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.

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A339609 Consider a triangle drawn on the perimeter of a triangular lattice with side length n. a(n) is the number of regions inside the triangle after drawing unit circles centered at each lattice point inside the triangle.

Original entry on oeis.org

0, 0, 4, 10, 22, 39, 61, 88, 120, 157, 199, 246, 298, 355, 417, 484, 556, 633, 715, 802, 894, 991, 1093, 1200, 1312, 1429, 1551, 1678, 1810, 1947, 2089, 2236, 2388, 2545, 2707, 2874, 3046, 3223, 3405, 3592, 3784, 3981, 4183, 4390, 4602, 4819, 5041, 5268, 5500, 5737
Offset: 1

Views

Author

Wesley Ivan Hurt, Dec 09 2020

Keywords

Links

Crossrefs

Cf. A005476, A339623 (square version).

Programs

  • Mathematica
    Join[{0, 0, 4}, Table[(5 n^2 - 21 n + 24)/2, {n, 4, 60}]]

Formula

a(n) = (5*n^2 - 21*n + 24)/2 for n >= 4, with a(1)=a(2)=0, a(3)=4.
a(n) = A005476(n-2)+1 for n >= 4. - Hugo Pfoertner, Dec 10 2020
From Stefano Spezia, Dec 10 2020: (Start)
G.f.: x^3*(4 - 2*x + 4*x^2 - x^3)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 4. (End)

A290168 If n is even then a(n) = n^2*(n+2)/8, otherwise a(n) = (n-1)*n*(n+1)/8.

Original entry on oeis.org

0, 0, 2, 3, 12, 15, 36, 42, 80, 90, 150, 165, 252, 273, 392, 420, 576, 612, 810, 855, 1100, 1155, 1452, 1518, 1872, 1950, 2366, 2457, 2940, 3045, 3600, 3720, 4352, 4488, 5202, 5355, 6156, 6327, 7220, 7410, 8400
Offset: 0

Views

Author

Jean-François Alcover and Paul Curtz, Jul 23 2017

Keywords

Comments

Bisection of a(n) [0, 2, 12, 36, 80, 150, 252, ...] is A011379.
Bisection [0, 3, 15, 42, 90, 165, 273, ...] is A059270.
Considering s(n) = [0, 0, 0, 0, 1, 1, 3, 3, 6, 6, 10, 10, 15, 15, ...] (triangular numbers repeated - see A008805), a(n) = n*s(n+2) holds.
Considering the first differences of a(n), b(n) = [0, 2, 1 , 9, 3, 21, 6, 38, 10, 60, 15, 87, ...], b(n) shows bisections A000217 and A005476. In addition, b(n) begins like A249264 up to 12th term, and is an alternation of 4 multiples of 3 and 2 not multiples; b(n) is also such that b(2n) + b(2n+1) = A049450(n).
Considering the second differences c(n), c(n) shows bisections A001105(n+1) and -A000384(n+1), c(n) has 3 consecutive terms multiples of 3 alternating with 3 not multiples; in addition, c(2n) + c(2n+1) = A000027(n).
Considering a(n)/c(n) = [0, 0, 1/4, -1/2, 2/3, -1, 9/8, -3/2, 8/5, -2, 25/12, -5/2, ...], it appears that it is A129194(n)/A022998(n+1) and -A026741(n)/A000034(n) alternating.

Links

Crossrefs

Cf. A000027, A000034, A000217, A000384, A001105, A005476, A008805, A011379, A022998, A026741, A049450, A059270, A129194, A135713, A161680, A249264.

Programs

  • Mathematica
    a[n_] := If[EvenQ[n], n^2*(n + 2)/8, (n - 1)*n*(n + 1)/8]; Table[a[n], {n, 0, 40}]
  • PARI
    a(n) = if(n%2==0, n^2*(n+2)/8, (n-1)*n*(n+1)/8) \\ Felix Fröhlich, Jul 23 2017

Formula

G.f.: x^2*(2 + x + 3*x^2)/((x-1)^4*(x+1)^3).
a(n) = (1/16)*(-1)^n*n*(1 + (-1)^(n+1) + 2*(1 + (-1)^n)*n + 2*(-1)^n*n^2).
Sum_{n>=2} 1/a(n) = 5 + Pi^2/6 - 8*log(2). - Amiram Eldar, Sep 17 2022

A361226 Square array T(n,k) = k*((1+2*n)*k - 1)/2; n>=0, k>=0, read by antidiagonals upwards.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 2, 5, 3, 0, 3, 9, 12, 6, 0, 4, 13, 21, 22, 10, 0, 5, 17, 30, 38, 35, 15, 0, 6, 21, 39, 54, 60, 51, 21, 0, 7, 25, 48, 70, 85, 87, 70, 28, 0, 8, 29, 57, 86, 110, 123, 119, 92, 36, 0, 9, 33, 66, 102, 135, 159, 168, 156, 117, 45
Offset: 0

Views

Author

Paul Curtz, Mar 05 2023

Keywords

Comments

The main diagonal is A002414.
The first upper diagonal is A160378(n+1).
The antidiagonals sums are A034827(n+2).
b(n) = (A034827(n+3) = 0, 2, 10, 30, 70, ...) - (A002414(n) = 0, 1, 9, 30, 70, ...) = 0, 1, 1, 0, 0, 5, 21, 56, ... .
b(n+2) = A299120(n). b(n+4) = A033275(n). b(n+4) - b(n) = A002492(n).

Examples

			The rows are
  0, 0,  1,  3,  6,  10,  15,  21, ...   = A161680
  0, 1,  5, 12, 22,  35,  51,  70, ...   = A000326
  0, 2,  9, 21, 38,  60,  87, 119, ...   = A005476
  0, 3, 13, 30, 54,  85, 123, 168, ...   = A022264
  0, 4, 17, 39, 70, 110, 159, 217, ...   = A022266
  ... .
Columns: A000004, A001477, A016813, A017197=3*A016777, 2*A017101, 5*A016873, 3*A017581, 7*A017017, ... (coefficients from A026741).
Difference between two consecutive rows: A000290. Hence A143844.
This square array read by antidiagonals leads to the triangle
  0
  0   0
  0   1   1
  0   2   5   3
  0   3   9  12   6
  0   4  13  21  22  10
  0   5  17  30  38  35  15
  ... .

Crossrefs

Cf. A000004, A000290, A000326, A001477, A002414, A005476, A016777, A016813, A016873, A017017, A017101, A017197, A017581, A022264, A022266, A026741, A034827, A160378, A161680, A360962.
Cf. A002492, A033275, A143844, A299120.

Programs

  • Mathematica
    T[n_, k_] := k*((2*n + 1)*k - 1)/2; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Mar 05 2023 *)
  • PARI
    a(n) = { my(row = (sqrtint(8*n+1)-1)\2, column = n - binomial(row + 1, 2)); binomial(column, 2) + column^2 * (row - column) } \\ David A. Corneth, Mar 05 2023
  • Python
    # Seen as a triangle:
from functools import cache
@cache
def Trow(n: int) -> list[int]:
    if n == 0: return [0]
    r = Trow(n - 1)
    return [r[k] + k * k if k < n else r[n - 1] + n - 1 for k in range(n + 1)]
for n in range(7): print(Trow(n)) # Peter Luschny, Mar 05 2023

Formula

Take successively sequences n*(n-1)/2, n*(3*n-1)/2, n*(5*n-1)/2, ... listed in the EXAMPLE section.
G.f.: y*(x + y)/((1 - y)^3*(1 - x)^2). - Stefano Spezia, Mar 06 2023
E.g.f.: exp(x+y)*y*(2*x + y + 2*x*y)/2. - Stefano Spezia, Feb 21 2024

A212267 Array A(i,j) read by antidiagonals: A(i,j) is the (2*i-1)-th derivative of tan(tan(tan(...tan(x)))) nested j times evaluated at 0.

Original entry on oeis.org

1, 1, 2, 1, 4, 16, 1, 6, 72, 272, 1, 8, 168, 2896, 7936, 1, 10, 304, 10672, 203904, 353792, 1, 12, 480, 26400, 1198080, 22112000, 22368256, 1, 14, 696, 52880, 4071040, 208521728, 3412366336, 1903757312, 1, 16, 952, 92912, 10373760, 976629760, 51874413568, 709998153728, 209865342976
Offset: 1

Views

Author

John M. Campbell, May 12 2012

Keywords

Comments

The determinant of the n X n such matrix has a closed form given in the Mathematica code below.
Rows appear to be given by polynomials (see formula section).

Examples

			Array A(i,j) begins:
.      1,        1,         1,         1,          1, ...
.      2,        4,         6,         8,         10, ...
.     16,       72,       168,       304,        480, ...
.    272,     2896,     10672,     26400,      52880, ...
.   7936,   203904,   1198080,   4071040,   10373760, ...
. 353792, 22112000, 208521728, 976629760, 3172514560, ...
Evaluate the (2*3-1)th derivate of tan(tan(tan(x))) at 0, which is 168. Thus A(3,3)=168.

Links

Crossrefs

Columns j=1-3 give: A000182, A003718, A003720.

Programs

  • Maple
    A:= (i, j)-> (D@@(2*i-1))(tan@@j)(0):
seq(seq(A(i, 1+d-i), i=1..d), d=1..8); # Alois P. Heinz, May 13 2012
  • Mathematica
    A[a_, b_] :=
  A[a, b] =
   Array[D[Nest[Tan, x, #2], {x, 2*#1 - 1}] /. x -> 0 &, {a, b}];
Print[A[7, 7] // MatrixForm];
Table2 = {};
k = 1;
While[k < 8, Table1 = {};
  i = 1;
  j = k;
  While[0 < j,
   AppendTo[Table1, First[Take[First[Take[A[7, 7], {i, i}]], {j, j}]]];
   j = j - 1;
   i = i + 1];
  AppendTo[Table2, Table1];
  k++];
Print[Flatten[Table2]];
Print[Table[Det[A[n, n]], {n, 1, 7}]];
Table[(2^(11/12 +
       1/2 (5 + 3 (-1 + n)) (-1 + n)) 3^(-(1/2) (-1 +
         n) n) Glaisher^3 \[Pi]^-n BarnesG[1/2 + n] BarnesG[1 + n] BarnesG[3/2 + n])/E^(1/4), {n, 1, 7}]

Formula

A(i,j) = ((d/dx)^(2i-1) tan^j(x))_{x=0}.
Third row: n*(5*n - 1)*4 = 8*A005476(n).
Fourth row: 8/3*n*(11 - 84*n + 175*n^2).

Extensions

More terms from Alois P. Heinz, May 13 2012
Previous Showing 21-24 of 24 results.

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