0, 25, 26, 9, 5, 25, 5, 1, 26, 25, 10, 1, 5, 25, 8, 25, 5, 1, 5, 9, 35, 25, 5, 1, 19, 25, 23, 1, 5, 1, 45, 25, 11, 25, 5, 25, 5, 25, 62, 1, 5, 49, 5, 13, 56, 65, 5, 1, 5, 25, 50, 17, 3, 25, 5, 65, 80, 25, 5, 1, 65
Offset: 1

Views

Author

Amarnath Murthy, Dec 22 2001

Keywords

Examples

			a(7) = 5 as 5^7 = 78125 = 7*11160 + 5.
		

Crossrefs

Cf. k^n mod n: A015910 (k=2), A066601 (k=3), A066602 (k=4), this sequence (k=5), A066604 (k=6), A066438 (k=7), A066439 (k=8), A066440 (k=9), A056969 (k=10), A066441 (k=11), A066442 (k=12), A116609 (k=13).

Programs

  • Maple
    seq(irem(5^n,n),n=1..85); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Table[PowerMod[5, n, n], {n, 85} ]
  • PARI
    a(n) = { lift(Mod(5, n)^n) } \\ Harry J. Smith, Mar 09 2010

Extensions

More terms from Robert G. Wilson v, Dec 27 2001

A066441 a(n) = 11^n mod n.

Original entry on oeis.org

0, 1, 2, 1, 1, 1, 4, 1, 8, 1, 0, 1, 11, 9, 11, 1, 11, 1, 11, 1, 8, 11, 11, 1, 1, 17, 26, 25, 11, 1, 11, 1, 11, 19, 16, 1, 11, 7, 5, 1, 11, 1, 11, 33, 26, 29, 11, 1, 18, 1, 5, 29, 11, 1, 11, 9, 20, 5, 11, 1, 11, 59, 8, 1, 46, 55, 11, 21, 20, 11, 11, 1, 11, 47, 26, 49, 44, 25, 11, 1
Offset: 1

Views

Author

Robert G. Wilson v, Dec 27 2001

Keywords

Crossrefs

Cf. k^n mod n: A015910 (k=2), A066601 (k=3), A066602 (k=4), A066603 (k=5), A066604 (k=6), A066438 (k=7), A066439 (k=8), A066440 (k=9), A056969 (k=10), this sequence (k=11), A066442 (k=12), A116609 (k=13).

Programs

  • Maple
    seq(irem(11^n,n),n=1..80); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Table[PowerMod[11, n, n], {n, 80} ]
  • PARI
    a(n) = { lift(Mod(11, n)^n) } \\ Harry J. Smith, Feb 14 2010

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

A114448 Array a(n,k) = n^k (mod k) read by antidiagonals (k>=1, n>=1).

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 2, 1, 0, 0, 2, 0, 3, 4, 1, 0, 1, 0, 1, 4, 3, 2, 1, 0, 0, 1, 0, 0, 4, 3, 0, 1, 0, 1, 2, 1, 1, 1, 4, 1, 8, 1, 0, 0, 0, 0, 2, 0, 5, 0, 0, 4, 1, 0, 1, 1, 1, 3, 1, 6, 1, 1, 9, 2, 1, 0, 0, 2, 0, 4, 4, 0, 0, 8, 6, 3, 4, 1, 0, 1, 0, 1, 0, 3, 1, 1, 0, 5, 4, 9, 2, 1
Offset: 1

Views

Author

Leroy Quet, Feb 14 2006

Keywords

Comments

Alternate description: triangular array a(n, k) = n^k (mod k) read by rows (n > 1, 0 < k < n). This is equivalent because a(n, k) = a(n-k, k). - David Wasserman, Jan 25 2007

Examples

			2^6 = 64 and 64 (mod 6) is 4. So a(2,6) = 4.
		

Crossrefs

Programs

  • Mathematica
    a[n_, k_] := Mod[n^k, k]; Table[a[n - k + 1, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 12 2012 *)

Extensions

More terms from David Wasserman, Jan 25 2007
Showing 1-7 of 7 results.