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 1-5 of 5 results.

A047361 Numbers that are congruent to {0, 1, 2, 3} mod 7.

Original entry on oeis.org

0, 1, 2, 3, 7, 8, 9, 10, 14, 15, 16, 17, 21, 22, 23, 24, 28, 29, 30, 31, 35, 36, 37, 38, 42, 43, 44, 45, 49, 50, 51, 52, 56, 57, 58, 59, 63, 64, 65, 66, 70, 71, 72, 73, 77, 78, 79, 80, 84, 85, 86, 87, 91, 92, 93, 94, 98, 99, 100, 101, 105, 106, 107, 108, 112
Offset: 1

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Author

Keywords

Comments

Nonnegative m for which floor(2*m/7) = 2*floor(m/7). [Bruno Berselli, Dec 03 2015]

Crossrefs

Programs

  • Magma
    [n : n in [0..150] | n mod 7 in [0..3]]; // Wesley Ivan Hurt, May 23 2016
  • Maple
    A047361:=n->(14*n-23-3*I^(2*n)-(3-3*I)*I^(-n)-(3+3*I)*I^n)/8: seq(A047361(n), n=1..100); # Wesley Ivan Hurt, May 23 2016
  • Mathematica
    Flatten[#+{0,1,2,3}&/@(7*Range[0,20])] (* Harvey P. Dale, Jan 17 2013 *)
  • PARI
    concat(0, Vec(x^2*(1+x+x^2+4*x^3)/((1+x)*(x^2+1)*(x-1)^2) + O(x^100))) \\ Altug Alkan, Dec 09 2015
    

Formula

a(n) = 7*floor(n/4) + (n mod 4), with offset 0 and a(0) = 0. - Gary Detlefs, Mar 09 2010
G.f.: x^2*(1+x+x^2+4*x^3) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Dec 04 2011
From Wesley Ivan Hurt, May 23 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (14*n-23-3*i^(2*n)-(3-3*i)*i^(-n)-(3+3*i)*i^n)/8, where i=sqrt(-1).
a(2k) = A047356(k), a(2k-1) = A047352(k). (End)
E.g.f.: (16 + 3*(sin(x) - cos(x)) + (7*x - 10)*sinh(x) + (7*x - 13)*cosh(x))/4. - Ilya Gutkovskiy, May 24 2016

Extensions

More terms from Wesley Ivan Hurt, May 23 2016

A047286 Numbers that are congruent to {1, 2, 3, 6} mod 7.

Original entry on oeis.org

1, 2, 3, 6, 8, 9, 10, 13, 15, 16, 17, 20, 22, 23, 24, 27, 29, 30, 31, 34, 36, 37, 38, 41, 43, 44, 45, 48, 50, 51, 52, 55, 57, 58, 59, 62, 64, 65, 66, 69, 71, 72, 73, 76, 78, 79, 80, 83, 85, 86, 87, 90, 92, 93, 94, 97, 99, 100, 101, 104, 106, 107, 108, 111
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n : n in [0..150] | n mod 7 in [1, 2, 3, 6]]; // Wesley Ivan Hurt, May 22 2016
  • Maple
    A047286:=n->(14*n-11+I^(2*n)+(1+3*I)*I^(-n)+(1-3*I)*I^n)/8: seq(A047286(n), n=1..100); # Wesley Ivan Hurt, May 22 2016
  • Mathematica
    Table[(14n-11+I^(2n)+(1+3I)*I^(-n)+(1-3I)*I^n)/8, {n, 80}] (* Wesley Ivan Hurt, May 22 2016 *)
    LinearRecurrence[{1, 0, 0, 1, -1}, {1, 2, 3, 6, 8}, 80] (* Vincenzo Librandi, May 24 2016 *)

Formula

G.f.: x*(1+x+x^2+3*x^3+x^4) / ( (1+x)*(1+x^2)*(x-1)^2 ). - R. J. Mathar, Oct 25 2011
From Wesley Ivan Hurt, May 22 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (14n-11+i^(2n)+(1+3i)*i^(-n)+(1-3i)*i^n)/8 where i=sqrt(-1).
a(2n) = A047276(n), a(2n-1) = A047356(n). (End)
E.g.f.: (4 + 3*sin(x) + cos(x) + (7*x - 6)*sinh(x) + (7*x - 5)*cosh(x))/4. - Ilya Gutkovskiy, May 23 2016

Extensions

More terms from Wesley Ivan Hurt, May 22 2016

A047338 Numbers that are congruent to {1, 2, 3, 4} mod 7.

Original entry on oeis.org

1, 2, 3, 4, 8, 9, 10, 11, 15, 16, 17, 18, 22, 23, 24, 25, 29, 30, 31, 32, 36, 37, 38, 39, 43, 44, 45, 46, 50, 51, 52, 53, 57, 58, 59, 60, 64, 65, 66, 67, 71, 72, 73, 74, 78, 79, 80, 81, 85, 86, 87, 88, 92, 93, 94, 95, 99, 100, 101, 102, 106, 107, 108, 109
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n : n in [0..150] | n mod 7 in [1, 2, 3, 4]]; // Wesley Ivan Hurt, May 23 2016
  • Maple
    A047338:=n->(14*n-15-3*I^(2*n)-(3-3*I)*I^(-n)-(3+3*I)*I^n)/8: seq(A047338(n), n=1..100); # Wesley Ivan Hurt, May 23 2016
  • Mathematica
    Table[(14n-15-3*I^(2n)-(3-3*I)*I^(-n)-(3+3*I)*I^n)/8, {n, 80}] (* Wesley Ivan Hurt, May 23 2016 *)

Formula

G.f.: x*(1+x+x^2+x^3+3*x^4) / ( (1+x)*(1+x^2)*(x-1)^2 ). - R. J. Mathar, Dec 04 2011
From Wesley Ivan Hurt, May 23 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (14n-15-3*i^(2n)-(3-3*i)*i^(-n)-(3+3*i)*i^n)/8 where i=sqrt(-1).
a(2n) = A047348(n), a(2n-1) = A047356(n). (End)
E.g.f.: (12 + 3*(sin(x) - cos(x)) + (7*x - 6)*sinh(x) + (7*x - 9)*cosh(x))/4. - Ilya Gutkovskiy, May 24 2016

Extensions

More terms from Wesley Ivan Hurt, May 23 2016

A047372 Numbers that are congruent to {1, 2, 3, 5} mod 7.

Original entry on oeis.org

1, 2, 3, 5, 8, 9, 10, 12, 15, 16, 17, 19, 22, 23, 24, 26, 29, 30, 31, 33, 36, 37, 38, 40, 43, 44, 45, 47, 50, 51, 52, 54, 57, 58, 59, 61, 64, 65, 66, 68, 71, 72, 73, 75, 78, 79, 80, 82, 85, 86, 87, 89, 92, 93, 94, 96, 99, 100, 101, 103, 106, 107, 108, 110
Offset: 1

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Crossrefs

Programs

Formula

From Bruno Berselli, Dec 01 2010: (Start)
G.f.: x*(1+x+x^2+2*x^3+2*x^4) / ((1-x)^2*(1+x+x^2+x^3)).
a(n) = (14*n+(3*i-1)*(-i)^n-(3*i+1)*i^n-(-1)^n-13)/8, i=sqrt(-1). (End)
From Wesley Ivan Hurt, Jun 04 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(2k) = A047385(k), a(2k-1) = A047356(k). (End)
E.g.f.: (8 + 3*sin(x) - cos(x) + (7*x - 6)*sinh(x) + 7*(x - 1)*cosh(x))/4. - Ilya Gutkovskiy, Jun 04 2016

A065450 Make an infinite chessboard from the squares in the first quadrant; sequence gives number of squares a knight can reach in n moves starting at the origin.

Original entry on oeis.org

1, 2, 10, 22, 37, 54, 76, 100, 129, 160, 196, 234, 277, 322, 372, 424, 481, 540, 604, 670, 741, 814, 892, 972, 1057, 1144, 1236, 1330, 1429, 1530, 1636, 1744, 1857, 1972, 2092, 2214, 2341, 2470, 2604, 2740, 2881, 3024, 3172, 3322, 3477, 3634, 3796, 3960
Offset: 0

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Author

Bodo Zinser, Nov 18 2001

Keywords

Comments

The first conjecture is true: Partial sums of A047356 = b(n) = (14*(n*(n+1)) + 2*n + 5 + 3*(-1)^n )/8, since A047356(n)=(14*n+1+3*(-1)^n)/4. And b(n) has g.f. (4*x^2 + 2*x + 1)/(-x^4 + 2*x^3 - 2*x + 1). The difference a(n) - b(n) = 0,2,2,0,0,0,0,0,0..., which has g.f. 2*x^2 + 2*x. Then (4*x^2 + 2*x + 1)/(-x^4 + 2*x^3 - 2*x + 1) + 2*x^2 + 2*x = (-2*x^6 + 2*x^5 + 4*x^4 - 4*x^3 + 2*x^2 + 4*x + 1)/(-x^4 + 2*x^3 - 2*x + 1). - Vim Wenders, Apr 16 2008

Crossrefs

Cf. A098498.

Formula

Conjectures: G.f.: [1+6x^2+4x^3-4x^4-2x^5+2x^6]/[(1+x)*(1-x)^3]. For n>3, partial sums of A047356. - Ralf Stephan, Mar 06 2004
The second conjecture "For n>3, partial sums of A047356" is also true. From the last possible move, we can either move back to the second last possible move or to b(n)=A047883(n) new squares. So a(n) = a(n-2)+b(n). For n>6, b(n)=7(n-1)+4=A017029(n-1). But a number of the form 7n+4 is naturally the sum of two consecutive terms in A047356 (4=1+3,11=3+8,18=8+10, ...). The conjecture follows. - Vim Wenders, Apr 12 2008

Extensions

More terms from Don Reble, Nov 28 2001
Showing 1-5 of 5 results.