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.

User: Colin Linzer

Colin Linzer's wiki page.

Colin Linzer has authored 11 sequences. Here are the ten most recent ones:

A377813 Decimal expansion of arctanh(phi-1).

Original entry on oeis.org

7, 2, 1, 8, 1, 7, 7, 3, 7, 5, 8, 9, 4, 0, 5, 1, 7, 1, 2, 4, 6, 6, 3, 8, 3, 7, 0, 1, 3, 6, 5, 5, 2, 6, 3, 4, 7, 0, 2, 7, 7, 6, 5, 0, 1, 5, 7, 8, 4, 9, 0, 7, 7, 9, 4, 9, 1, 5, 2, 7, 2, 5, 3, 2, 6, 0, 2, 4, 5, 8, 0, 1, 4, 1, 2, 3, 3
Offset: 0

Author

Colin Linzer, Nov 08 2024

Keywords

Comments

arctanh(phi-1) is the solution for real valued x in tanh(x) = d/dx tanh(x).
arctanh(phi-1) is the solution for real valued x in cosh(x) * sinh(x) = 1. - Colin Linzer, Nov 22 2024

Examples

			0.721817737589405171246638370136552634702...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[ArcTanh[GoldenRatio - 1], 10, 120][[1]] (* Amiram Eldar, Nov 12 2024 *)
  • PARI
    atanh((1+sqrt(5))/2-1)

Formula

Equals (1/2)*log(2+sqrt(5)).
Equals (3/2)*log(phi).
Equals arccosh(sqrt(phi)).
Equals arcsinh(sqrt(phi-1)).
Equals f(phi-1) with f(x) = (1/2)*log((2-x+2*sqrt(1-x))/x), a branch of the converse function of the derivative of tanh(x).
Equals 3*A202541. - Hugo Pfoertner, Nov 12 2024
Equals arcsinh(2)/2. - Colin Linzer, Nov 22 2024

A371627 The x-coordinate of the point where x + y = n, x is an integer and x/y is as close as possible to 1/phi.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 26, 26, 26, 27, 27, 27
Offset: 1

Author

Colin Linzer, Mar 29 2024

Keywords

Comments

Each term is equal to or one greater than the previous term.
The average run length approaches 1+phi.
a(n) = x = either ceiling or floor of n/phi^2, according to which minimizes abs(x/(n-x) - phi).
The 4 following statements are equivalent for any positive integer n and any function f(x) such that for all real x, x-1
a(n) != A371626(n);
A371625(n) != y(n);
a(n) != n-f(n/phi) xor A371626(n) != n-f(n/phi);
A371625(n) != f(n/phi) xor y(n) != f(n/phi).

Examples

			For n=4, the possibilities are (0,4), (1,3), (2,2), and (3,1). 1/3 is the closest to 1/phi out of them, so a(4)=1.
		

Crossrefs

Cf. A094214 (1/phi), A371625 (with phi).

Extensions

Elements referring to sequences that were not submitted removed by Peter Munn, Aug 04 2025

A371626 The y-coordinate of the point where x + y = n, x is an integer and x/y is as close as possible to phi (by absolute difference).

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 26, 26, 26, 27, 27, 28
Offset: 1

Author

Colin Linzer, Mar 29 2024

Keywords

Comments

a(n) = k = n - either ceiling or floor of n/phi, according to which minimizes abs((n-k)/k - phi).
Each term is equal to or one greater than the previous term.
The average run length approaches 1+phi.
The 2 following statements are equivalent for any positive integer n and any function f(x) such that for any real x, f(x) equals a integer within the range (x-1,x+1):
a(n) != A371627(n);
a(n) != n-f(n/phi) xor A371627(n) != n-f(n/phi).
Let s(n) = (phi*n - 1 - sqrt(1+(n^2)*(phi^-4)))/2.
Floor(s(n)) equals the number of times that a(n) swapped from being equal to n-floor(n/phi) to being equal to n-ceiling(n/phi) when n is extended to the reals.
This is true because s(n) is the solution to the equation n = (phi/4) * (phi(2r+1) + sqrt((2r+1)^2 / phi^4 + 4/phi)) solved for w. The equation gives the n-value of w-th swap from a(n) = n-floor(n/phi) to a(n) = n-ceiling(n/phi).
s(n) is asymptotic to n/phi - 1/2.
Floor(s(n)) != floor(n/phi - 1/2) <-> a(n) != round(n).
Floor(n/phi) equals the number of that a(n) swapped from being equal to n-ceiling(n/phi) to being equal to n-floor(n/phi) when n is extended to the reals.

Examples

			For n=5, the possibilities are (0,5), (1,4), (2,3), (3,2) and (4,1). Of those, 3/2 is the closest to phi, so a(5)=3.
		

Crossrefs

Cf. A001622 (phi), A371625 (x_coordinate).

Formula

a(n) = n - A371625(n).
Let s(n) = (phi*n - 1 - sqrt(1+(n^2) / phi^4))/2.
Floor(s(n))+floor(n/phi) is even -> a(n) = n-ceiling(n/phi) = (n mod 1) + floor(n/phi^2).
Floor(s(n))+floor(n/phi) is odd -> a(n) = n-floor(n/phi) = (n mod 1) + ceiling(n/phi^2).
a(n) = -a(-n).

Extensions

Elements referring to sequences that were not submitted removed by Peter Munn, Aug 04 2025

A371625 The x-coordinate of the point (x,y) where x + y = n, x is an integer, and x/y is as close as possible to phi (by absolute difference).

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 5, 5, 6, 7, 7, 8, 9, 9, 10, 10, 11, 12, 12, 13, 14, 14, 15, 15, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 23, 24, 25, 25, 26, 27, 27, 28, 28, 29, 30, 30, 31, 32, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 41, 42, 43
Offset: 1

Author

Colin Linzer, Mar 29 2024

Keywords

Comments

a(n) = x = either ceiling or floor of n/phi, according to which minimizes abs(x/(n-x) - phi).
Each term is equal to or one greater than the previous term.
The average run length approaches phi.
The 2 following statements are equivalent for any real n and any function f(x) such that for any real x, f(x) equals an integer within the range (x-1,x+1) (e.g., round(x), ceiling(x), floor(x)):
A371626(n) != A371627(n);
A371626(n) != n-f(n/phi) xor A371627(n) != n-f(n/phi).
Let s(n) = (phi*n - 1 - sqrt(1+(n^2)*(phi^-4)))/2.
Floor(s(n)) equals the number of times that a(n) swapped from being equal to floor(n/phi) to being equal to ceiling(n/phi) when n is extended to the reals.
This is true because s(n) is the solution to the equation n = (phi/4)(phi(2w+1)+sqrt((2w+1)^2 * phi^-4 + 4/phi)) solved for w. The equation gives the n-value of w-th swap from a(n) = floor(n/phi) to a(n) = ceiling(n/phi).
s(n) is asymptotic to n/phi - 1/2.
floor(s(n)) != floor(n/phi - 1/2) <-> a(n) != round(n).
Floor(n/phi) equals the number of times that a(n) swapped from being equal to ceiling(n/phi) to being equal to floor(n/phi) when n is extended to the reals.

Examples

			For n=5, the possibilities are (0,5), (1,4), (2,3), (3,2) and (4,1). Of those, 3/2 is the closest to phi, so a(5)=3.
		

Crossrefs

Cf. A001622 (phi), A371626 (y_coordinate), A371627 (with 1/phi), A002163 (sqrt(5)).

Formula

a(n) = n - A371626(n).
a(n) = ceiling(n/phi) if floor(s(n)) + floor(n/phi) is even.
a(n) = floor(n/phi) if floor(s(n)) + floor(n/phi) is odd.
a(n) = ceiling(n/phi) - (floor(s(n))+floor(n/phi) mod 2).
a(n) = round(n/phi) + floor(s(n)) - floor(n/phi+1/2).

A370892 Least positive integer k such that 4 numbers starting with k in an arithmetic progression with difference n have a solution in the game 24.

Original entry on oeis.org

3, 1, 1, 2, 1, 1, 1, 1, 2, 3, 2, 1, 1, 2, 2, 3, 2, 12, 9, 12, 2, 6, 2, 2, 12, 2, 2, 12, 4, 12, 5, 12, 4, 12, 12, 10, 12, 12, 12, 12, 8, 12, 12, 12, 2, 12, 2, 1, 8, 12, 2, 12, 12, 12, 12, 12, 12, 12, 12, 12, 8, 12, 12, 12, 12, 12, 12, 12, 12, 6, 5, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12
Offset: 0

Author

Colin Linzer, Mar 04 2024

Keywords

Comments

24 is a game in which, given 4 four numbers, you must find an expression that evaluates to 24 and contains each of those four numbers exactly once, no other numbers and only the operators of addition, subtraction, multiplication and division.
No term exceeds 12 as for any arithmetic progression 12, 12+n, 12+2n, 12+3n, there is the solution (12)+(12+n)+(12+2n)-(12+3n) = 24.
a(n) = 12 for all n > 144. - Robert Israel, Mar 06 2024

Examples

			For n=4, the arithmetic sequence 1, 5, 9, 13 has the solution (1+5)*(13-9) = 24 so a(4)=1.
		

Programs

  • Maple
    nv:= proc(V1,V2)
      local a1,a2;
      {seq(seq(op([a1+a2,a1*a2,a1-a2,a2-a1]),a1 = V1),a2 = V2),
           seq(seq(a1/a2,a1 = V1),a2=V2 minus {0}),
           seq(seq(a2/a1,a1 = V1 minus {0}),a2 = V2)}
    end proc:
    s24:= proc(a,b,c,d) local t;
      for t in combinat:-permute([a,b,c,d]) do
        if member(24, nv(nv(nv({t[1]},{t[2]}),{t[3]}),{t[4]})) then return true fi;
        if member(24, nv(nv({t[1]},{t[2]}),nv({t[3]},{t[4]}))) then return true fi;
      od;
      false
    end proc:
    f:= proc(n) local k;
      for k from 1 do
         if s24(k, k+n, k+2*n, k+3*n) then return k fi;
      od;
    end proc:
    map(f, [$0..100]); # Robert Israel, Mar 05 2024

A367329 The y-coordinate of the point where x + y = n, x & y are integers and x/y is as close as possible to e.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 19, 19, 19, 19, 20, 20
Offset: 1

Author

Colin Linzer, Nov 14 2023

Keywords

Comments

a(n) is nondecreasing; lim_{n->oo} a(n) = oo.
Swapping the x and y coordinate of the sequence does not yield the sequence defined as the point where x + y = n, x and y are integers and x/y is as close as possible to 1/e even when excluding terms that would lead to a division by 0.

Examples

			For n = 3, the possible points are (0,3), (1,2), (2,1), as any negative fraction would would be further from e than 0/3. The closest fraction to e out of these is 2/1 so a(3) = 1.
		

Crossrefs

Cf. A001113 (e), A367328 (x-coordinate), A007677.

A367328 The x-coordinate of the point where x + y = n, x and y are integers and x/y is as close as possible to e.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 20, 20, 21, 22, 23, 23, 24, 25, 26, 26, 27, 28, 28, 29, 30, 31, 31, 32, 33, 34, 34, 35, 36, 37, 37, 38, 39, 39, 40, 41, 42, 42, 43, 44, 45, 45, 46, 47, 48, 48, 49, 50
Offset: 1

Author

Colin Linzer, Nov 14 2023

Keywords

Comments

a(n) is nondecreasing; lim_{n->oo} a(n) = oo.
Swapping the x and y coordinate of the sequence does not yield the sequence defined as the point where x + y = n, x and y are integers and x/y is as close as possible to 1/e even when excluding terms that would lead to a division by 0.

Examples

			For n = 3, the possible points are (0,3), (1,2), (2,1) as any negative value would would be further from e than 0/3. The closest value to e out of these is 2/1 so a(3) = 2.
		

Crossrefs

Cf. A001113 (e), A367329 (y-coordinate), A007676.

Formula

a(n) is always either ceiling(n*e/(1 + e)) or floor(n*e/(1 + e)) = A076538(n).

A367194 The y-coordinate of the point where x + y = n, x and y are integers and x/y is as close as possible to Pi.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18
Offset: 1

Author

Colin Linzer, Nov 13 2023

Keywords

Comments

a(n) is nondecreasing; lim_{n->oo} a(n) = oo.
Swapping the x and y coordinate of the sequence does not yield the sequence defined as the point where x + y = n, x and y are integers and x/y is as close as possible to 1/Pi even when excluding terms that would lead to a division by 0.

Examples

			For n = 3, the possible fractions are (0,3), (1,2), (2,1) as any negative values would would be further from Pi than 0/3. The closest fraction to Pi out of these is 2/1 so a(3) = 1.
		

Crossrefs

Cf. A367193 (x-coordinate), A000796, A002486.

Programs

  • Maple
    f:= proc(n) local x;
       x:= floor(n/(1+Pi));
       if x = 0 then return 1 fi;
       if is((n-x)/x + (n-x-1)/(x+1) < 2*Pi) then x else x+1 fi
    end proc:
    map(f, [$1..100]); # Robert Israel, Nov 13 2023

Formula

a(n) is always either ceiling(n/(1+Pi)) or floor(n/(1+Pi)).

A367193 The x-coordinate of the point where x + y = n, x and y are integers and x/y is as close as possible to Pi.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 5, 6, 7, 7, 8, 9, 10, 11, 11, 12, 13, 14, 14, 15, 16, 17, 17, 18, 19, 20, 20, 21, 22, 23, 23, 24, 25, 26, 27, 27, 28, 29, 30, 30, 31, 32, 33, 33, 34, 35, 36, 36, 37, 38, 39, 39, 40, 41, 42, 42, 43, 44, 45, 45, 46, 47, 48, 49, 49, 50, 51, 52
Offset: 1

Author

Colin Linzer, Nov 09 2023

Keywords

Comments

a(n) is nondecreasing; lim_{n->oo} a(n) = oo.
Swapping the x and y coordinate of the sequence does not yield the sequence defined as the point where x + y = n, x and y are integers and x/y is as close as possible to 1/Pi even when excluding terms that would lead to a division by 0.

Examples

			For n = 3, the possible fractions are (0,3), (1,2), (2,1) as any negative values would would be further from Pi than 0/3. The closest fraction to Pi out of these is 2/1 so a(3) = 2.
		

Crossrefs

Cf. A367194 (y-coordinate), A000796, A002485.

Formula

a(n) is either ceiling(n*Pi/(1+Pi)) or floor(n*Pi/(1+Pi)).
a(n) = round((2*n*Pi + n - sqrt(Pi^2 + 2*Pi + n^2 + 1))/(2*Pi + 2)). - Jon E. Schoenfield, Nov 17 2023

Extensions

Corrected information and made it in line with A367194.

A365553 Starting with a plane on which two parallel lines and two additional lines have been drawn such that the four lines form two noncongruent isosceles triangles, a(n) is the total number of intersections on the plane after the n-th step, where each step consists of drawing lines that connect every intersection of two lines. If more than 2 lines intersect at the same point it is only counted once.

Original entry on oeis.org

5, 6, 8, 20, 861
Offset: 1

Author

Colin Linzer, Sep 08 2023

Keywords

Comments

Definition is better understood when viewing the linked Desmos graph.
The parity of a term is the parity of the number of intersections along the median from the base of the isosceles triangles. This is because the median is an axis of symmetry of the plane, the number of intersections not on it must be even.

Crossrefs

Formula

A recursive formula for an upper bound:
a(n+1) <= (a(n)^4 - a(n)^2)/8 + (a(n) - a(n)^3)/4 which is equivalent to
a(n+1) <= binomial(binomial(a(n),2),2) (proven).
The proof of the above formula comes from the fact that if there are o points on a graph, then at most (o^2-o)/2 lines that can be drawn between them. If there are m lines on a graph, then there are at most (m^2-m)/2 intersections between them; substituting and simplifying leads to the former upper limit.