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.

A047263 Numbers that are congruent to {0, 1, 2, 4, 5} mod 6.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 64, 65, 66, 67, 68, 70, 71, 72, 73, 74, 76, 77, 78, 79, 80, 82, 83, 84, 85
Offset: 1

Views

Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Complement of A016945. - R. J. Mathar, Feb 25 2008
Nonnegative integers m such that floor(2*m^2/12) = 2*floor(m^2/12). See the Crossrefs field of A265187 for similar sequences. - Bruno Berselli, Dec 08 2015
Also, numbers k such that Fibonacci(k) mod 4 = 0, 1 or 3. - Bruno Berselli, Oct 17 2017

Links

Crossrefs

Cf. A016945, A265187.

Programs

  • Magma
    [n : n in [0..100] | n mod 6 in [0, 1, 2, 4, 5]]; // Wesley Ivan Hurt, Aug 16 2016
  • Maple
    for n from 0 to 200 do if n mod 6 <> 3 then printf(`%d,`,n) fi: od:
A047263:=n->6*floor(n/5)+[0, 1, 2, 4, 5][(n mod 5)+1]: seq(A047263(n), n=0..100); # Wesley Ivan Hurt, Aug 16 2016
  • Mathematica
    Select[Range[0,100], Mod[#,6]!=3&] (* Harvey P. Dale, May 17 2011 *)
LinearRecurrence[{1,0,0,0,1,-1},{0,1,2,4,5,6},90] (* Harvey P. Dale, Oct 05 2014 *)
  • PARI
    first(n) = {select(x->(x%6!=3), vector(6*n\5, i, i-1))} \\ David A. Corneth, Oct 17 2017

Formula

O.g.f.: x*(x^2+1)*(x^2+x+1)/((x-1)^2*(x^4+x^3+x^2+x+1)). - R. J. Mathar, Feb 25 2008
a(n) = a(n-5) + 6 for n > 5. - R. J. Mathar, Feb 25 2008
a(n) = a(n-1) + a(n-5) - a(n-6) for n > 6. - Harvey P. Dale, Oct 05 2014
From Wesley Ivan Hurt, Aug 16 2016: (Start)
a(n) = n + floor((n-4)/5).
a(n) = (6*n - 4 - ((n+1) mod 5))/5.
a(5k) = 6k-1, a(5k-1) = 6k-2, a(5k-2) = 6k-4, a(5k-3) = 6k-5, a(5k-4) = 6k-6. (End)
Sum_{n>=2} (-1)^n/a(n) = log(2+sqrt(3))/sqrt(3) - log(2)/6. - Amiram Eldar, Dec 17 2021

Extensions

More terms from James Sellers, Feb 19 2001

A265188 Nonnegative m for which 3*floor(m^2/11) = floor(3*m^2/11).

Original entry on oeis.org

0, 1, 5, 6, 10, 11, 12, 16, 17, 21, 22, 23, 27, 28, 32, 33, 34, 38, 39, 43, 44, 45, 49, 50, 54, 55, 56, 60, 61, 65, 66, 67, 71, 72, 76, 77, 78, 82, 83, 87, 88, 89, 93, 94, 98, 99, 100, 104, 105, 109, 110, 111, 115, 116, 120, 121, 122, 126, 127, 131, 132, 133, 137, 138, 142
Offset: 1

Views

Author

Bruno Berselli, Dec 04 2015

Keywords

Comments

See the second comment in A265187.
Also, nonnegative m congruent to 0, 1, 5, 6 or 10 (mod 11).
Primes in sequence: 5, 11, 17, 23, 43, 61, 67, 71, 83, 89, 109, 127, ...

Links

Crossrefs

Cf. A265187.
Cf. similar sequences provided by 3*floor(n^2/h) = floor(3*n^2/h): A005843 (h=2), A008585 (h=3), A001477 (h=4), A008854 (h=5), A047266 (h=6), A047299 (h=7), A042965 (h=8), A265227 (h=9), A054967 (h=10), this sequence (h=11), A047266 (h=12).

Programs

  • Magma
    [n: n in [0..150] | 3*Floor(n^2/11) eq Floor(3*n^2/11)];
  • Mathematica
    Select[Range[0, 150], 3 Floor[#^2/11] == Floor[3 #^2/11] &]
Select[Range[0, 150], MemberQ[{0, 1, 5, 6, 10}, Mod[#, 11]] &]
LinearRecurrence[{1, 0, 0, 0, 1, -1}, {0, 1, 5, 6, 10, 11}, 70]
  • PARI
    is(n) = 3*(n^2\11) == (3*n^2)\11 \\ Anders Hellström, Dec 05 2015
  • PARI
    concat(0, Vec(x^2*(1 + 4*x + x^2 + 4*x^3 + x^4)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)) + O(x^100))) \\ Michel Marcus, Dec 05 2015
  • Sage
    [n for n in (0..150) if 3*floor(n^2/11) == floor(3*n^2/11)]

Formula

G.f.: x^2*(1 + 4*x + x^2 + 4*x^3 + x^4)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n-1) + a(n-5) - a(n-6), n>6.

A112654 Numbers k such that k^3 == k (mod 11).

Original entry on oeis.org

0, 1, 10, 11, 12, 21, 22, 23, 32, 33, 34, 43, 44, 45, 54, 55, 56, 65, 66, 67, 76, 77, 78, 87, 88, 89, 98, 99, 100, 109, 110, 111, 120, 121, 122, 131, 132, 133, 142, 143, 144, 153, 154, 155, 164, 165, 166, 175, 176, 177, 186, 187, 188, 197, 198, 199, 208, 209
Offset: 1

Views

Author

Jeremy Gardiner, Dec 28 2005

Keywords

Comments

Nonnegative integers m such that floor(k*m^2/11) = k*floor(m^2/11), where k can assume the values from 4 to 10. See the second comment in A265187. - Bruno Berselli, Dec 03 2015

Examples

			a(3) = 11 because 11^3 = 1331 == 0 (mod 11) and 11 == 0 (mod 11).

Links

Programs

  • Maple
    m = 11 for n = 1 to 300 if n^3 mod m = n mod m then print n; next n
  • Mathematica
    Select[Range@ 209, Mod[#, 11] == Mod[#^3, 11] &] (* Michael De Vlieger, Dec 03 2015 *)
Select[Range[0,250],PowerMod[#,3,11]==Mod[#,11]&] (* Harvey P. Dale, May 15 2016 *)

Formula

From Colin Barker, Apr 11 2012: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4).
G.f.: x^2*(1+9*x+x^2)/((1-x)^2*(1+x+x^2)). (End)
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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