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: A. Timothy Royappa

A. Timothy Royappa's wiki page.

A. Timothy Royappa has authored 12 sequences. Here are the ten most recent ones:

A373732 a(n) = floor(4*n^2/sqrt(3)).

Original entry on oeis.org

0, 2, 9, 20, 36, 57, 83, 113, 147, 187, 230, 279, 332, 390, 452, 519, 591, 667, 748, 833, 923, 1018, 1117, 1221, 1330, 1443, 1561, 1683, 1810, 1942, 2078, 2219, 2364, 2514, 2669, 2829, 2992, 3161, 3334, 3512, 3695, 3882, 4073, 4270, 4471, 4676, 4886, 5101, 5320, 5544, 5773
Offset: 0

Author

A. Timothy Royappa, Jun 17 2024

Keywords

Comments

Maximum number of equilateral triangles with unit side, possibly cut into pieces, that can fit into a square of side n without overlapping.
The area of an equilateral triangle with unit side is sqrt(3)/4 (A120011), which gives the number a(n) of such triangles in a square of side n as at most floor(n^2/(sqrt(3)/4)).

Examples

			At most 9 unit equilateral triangles can fit into a square of side 2, so a(2) = 9.
		

Crossrefs

Cf. A120011.

Formula

a(n) = floor(4*n^2/sqrt(3)).

A372667 Norm i^2+j^2+k^2 of (i,j,k) for 0 <= k <= j <= i.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 9, 12, 9, 10, 11, 13, 14, 17, 18, 19, 22, 27, 16, 17, 18, 20, 21, 24, 25, 26, 29, 34, 32, 33, 36, 41, 48, 25, 26, 27, 29, 30, 33, 34, 35, 38, 43, 41, 42, 45, 50, 57, 50, 51, 54, 59, 66, 75, 36, 37, 38, 40, 41, 44, 45, 46, 49, 54, 52, 53
Offset: 0

Author

A. Timothy Royappa, May 09 2024

Keywords

Comments

In crystallography, these triples (i,j,k) can be interpreted as Miller indices, which can be sorted into a list: (0 0 0), (1 0 0), (1 1 0), (1 1 1), (2 0 0), (2 1 0), (2 1 1), (2 2 0), (2 2 1), (2 2 2), (3 0 0), (3 1 0), (3 1 1), (3 2 0), etc.

Examples

			The first few triples are:
   0, 0, 0
   1, 0, 0
   1, 1, 0
   1, 1, 1
   2, 0, 0
   2, 1, 0
   2, 1, 1
   2, 2, 0
   2, 2, 1
   2, 2, 2
   3, 0, 0
   ...
		

References

  • C. Suryanarayana and M. Grant Norton, X-Ray Diffraction - A Practical Approach, Springer Science + Business Media, 1998, p. 83.

Crossrefs

The table of triples forms A331195.
Cf. A070770, A069011 (2-dimensional analog), A004215 (complement to this sequence)

Programs

  • Maple
    a:=[];
    for i from 0 to 10 do for j from 0 to i do for k from 0 to j do
    a:=[op(a),i^2+j^2+k^2]; od: od: od: a; # N. J. A. Sloane, Jun 03 2024
  • Python
    print([i**2 + j**2 + k**2 for i in range(7) for j in range(i+1) for k in range(j+1)]) # Andrey Zabolotskiy, May 09 2024

Extensions

More terms from Andrey Zabolotskiy, May 09 2024

A360792 Integer portion of area of inscribed circle in a regular polygon having n sides of unit length.

Original entry on oeis.org

0, 0, 1, 2, 3, 4, 5, 7, 9, 10, 12, 15, 17, 19, 22, 25, 28, 31, 34, 37, 41, 45, 49, 53, 57, 61, 66, 71, 75, 80, 86, 91, 96, 102, 108, 114, 120, 126, 133, 139, 146, 153, 160, 167, 175, 182, 190, 198, 206, 214, 223, 231, 240, 249, 258, 267, 276, 285, 295, 305
Offset: 3

Author

A. Timothy Royappa, Feb 20 2023

Keywords

Comments

Asymptotically equivalent to the area of a circle with circumference n, which is Pi*r^2 with r = n/(2*Pi), thus a(n) ~ n^2 / (4*Pi). (The same follows from taking tan(x) = x in the FORMULA.) Indeed, the integer part of this expression is equal to a(n) or a(n)+1, for all n. - M. F. Hasler, Apr 03 2025

Examples

			For n = 5, the circle inscribed in a regular pentagon with sides of unit length has area (Pi/4)*cot(Pi/5)^2 = 1.4878796365..., so a(5) = floor(1.4878796365...) = 1.
		

Crossrefs

Programs

  • Maple
    a:= n-> floor(Pi/(2*tan(Pi/n))^2):
    seq(a(n), n=3..65);  # Alois P. Heinz, Feb 20 2023
  • Mathematica
    a[n_] := Floor[(Pi/4)*Cot[Pi/n]^2]; Array[a, 60, 3] (* Amiram Eldar, Feb 24 2023 *)
  • PARI
    a(n) = floor((Pi/4)/tan(Pi/n)^2) \\ Andrew Howroyd, Feb 20 2023
    
  • PARI
    apply( {A360792(n)=Pi/4\tan(Pi/n)^2}, [3..62]) \\ M. F. Hasler, Apr 03 2025

Formula

a(n) = floor((Pi/4)*(cot(Pi/n)^2)).

Extensions

More terms from Andrew Howroyd, Feb 20 2023

A306396 Consider the numbers in A024796, numbers expressible in more than one way as i^2 + j^2 + k^2, where 1 <= i <= j <= k; sequence number of ways these numbers can be expressed.

Original entry on oeis.org

2, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 3, 3, 2, 2, 2, 3, 3, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 2, 3, 4, 3, 2, 4, 2, 2, 2, 2, 4, 2, 3, 3, 2, 4, 2, 2, 2, 4, 3, 2, 2, 3, 2, 4, 3, 3, 2, 2, 3, 2, 3, 3, 3, 2, 4, 5, 2, 2, 4, 4, 2, 2, 5, 6, 2, 2, 2, 2, 3, 2, 2, 3, 3, 3, 3, 2, 3, 2, 2, 5, 3, 4, 2, 3, 2, 3, 3, 4, 3, 4, 2, 4, 2, 4, 4, 4, 3, 2, 4, 2, 3, 5, 2, 5, 4, 2
Offset: 1

Author

A. Timothy Royappa, Feb 12 2019

Keywords

Comments

Number of accidental degeneracies in the quantum mechanical 3-D "particle-in-a-box" model.

Examples

			The fourth term in A024796 is 41, which can be expressed in two ways as the sum of three nonzero squares (1^2 + 2^2 + 6^2 or 3^2 + 4^2 + 4^2), so a(4) = 2.
		

Crossrefs

Programs

  • Mathematica
    r[n_] := Length@ IntegerPartitions[n, {3}, Range[Sqrt[n]]^2]; Select[ Array[r, 300], # > 1 &] (* Giovanni Resta, Feb 21 2020 *)

Formula

a(n) = A025427(A024796(n)).

Extensions

Offset changed to 1 by Jinyuan Wang, Feb 20 2020

A210032 a(n)=n for n=1,2,3 and 4; a(n)=5 for n >= 5.

Original entry on oeis.org

1, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
Offset: 1

Author

A. Timothy Royappa, Mar 16 2012

Keywords

Comments

In atomic spectroscopy, a(n) is the number of D term symbols with spin multiplicity equal to n, i.e., there is one singlet-D term (n=1), and there are two doublet-D terms (n=2), three triple-D terms (n=3), four quartet-D terms (n=4) and five terms for every other D term of multiplicity 5 or higher (n >= 5).
Decimal expansion of 11111/9000. - Arkadiusz Wesolowski, Mar 29 2012

Crossrefs

Programs

Formula

a(n) = min(n,5). - Wesley Ivan Hurt, Apr 16 2014
From Elmo R. Oliveira, Jun 26 2024: (Start)
G.f.: x*(1+x+x^2+x^3+x^4)/(1-x) = x*(1-x^5)/(1-x)^2.
a(n) = 1 + A158411(n-1) = A101272(n+1) - 1 = A168093(n-1) - 2. (End)

A188385 Highest exponent in the prime factorization of n^n.

Original entry on oeis.org

0, 2, 3, 8, 5, 6, 7, 24, 18, 10, 11, 24, 13, 14, 15, 64, 17, 36, 19, 40, 21, 22, 23, 72, 50, 26, 81, 56, 29, 30, 31, 160, 33, 34, 35, 72, 37, 38, 39, 120, 41, 42, 43, 88, 90, 46, 47, 192, 98, 100, 51, 104, 53, 162, 55, 168, 57, 58, 59, 120, 61, 62, 126, 384
Offset: 1

Author

A. Timothy Royappa, Mar 29 2011

Keywords

Examples

			For n = 1, 1^1 = 1, giving a(1) = 0.
For n = 12, 12^12 = 8916100448256 = (2^24)(3^12), giving a(12) = 24.
		

Crossrefs

Programs

  • Mathematica
    Join[{0}, Table[n*Max[Last /@ FactorInteger[n]], {n, 2, 100}]] (* T. D. Noe, Mar 30 2011 *)
  • PARI
    a(n) = if (n==1, 0, n*vecmax(factor(n)[,2])); \\ Michel Marcus, Dec 08 2020

Formula

a(n) = n * A051903(n). - Franklin T. Adams-Watters, Mar 29 2011
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = A033150 = 1.705211... . - Amiram Eldar, Jan 05 2024

A163581 Number of zeros of sin(x) in integer intervals starting with (0,1).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0
Offset: 0

Author

A. Timothy Royappa, Jul 31 2009

Keywords

Examples

			For n = 0, 1 and 2, sin(x) has no zeros in the intervals (0,1), (1,2) and (2,3), respectively, so a(0), a(1) and a(2) are all zero. For n = 3, sin(x) has a zero in the interval (3,4) at x = pi, so a(3) = 1.
		

Crossrefs

Cf. A163584.

Formula

a(n) = 1 if n < m*pi < (n+1) for any positive integer m; a(n) = 0 otherwise.
For n>0, a(A022844(n)) = 1. - Michel Marcus, Aug 07 2013

A163584 Number of singularities of tan(x) in integer intervals starting with (0,1).

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0
Offset: 0

Author

A. Timothy Royappa, Jul 31 2009

Keywords

Examples

			a(0) = 0 because tan(x) has no singularities in the interval (0,1).
a(1) = 1 because tan(x) has one singularity in the interval (1,2).
		

Crossrefs

Cf. A163581.

Formula

For n>0, a(A062389(n)) = 1. - Michel Marcus, Aug 07 2013

Extensions

More terms from Michel Marcus, Aug 07 2013

A154392 Number of zeros of sin(x^2) in integer intervals starting with (0,1).

Original entry on oeis.org

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

Author

A. Timothy Royappa, Jan 08 2009

Keywords

Comments

First differences of A032616. - R. J. Mathar, Jan 12 2009

Examples

			For n = 1, there are no zeros of sin(x^2) in (0,1), so a(1) = 0. For n = 2, there is one zero of sin(x^2) in (1,2) at x = sqrt(pi), or x ~ 1.772, so a(2) = 1.
		

Crossrefs

Cf. A032616.

Programs

  • Mathematica
    Join[{0},Table[CountRoots[Sin[x^2],{x,k,k+1}],{k,20}]] (* Harvey P. Dale, Dec 24 2018 *)

Extensions

More terms from Alois P. Heinz, Feb 20 2023

A125089 First nonzero digit of solution to log_n(z) = -z, where log_n stands for the base-n logarithm.

Original entry on oeis.org

6, 5, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
Offset: 2

Author

A. Timothy Royappa, Apr 08 2008, Apr 14 2008

Keywords

Examples

			a(2) = 6 because 6 is the first digit of the solution (z = 0.641...) to the generating equation log_2(z) = -z.
		

Formula

First nonzero digit of W(log(n))/log(n) for integer n > 1. Here W is the Lambert W function.