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.

A372784 a(n) = tau(4*n) = A000005(4*n).

Original entry on oeis.org

3, 4, 6, 5, 6, 8, 6, 6, 9, 8, 6, 10, 6, 8, 12, 7, 6, 12, 6, 10, 12, 8, 6, 12, 9, 8, 12, 10, 6, 16, 6, 8, 12, 8, 12, 15, 6, 8, 12, 12, 6, 16, 6, 10, 18, 8, 6, 14, 9, 12, 12, 10, 6, 16, 12, 12, 12, 8, 6, 20, 6, 8, 18, 9, 12, 16, 6, 10, 12, 16, 6, 18, 6, 8, 18, 10
Offset: 1

Views

Author

Vaclav Kotesovec, May 13 2024

Keywords

Links

Crossrefs

Cf. A193553, A366872.
Cf. A000005, A099777, A372713, A372785, A372786, A372787, A372788, A372789, A372790, A372791, A372792.

Programs

  • Mathematica
    Table[DivisorSigma[0, 4*n], {n, 1, 150}]

Formula

For n > 1, a(n) = A366872(n-2).
Sum_{k=1..n} a(k) ~ (8*n*(log(n) + 2*gamma - 1) + n*4*log(2)) / 4, where gamma is the Euler-Mascheroni constant A001620.

A383168 Triangle T(n,k) read by rows: For closed chains of identical regular m-gons with connecting inner vertices lying n vertices apart, the n-th row lists the possible m in ascending order; n>=0, 1<=k<=d(8+4n).

Original entry on oeis.org

5, 6, 8, 12, 7, 8, 9, 10, 12, 18, 9, 10, 12, 16, 24, 11, 12, 14, 15, 20, 30, 13, 14, 15, 16, 18, 20, 24, 36, 15, 16, 18, 21, 28, 42, 17, 18, 20, 24, 32, 48, 19, 20, 21, 22, 24, 27, 30, 36, 54, 21, 22, 24, 25, 28, 30, 40, 60, 23, 24, 26, 33, 44, 66
Offset: 1

Views

Author

Manfred Boergens, Apr 18 2025

Keywords

Comments

Consider j identical regular m-gons, assembled into a circular closed chain. Two neighboring polygons share an edge and two vertices, the "inner" one lying in the interior of the chain. The interior is a j-pointed star with equal edges.
n is introduced in order to partition the set of chains into finite subsets. Two neighboring star points are separated by n vertices; there the star has reflex angles. (With n=0, regular polygons are considered as stars with no reflex angles.)
For every m > 4 there exists a chain of m-gons.
A366872 gives the number of row elements.
This sequence is interconnected with A383169. For each n there are finitely many pairs (m,j) for j m-gons building closed chains. m are given by T(n,k) and the corresponding j are given by A383169(n,k).
j = 2 + (8+4n)/(m-4-2n).
m = 4 + 2n + (8+4n)/(j-2).
These two equations allow a computation of T(n,k) and A383169(n,k) from each other, see Formula.

Examples

			Triangle begins:
  5,  6,  8, 12;
  7,  8,  9, 10, 12, 18;
  9, 10, 12, 16, 24;
 11, 12, 14, 15, 20, 30;
 13, 14, 15, 16, 18, 20, 24, 36;
 15, 16, 18, 21, 28, 42;
 17, 18, 20, 24, 32, 48;
 19, 20, 21, 22, 24, 27, 30, 36, 54;
 21, 22, 24, 25, 28, 30, 40, 60;
 23, 24, 26, 33, 44, 66;
 25, 26, 27, 28, 30, 32, 36, 40, 48, 72;
 ...
The third row T(2,.) asserts that regular 9-gons, 10-gons, 12-gons, 16-gons and 24-gons are the only regular polygons which can be assembled to a closed chain with connecting inner vertices lying 2 vertices apart.

Links

Crossrefs

Cf. A047244, A366872, A383169.

Programs

  • Mathematica
    Table[4 + 2*n + Divisors[8 + 4 n], {n, 0, 10}]//Flatten

Formula

T(n,k) = 4+2n + (k-th divisor of 8+4n in ascending order).
T(n,k) = 4+2n + A027750(8+4n,k).
T(n,k) = 4+2n + (8+4n)/(A383169(n,k)-2).
A383169(n,k) = 2 + (8+4n)/(T(n,k)-4-2n).
T(n,1) = 5+2n.
T(n,2) = 6+2n.
T(n,2) = A383169(n,2).
T(n,3) = 7+2n if n=1 mod 3, else = 8+2n.
T(n,3) = A047244(5+n).
T(n,d(8+4n)) = 12+6n (last row elements).
T(n,d(8+4n)-1) = 8+4n (second to last row elements).
T(n,d(8+4n)-2) = (10/3)*(2+n) if n=1 mod 3, else = 3*(2+n) (third last row elements).

A383169 Triangle T(n,k) read by rows: For closed chains of j identical regular polygons with connecting inner vertices lying n vertices apart, the n-th row lists the possible j in descending order; n>=0, 1<=k<=d(8+4n).

Original entry on oeis.org

10, 6, 4, 3, 14, 8, 6, 5, 4, 3, 18, 10, 6, 4, 3, 22, 12, 7, 6, 4, 3, 26, 14, 10, 8, 6, 5, 4, 3, 30, 16, 9, 6, 4, 3, 34, 18, 10, 6, 4, 3, 38, 20, 14, 11, 8, 6, 5, 4, 3, 42, 22, 12, 10, 7, 6, 4, 3, 46, 24, 13, 6, 4, 3, 50, 26, 18, 14, 10, 8, 6, 5, 4, 3
Offset: 1

Views

Author

Manfred Boergens, Apr 18 2025

Keywords

Comments

Consider j identical regular m-gons, assembled into a circular closed chain. Two neighboring polygons share an edge and two vertices, the "inner" one lying in the interior of the chain. The interior is a j-pointed star with equal edges.
n is introduced in order to partition the set of chains into finite subsets. Two neighboring star points are separated by n vertices; there the star has reflex angles. (With n=0, regular polygons are considered as stars with no reflex angles.)
For every j > 2 there exists a chain with exactly j polygons.
A366872 gives the number of row elements.
The descending order in the definition was chosen with respect to the interconnection with A383168. For each n there are finitely many pairs (m,j) for j m-gons building closed chains. j are given by T(n,k) and the corresponding m are given by A383168(n,k).
m = 4 + 2n + (8+4n)/(j-2).
j = 2 + (8+4n)/(m-4-2n).
These two equations allow a computation of T(n,k) and A383168(n,k) from each other, see Formula.

Examples

			Triangle begins:
 10,  6,  4,  3;
 14,  8,  6,  5,  4, 3;
 18, 10,  6,  4,  3;
 22, 12,  7,  6,  4, 3;
 26, 14, 10,  8,  6, 5, 4, 3;
 30, 16,  9,  6,  4, 3;
 34, 18, 10,  6,  4, 3;
 38, 20, 14, 11,  8, 6, 5, 4, 3;
 42, 22, 12, 10,  7, 6, 4, 3;
 46, 24, 13,  6,  4, 3;
 50, 26, 18, 14, 10, 8, 6, 5, 4, 3;
 ...
The third row T(2,.) asserts that closed chains of identical regular polygons with connecting inner vertices lying 2 vertices apart can only be assembled with 18, 10, 6, 4 or 3 polygons.

Links

Crossrefs

Cf. A366872, A383168.

Programs

  • Mathematica
    Table[2 + Sort[Divisors[8 + 4 n], Greater], {n, 0, 10}]//Flatten

Formula

T(n,k) = 2 + (k-th divisor of 8+4n in descending order).
T(n,k) = 2 + A027750(8+4n,A000005(8+4n)-k+1).
T(n,k) = 2 + (8+4n)/(A383168(n,k)-4-2n).
A383168(n,k) = 4 + 2n + (8+4n)/(T(n,k)-2).
T(n,1) = 10 + 4n.
T(n,2) = 6 + 2n.
T(n,2) = A383168(n,2).
T(n,3) = (2/3)*(7+2n) if n=1 mod 3, else = 4+n.
T(n,d(8+4n)) = 3 (last row elements).
T(n,d(8+4n)-1) = 4 (second to last row elements).
T(n,d(8+4n)-2) = 5 if n=1 mod 3, else = 6 (third last row elements).
Showing 1-3 of 3 results.

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