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.

A047517 Numbers that are congruent to {0, 1, 3, 4, 6, 7} mod 8.

Original entry on oeis.org

0, 1, 3, 4, 6, 7, 8, 9, 11, 12, 14, 15, 16, 17, 19, 20, 22, 23, 24, 25, 27, 28, 30, 31, 32, 33, 35, 36, 38, 39, 40, 41, 43, 44, 46, 47, 48, 49, 51, 52, 54, 55, 56, 57, 59, 60, 62, 63, 64, 65, 67, 68, 70, 71, 72, 73, 75, 76, 78, 79, 80, 81, 83, 84, 86, 87, 88, 89
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A047428, A047585.

Programs

  • Magma
    [n: n in [0..110] | n mod 8 in [0,1,3,4,6,7]]; // Vincenzo Librandi, May 30 2016
  • Maple
    A047424:=n->(24*n-21-3*cos(n*Pi)+2*sqrt(3)*cos((1+4*n)*Pi/6)+6*sin((1-2*n)* Pi/6))/18: seq(A047424(n), n=1..100); # Wesley Ivan Hurt, Jun 16 2016
  • Mathematica
    LinearRecurrence[{1, 0, 0, 0, 0, 1, -1}, {0, 1, 3, 4, 6, 7, 8}, 50] (* G. C. Greubel, May 30 2016 *)
Select[Range[0,200], MemberQ[{0, 1, 3, 4, 6, 7}, Mod[#, 8]] &] (* Vincenzo Librandi, May 30 2016 *)
  • PARI
    my(x='x+O('x^50)); concat([0], Vec(x^2*(x^5 + x^4 + 2*x^3 + x^2 + 2*x + 1)/(x^7 - x^6 - x + 1))) \\ G. C. Greubel, Oct 29 2017

Formula

From Chai Wah Wu, May 30 2016: (Start)
a(n) = a(n-1) + a(n-6) - a(n-7), for n > 7.
G.f.: x^2*(x^5 + x^4 + 2*x^3 + x^2 + 2*x + 1)/(x^7 - x^6 - x + 1). (End)
From Wesley Ivan Hurt, Jun 16 2016: (Start)
a(n) = (24*n-21-3*cos(n*Pi)+2*sqrt(3)*cos((1+4*n)*Pi/6)+6*sin((1-2*n)*Pi/6))/18.
a(6k) = 8k-1, a(6k-1) = 8k-2, a(6k-2) = 8k-4, a(6k-3) = 8k-5, a(6k-4) = 8k-7, a(6k-5) = 8k-8. (End)
Sum_{n>=2} (-1)^n/a(n) = (2-sqrt(2))*Pi/16 + (6-3*sqrt(2))*log(2)/16 + 3*sqrt(2)*log(sqrt(2)+2)/8. - Amiram Eldar, Dec 27 2021

A047503 Numbers that are congruent to {0, 2, 3, 4, 5, 7} mod 8.

Original entry on oeis.org

0, 2, 3, 4, 5, 7, 8, 10, 11, 12, 13, 15, 16, 18, 19, 20, 21, 23, 24, 26, 27, 28, 29, 31, 32, 34, 35, 36, 37, 39, 40, 42, 43, 44, 45, 47, 48, 50, 51, 52, 53, 55, 56, 58, 59, 60, 61, 63, 64, 66, 67, 68, 69, 71, 72, 74, 75, 76, 77, 79, 80, 82, 83, 84, 85, 87
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A047424, A047587.

Programs

  • Magma
    [n : n in [0..100] | n mod 8 in [0, 2, 3, 4, 5, 7]]; // Wesley Ivan Hurt, Jun 16 2016
  • Maple
    A047503:=n->(24*n-21+3*cos(n*Pi)+2*sqrt(3)*cos((1+4*n)*Pi/6)-6*sin((1-2*n)*Pi/6))/18: seq(A047503(n), n=1..100); # Wesley Ivan Hurt, Jun 16 2016
  • Mathematica
    Select[Range[0, 100], MemberQ[{0, 2, 3, 4, 5, 7}, Mod[#, 8]] &] (* Wesley Ivan Hurt, Jun 16 2016 *)
LinearRecurrence[{1,0,0,0,0,1,-1},{0,2,3,4,5,7,8},100] (* Harvey P. Dale, Dec 25 2023 *)

Formula

G.f.: x^2*(2+x+x^2+x^3+2*x^4+x^5) / ( (1+x)*(1+x+x^2)*(x^2-x+1)*(x-1)^2 ). - R. J. Mathar, Nov 06 2015
From Wesley Ivan Hurt, Jun 16 2016: (Start)
a(n) = a(n-1) + a(n-6) - a(n-7) for n>7.
a(n) = (24*n-21+3*cos(n*Pi)+2*sqrt(3)*cos((1+4*n)*Pi/6)-6*sin((1-2*n)*Pi/6))/18.
a(6k) = 8k-1, a(6k-1) = 8k-3, a(6k-2) = 8k-4, a(6k-3) = 8k-5, a(6k-4) = 8k-6, a(6k-5) = 8k-8. (End)
Sum_{n>=2} (-1)^n/a(n) = log(2)/8 - sqrt(2)*Pi/16 - sqrt(2)*log(99-70*sqrt(2))/16. - Amiram Eldar, Dec 27 2021

A180636 Positive integers that are divisible by neither 8k-1 nor 8k+1, for all k > 0.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 10, 11, 12, 13, 16, 19, 20, 22, 24, 26, 29, 32, 37, 38, 40, 43, 44, 48, 52, 53, 58, 59, 61, 64, 67, 74, 76, 80, 83, 86, 88, 96, 101, 104, 106, 107, 109, 116, 118, 122, 128, 131, 134, 139, 148, 149, 152, 157, 160, 163, 166, 172, 173, 176, 179, 181, 192
Offset: 1

Views

Author

Glenn G. Chappell, Sep 13 2010

Keywords

Links

Crossrefs

Cf. A047424. - Robert G. Wilson v, Oct 06 2010

Programs

  • Mathematica
    fQ[n_] := Union[ MemberQ[{1, 7}, # ] & /@ Union@ Mod[ Rest@ Divisors@ n, 8]] == {False}; fQ[1] = True; Select[ Range@ 200, fQ] (* Robert G. Wilson v, Oct 06 2010 *)
  • Python
    # Works in Python 2 or 3
import itertools
for n in itertools.count(1):
    for k in range(1, 2+n//8):
        if n%(8*k-1)==0 or n%(8*k+1)==0:
            break
    else:
        print(n)

Extensions

More terms from Robert G. Wilson v, Oct 06 2010
Showing 1-3 of 3 results.

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