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 1-6 of 6 results.

A047578 Numbers that are congruent to {2, 5, 6, 7} mod 8.

Original entry on oeis.org

2, 5, 6, 7, 10, 13, 14, 15, 18, 21, 22, 23, 26, 29, 30, 31, 34, 37, 38, 39, 42, 45, 46, 47, 50, 53, 54, 55, 58, 61, 62, 63, 66, 69, 70, 71, 74, 77, 78, 79, 82, 85, 86, 87, 90, 93, 94, 95, 98, 101, 102, 103, 106, 109, 110, 111, 114, 117, 118, 119, 122, 125
Offset: 1

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Author

Keywords

Crossrefs

Programs

Formula

G.f.: x*(1+x)*(x^2-x+2) / ((1+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
a(n) = 2*n - cos(Pi*n/2). - Wesley Ivan Hurt, Oct 22 2013
From Wesley Ivan Hurt, May 20 2016: (Start)
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4) for n > 4.
a(n) = (4*n - i^(-n) - i^n)/2 where i=sqrt(-1).
a(2n) = A047550(n), a(2n-1) = A016825(n-1). (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(2)*Pi/8 - log(2)/4. - Amiram Eldar, Dec 26 2021

Extensions

More terms from Wesley Ivan Hurt, May 20 2016

A214863 Numbers n such that n XOR 11 = n - 11.

Original entry on oeis.org

11, 15, 27, 31, 43, 47, 59, 63, 75, 79, 91, 95, 107, 111, 123, 127, 139, 143, 155, 159, 171, 175, 187, 191, 203, 207, 219, 223, 235, 239, 251, 255, 267, 271, 283, 287, 299, 303, 315, 319, 331, 335, 347, 351, 363
Offset: 1

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Author

Brad Clardy, Mar 09 2013

Keywords

Comments

Links to sequences of the form n XOR m = n - m are found below with the value of m specified.

Crossrefs

Cf. A005408 (m=1), A042964 (m=2), A131098 (m=3), A047566 (m=4), A047550 (m=5), A047589 (m=6), A004771 (m=7), A115419 (m=8), A214865 (m=9), A214864 (m=10), A133894 (m=12), A125169 (m=15).
Cf. also A016825, A168392.

Programs

  • Magma
    XOR := func;
    m:=11;
    for n in [1 .. 500] do
          if (XOR(n, m) eq n-m) then n; end if;
    end for;
  • Mathematica
    Select[Range[400],BitXor[#,11]==#-11&] (* or *) LinearRecurrence[{1,1,-1},{11,15,27},50] (* Harvey P. Dale, Jun 05 2021 *)

Formula

a(n)= 1+8*n-2*(-1)^n.
a(n)=A016825(n) + A168392(n) + for n>0.
G.f. x*(11+4*x+x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Mar 10 2013

A047575 Numbers that are congruent to {0, 5, 6, 7} mod 8.

Original entry on oeis.org

0, 5, 6, 7, 8, 13, 14, 15, 16, 21, 22, 23, 24, 29, 30, 31, 32, 37, 38, 39, 40, 45, 46, 47, 48, 53, 54, 55, 56, 61, 62, 63, 64, 69, 70, 71, 72, 77, 78, 79, 80, 85, 86, 87, 88, 93, 94, 95, 96, 101, 102, 103, 104, 109, 110, 111, 112, 117, 118, 119, 120, 125
Offset: 1

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Author

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Programs

  • Magma
    [n : n in [0..150] | n mod 8 in [0, 5, 6, 7]]; // Wesley Ivan Hurt, May 29 2016
  • Maple
    A047575:=n->(4*n-1+I^(2*n)-(1+I)*I^(-n)-(1-I)*I^n)/2: seq(A047575(n), n=1..100); # Wesley Ivan Hurt, May 29 2016
  • Mathematica
    Select[Range[0,120], MemberQ[{0,5,6,7}, Mod[#,8]]&] (* Harvey P. Dale, Jun 30 2011 *)

Formula

From Wesley Ivan Hurt, May 29 2016: (Start)
G.f.: x^2*(5+x+x^2+x^3) / ((x-1)^2*(1+x+x^2+x^3)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (4*n-1+i^(2*n)-(1+i)*i^(-n)-(1-i)*i^n)/2 where i=sqrt(-1).
a(2k) = A047550(k), a(2k-1) = A047451(k). (End)
E.g.f.: 1 - sin(x) - cos(x) - sinh(x) + 2*x*exp(x). - Ilya Gutkovskiy, May 30 2016
Sum_{n>=2} (-1)^n/a(n) = 5*log(2)/8 - (2*sqrt(2)-1)*Pi/16. - Amiram Eldar, Dec 23 2021

A047576 Numbers that are congruent to {1, 5, 6, 7} mod 8.

Original entry on oeis.org

1, 5, 6, 7, 9, 13, 14, 15, 17, 21, 22, 23, 25, 29, 30, 31, 33, 37, 38, 39, 41, 45, 46, 47, 49, 53, 54, 55, 57, 61, 62, 63, 65, 69, 70, 71, 73, 77, 78, 79, 81, 85, 86, 87, 89, 93, 94, 95, 97, 101, 102, 103, 105, 109, 110, 111, 113, 117, 118, 119, 121, 125
Offset: 1

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Author

Keywords

Crossrefs

Programs

Formula

From Wesley Ivan Hurt, May 29 2016: (Start)
G.f.: x*(1+4*x+x^2+x^3+x^4) / ((x-1)^2*(1+x+x^2+x^3)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (8*n-1+i^(2*n)-(2+i)*i^(-n)-(2-i)*i^n)/4 where i=sqrt(-1).
a(2k) = A047550(k), a(2k-1) = A047452(k). (End)
E.g.f.: (2 - sin(x) - 2*cos(x) - sinh(x) + 4*x*exp(x))/2. - Ilya Gutkovskiy, May 30 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*sqrt(2)*Pi/16 - (sqrt(2)+2)*log(2)/16 + sqrt(2)*log(sqrt(2)+2)/8. - Amiram Eldar, Dec 24 2021

A047582 Numbers that are congruent to {3, 5, 6, 7} mod 8.

Original entry on oeis.org

3, 5, 6, 7, 11, 13, 14, 15, 19, 21, 22, 23, 27, 29, 30, 31, 35, 37, 38, 39, 43, 45, 46, 47, 51, 53, 54, 55, 59, 61, 62, 63, 67, 69, 70, 71, 75, 77, 78, 79, 83, 85, 86, 87, 91, 93, 94, 95, 99, 101, 102, 103, 107, 109, 110, 111, 115, 117, 118, 119, 123, 125
Offset: 1

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Author

Keywords

Crossrefs

Programs

Formula

From Wesley Ivan Hurt, May 29 2016: (Start)
G.f.: x*(3+2*x+x^2+x^3+x^4) / ((x-1)^2*(1+x+x^2+x^3)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (8*n+1-i^(2*n)-(2-i)*i^(-n)-(2+i)*i^n)/4 where i=sqrt(-1).
a(2k) = A047550(k), a(2k-1) = A047398(k). (End)
E.g.f.: (2 + sin(x) - 2*cos(x) + sinh(x) + 4*x*exp(x))/2. - Ilya Gutkovskiy, May 30 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = (3*sqrt(2)-2)*Pi/16 - log(2)/8 + sqrt(2)*log(3-2*sqrt(2))/16. - Amiram Eldar, Dec 26 2021

A343007 Relative position of the average value between two consecutive partial sums of the Leibniz formula for Pi.

Original entry on oeis.org

6, 13, 26, 41, 62, 85, 114, 145, 182, 221, 266, 313, 366, 421, 482, 545, 614, 685, 762, 841, 926, 1013, 1106, 1201, 1302, 1405, 1514, 1625, 1742, 1861, 1986, 2113, 2246, 2381, 2522, 2665, 2814, 2965, 3122, 3281, 3446, 3613, 3786, 3961, 4142, 4325, 4514, 4705
Offset: 1

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Author

Raphael Ranna, Apr 02 2021

Keywords

Comments

Define L(n) to be the n-th partial sum of the Leibniz formula Pi = 4 - 4/3 + 4/5 - 4/7 + ..., i.e., L(n) = Sum_{j=1..n} 4*(-1)^(j+1)/(2*j-1). For every positive integer n, L(n+1) is closer to Pi than L(n) is. If we let V be the average of the two consecutive partial sums L(n) and L(n+1), then the partial sums that lie closest to V are L(a(n)-1) and L(a(n)+1) (one of which is above V, the other below).

Examples

			The first several partial sums are as follows:
  n      L(n)
  -  ------------
  1  4.0000000000
  2  2.6666666...
  3  3.4666666...
  4  2.8952380...
  5  3.3396825...
  6  2.9760461...
  7  3.2837384...
  8  3.0170718...
.
For n=1, the average of the partial sums L(1) and L(2) is V = (L(1) + L(2))/2 = (4 + 2.6666666...)/2 = 3.3333333...; the two partial sums closest to V are L(5)=3.3396825... and L(7)=3.2837384..., and V lies in the interval between them, so a(1)=6.
The formula as it is written works for all data in the sequence, but it needs to be proven that it works for all possible integer values of n.
		

Crossrefs

Programs

  • Mathematica
    Rest@ CoefficientList[Series[x (6 + x + x^3)/((1 + x) (1 - x)^3), {x, 0, 48}], x] (* Michael De Vlieger, Apr 05 2021 *)

Formula

a(1) = 6; a(n) = a(n-1) + r(n), where r(n) = A047550(n) = 4*n - (-1)^n.
G.f.: x*(6 + x + x^3)/((1 + x)*(1 - x)^3). - Jinyuan Wang, Apr 03 2021
From Stefano Spezia, Apr 03 2021: (Start)
a(n) = (3 + (-1)^(n+1) + 4*n + 4*n^2)/2.
a(2*n) = A102083(n).
a(2*n-1) = A254527(n). (End)
Showing 1-6 of 6 results.