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.

A287867 a(n) = floor(n/2) - A287864(n).

Original entry on oeis.org

-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
Offset: 1

Views

Author

N. J. A. Sloane, Jun 05 2017

Keywords

Comments

It would be nice to know how fast this sequence grows.

Links

Crossrefs

Cf. A287864, A274933, A287866.

A274650 Triangle read by rows: T(n,k), (0 <= k <= n), in which each term is the least nonnegative integer such that no row, column, diagonal, or antidiagonal contains a repeated term.

Original entry on oeis.org

0, 1, 2, 3, 0, 1, 2, 4, 3, 5, 5, 1, 0, 2, 3, 4, 3, 5, 1, 6, 7, 6, 7, 2, 0, 5, 4, 8, 8, 5, 9, 4, 7, 2, 10, 6, 7, 10, 8, 3, 0, 6, 9, 5, 4, 11, 6, 12, 7, 1, 8, 3, 10, 9, 13, 9, 8, 4, 11, 2, 0, 1, 12, 6, 7, 10, 10, 11, 7, 12, 4, 3, 2, 9, 8, 14, 13, 15, 12, 9, 10, 6, 8, 1, 0, 11, 7, 4, 16, 14, 17
Offset: 0

Views

Author

Omar E. Pol, Jul 02 2016

Keywords

Comments

Similar to A274528, but the triangle here is a right triangle.
The same rule applied to an equilateral triangle gives A274528.
By analogy, the offset for both rows and columns is the same as the offset of A274528.
We construct the triangle by reading from left to right in each row, starting with T(0,0) = 0.
Presumably every diagonal and every column is also a permutation of the nonnegative integers, but the proof does not seem so straightforward. Of course neither the rows nor the antidiagonals are permutations of the nonnegative integers, since they are finite in length.
Omar E. Pol's conjecture that every column and every diagonal of the triangle is a permutation of the nonnegative integers is true: see the link. - N. J. A. Sloane, Jun 07 2017
It appears that the numbers generally appear for the first time in or near the right border of the triangle.
Theorem 1: the middle diagonal gives A000004 (the zero sequence).
Proof: T(0,0) = 0 by definition. For the next rows we have that in row 1 there are no zeros because the first term belongs to a column that contains a zero and the second term belongs to a diagonal that contains a zero. In row 2 the unique zero is T(2,1) = 0 because the preceding term belongs to a column that contains a zero and the following term belongs to a diagonal that contains a zero. Then we have two recurrences for all rows of the triangle:
a) If T(n,k) = 0 then row n+1 does not contain a zero because every term belongs to a column that contains a zero or it belongs to a diagonal that contains a zero.
b) If T(n,k) = 0 the next zero is T(n+2,k+1) because every preceding term in row n+2 is a positive integer because it belongs to a column that contains a zero and, on the other hand, the column, the diagonal and the antidiagonal of T(n+2,k+1) do not contain zeros.
Finally, since both T(n,k) = 0 and T(n+2,k+1) = 0 are located in the middle diagonal, the other terms of the middle diagonal are zeros, or in other words: the middle diagonal gives A000004 (the zero sequence). QED
Theorem 2: all zeros are in the middle diagonal.
Proof: consider the first n rows of the triangle. Every element located above or at the right-hand side of the middle diagonal must be positive because it belongs to a diagonal that contains one of the zeros of the middle diagonal. On the other hand every element located below the middle diagonal must be positive because it belongs to a column that contains one of the zeros of the middle diagonal, hence there are no zeros outside of the middle diagonal, or in other words: all zeros are in the middle diagonal. QED
From Hartmut F. W. Hoft, Jun 12 2017: (Start)
T(2k,k) = 0, for all k >= 0, and T(n,{(n-1)/2,(n+2)/2,(n-2)/2,(n+1)/2}) = 1, for all n >= 0 with n mod 8 = {1,2,4,5} respectively, and no 0's or 1's occur in other positions. The triangle of positions of 0's and 1's for this sequence is the triangle in the Comment section of A274651 with row and column indices and values shifted down by one.
The sequence of rows containing 1's is A047613 (n mod 8 = {1,2,4,5}), those containing only 1's is A016813 (n mod 8 = {1,5}), those containing both 0's and 1's is A047463 (n mod 8 = {2,4}), those containing only 0's is A047451 (n mod 8 = {0,6}), and those containing neither 0's nor 2's is A004767 (n mod 8 = {3,7}).
(End)

Examples

			Triangle begins:
   0;
   1,  2;
   3,  0,  1;
   2,  4,  3,  5;
   5,  1,  0,  2,  3;
   4,  3,  5,  1,  6,  7;
   6,  7,  2,  0,  5,  4,  8;
   8,  5,  9,  4,  7,  2, 10,  6;
   7, 10,  8,  3,  0,  6,  9,  5,  4;
  11,  6, 12,  7,  1,  8,  3, 10,  9, 13;
   9,  8,  4, 11,  2,  0,  1, 12,  6,  7, 10;
  10, 11,  7, 12,  4,  3,  2,  9,  8, 14, 13, 15;
  12,  9, 10,  6,  8,  1,  0, 11,  7,  4, 16, 14, 17;
  ...
From _Omar E. Pol_, Jun 07 2017: (Start)
The triangle may be reformatted as an isosceles triangle so that the zero sequence (A000004) appears in the central column (but note that this is NOT the way the triangle is constructed!):
.
.                    0;
.                  1,  2;
.                3,  0,  1;
.              2,  4,  3,  5;
.            5,  1,  0,  2,  3;
.          4,  3,  5,  1,  6,  7;
.        6,  7,  2,  0,  5,  4,  8;
.     8,   5,  9,  4,  7,  2, 10,  6;
.   7,  10,  8,  3,  0,  6,  9,  5,  4;
...
(End)

Links

Crossrefs

Cf. A000004 (middle diagonal).
Cf. A046092 (indices of the zeros).
Every diagonal and every column of the right triangle is a permutation of A001477.
The left and right edges of the right triangle give A286294 and A286295.
Cf. A274651 is the same triangle but with 1 added to every entry.
Other sequences of the same family are A269526, A274528, A274820, A274821, A286297, A288530, A288531.
Sequences mentioned in N. J. A. Sloane's proof are A000170, A274616 and A287864.
Cf. A288384.
Cf. A004767, A016813, A047451, A047463, A047613. - Hartmut F. W. Hoft, Jun 12 2017
See A308179, A308180 for a very similar triangle.

Programs

  • Mathematica
    (* function a274651[] is defined in A274651 *)
(* computation of rows 0 ... n-1 *)
a274650[n_] := a274651[n]-1
Flatten[a274650[13]] (* data *)
TableForm[a274650[13]] (* triangle *)
(* Hartmut F. W. Hoft, Jun 12 2017 *)
  • PARI
    See Links section.

Formula

T(n,k) = A274651(n+1,k+1) - 1.

A274651 Triangle read by rows: T(n,k), (1<=k<=n), in which each term is the least positive integer such that no row, column, diagonal, or antidiagonal contains a repeated term.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Jul 02 2016

Keywords

Comments

Analog of A269526, but note that this is a right triangle.
The same rule applied to an equilateral triangle gives A269526.
We construct the triangle by reading from left to right in each row, starting with T(1,1) = 1.
Presumably every diagonal and every column is also a permutation of the positive integers, but the proof does not seem so straightforward. Of course, neither the rows nor the antidiagonals are permutations of the positive integers, since they are finite in length.
Omar E. Pol's conjecture that every column and every diagonal of the triangle is a permutation of the positive integers is true: see the link in A274650 (duplicated below). - N. J. A. Sloane, Jun 07 2017
It appears that the numbers generally appear for the first time in or near the right border of the triangle.
Theorem 1: the middle diagonal gives A000012 (the all 1's sequence).
Theorem 2: all 1's are in the middle diagonal.
For the proofs of the theorems 1 and 2 see the proofs of the theorems 1 and 2 of A274650 since both sequences are essentially the same.
From Bob Selcoe, Feb 15 2017: (Start)
The columns and diagonal are permutations of the natural numbers. The proofs are essentially the same as the proofs given for the columns and rows (respectively) in A269526.
All coefficients j <= 4 eventually populate in a repeating pattern toward the "middle diagonal" (i.e., relatively near the 1's); this is because we can build the triangle by j in ascending order; that is, we can start by placing all the 1's in the proper cells, then add the 2's, 3's, 4's, 5's, etc. So for i >= 0: since the 1's appear at T(1+2i, 1+i), the 2's appear at T(2+8i, 1+4i), T(3+8i, 3+4i), T(5+8i, 2+4i) and T(6+8i, 4+4i). Accordingly, after the first five 3's appear (at T(2,2), T(4,1), T(5,4), T(7,3) and T(8,6)), the remaining 3's appear at T(11+8i, 5+4i), T(12+8i, 7+4i), T(16+8i, 8+4i) and T(17+8i, 10+4i). Similarly for 4's, after the first 21 appearances, 4's appear at T(44+8i, 21+4i), T(45+8i, 24+4i), T(47+8i, 23+4i) and T(48+8i, 26+4i). So starting at T(41,21), this 16-coefficient pattern repeats at T(41+8i, 21+4i):
n/k 21 22 23 24 25 26
41 1 3
42 2
43 3 1 2
44 4 3
45 2 1 4
46 2
47 4 1
48 3 4
where the next 1 appears at T(49,25), and the pattern repeats at that point from the top left (so T(49,26) = 3, T(50,25) = 2, etc.).
Conjecture: as n gets sufficiently large, all coefficients j>4 will appear in a repeating pattern, populating all rows and diagonals around smaller j's near the "middle diagonal" (while I can offer no formal proof, it appears very likely that this is the case). (End)
From Hartmut F. W. Hoft, Jun 12 2017: (Start)
T(2k+1,k+1) = 1, for all k>=0, and T(n,{n/2,(n+3)/2,(n-1)/2,(n+2)/2}) = 2, for all n>=1 with mod(n,8) = {2,3,5,6} respectively, and no 1's or 2's occur in other positions.
Proof by (recursive) picture:
Positions in the triangle that are empty and those containing the dots of the guiding diagonals contain numbers larger than two.
n\k 1 2 3 4 5 6 7 8 10 12 14 16 18 20 22 24
1 |1
2 |2 .
3 | 1 2
4 | .
5 | 2 1 .
6 | 2 .
7 | 1 .
8 |____________.
9 | |1 .
10 | |2 . .
11 | | 1 2 .
12 | | . .
13 | | 2 1 . .
14 | | 2 . .
15 | | 1 . .
16 |_____|______________._____.
17 | | |1 . .
18 | | |2 . . .
19 | | | 1 2 . .
20 | | | . . .
21 | | | 2 1 . . .
22 | | | 2 . . .
23 | | | 1 . . .
24 |_____|_______|____._______._____._______.
1 2 3 4 5 6 7 8 12 16 20 24
Consider the center of the triangle. In each octave of rows the columns in the first central quatrain contain a 1 and a 2, and the diagonals in the second central quatrain contain a 1 and a 2. Therefore, no 1's or 2's can occur in the respective downward quatrains of leading columns and trailing diagonals.
The sequence of rows containing 2's is A047447 (n mod 8 = {2,3,5,6}), those containing only 2's is A016825 (n mod 8 = {2,6}), those containing both 1's and 2's is A047621 (n mod 8 = {3,5}), those containing only 1's is A047522 (n mod 8 = {1,7}), and those containing neither 1's nor 2's is A008586 (n mod 8 = {0,4}).
(End)

Examples

			Triangle begins:
   1;
   2,  3;
   4,  1,  2;
   3,  5,  4,  6;
   6,  2,  1,  3,  4;
   5,  4,  6,  2,  7,  8;
   7,  8,  3,  1,  6,  5,  9;
   9,  6, 10,  5,  8,  3, 11,  7;
   8, 11,  9,  4,  1,  7, 10,  6,  5;
  12,  7, 13,  8,  2,  9,  4, 11, 10, 14;
  10,  9,  5, 12,  3,  1,  2, 13,  7,  8, 11;
  11, 12,  8, 13,  5,  4,  3, 10,  9, 15, 14, 16;
  13, 10, 11,  7,  9,  2,  1, 12,  8,  5, 17, 15, 18;
  ...
From _Omar E. Pol_, Jun 07 2017: (Start)
The triangle may be reformatted as an isosceles triangle so that the all 1's sequence (A000012) appears in the central column (but note that this is NOT the way the triangle is constructed!):
.
.                  1;
.                2,  3;
.              4,  1,  2;
.            3,  5,  4,  6;
.          6,  2,  1,  3,  4;
.        5,  4,  6,  2,  7,  8;
.      7,  8,  3,  1,  6,  5,  9;
.    9,  6, 10,  5,  8,  3, 11,  7;
.  8, 11,  9,  4,  1,  7, 10,  6,  5;
...
(End)

Links

Crossrefs

Cf. A001844 (indices of the 1's).
Cf. A000012 (middle diagonal).
Every diagonal and every column of the right triangle is a permutation of A000027.
Cf. A274650 is the same triangle but with every entry minus 1.
Other sequences of the same family are A269526, A274528, A274820, A274821, A286297, A288530, A288531.
Sequences mentioned in N. J. A. Sloane's proof are A000170, A274616 and A287864.
Cf. A008586, A016825, A047447, A047522, A047621. - Hartmut F. W. Hoft, Jun 12 2017

Programs

  • Mathematica
    f[1,1] = 1; (* for 1 < n and 1 <= k <= n *)
f[n_,k_] := f[n,k] = Module[{vals=Sort[Join[Map[f[n, #]&, Range[1, k-1]], Map[f[#, k]&, Range[k, n-1]], Map[f[n-k+#, #]&, Range[1, k-1]], Map[f[n-#, k+#]&, Range[1, Floor[(n-k)/2]]]]], c}, c=Complement[Range[1, Last[vals]], vals]; If[c=={}, Last[vals]+1, First[c]]]
(* computation of rows 1 ... n of triangle *)
a274651[n_] := Prepend[Table[f[i, j], {i, 2, n}, {j, 1, i}], {1}]
Flatten[a274651[13]] (* data *)
TableForm[a274651[13]] (* triangle *)
(* Hartmut F. W. Hoft, Jun 12 2017 *)

Formula

T(n,k) = A274650(n-1,k-1) + 1.

A274616 Maximal number of non-attacking queens on a right triangular board with n cells on each side.

Original entry on oeis.org

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

Views

Author

Rob Pratt and N. J. A. Sloane, Jul 01 2016

Keywords

Comments

This sequence was mentioned by R. K. Guy in the first comment in A004396.

Examples

			n=3:
OOX
XO
O
n=4:
OOOX
OXO
OO
O
n=5:
OOOOX
OOXO
XOO
OO
O

References

  • Paul Vanderlind, Richard K. Guy, and Loren C. Larson, The Inquisitive Problem Solver, MAA, 2002. See Problem 252, pages 67, 87, 198 and 276.

Links

Crossrefs

Cf. A000170, A004396, A287864.

Programs

  • Mathematica
    CoefficientList[Series[x*(1 +x^2 -x^3)*(1 +x^4)/((1-x)^2*(1+x+x^2)), {x, 0, 50}], x] (* G. C. Greubel, Jul 03 2016 *)
  • PARI
    concat(0, Vec(x*(1+x^2-x^3)*(1+x^4)/((1-x)^2*(1+x+x^2)) + O(x^100))) \\ Colin Barker, Jul 02 2016

Formula

Except for n=4, this is round(2n/3).
From Colin Barker, Jul 02 2016: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>5.
G.f.: x*(1+x^2-x^3)*(1+x^4)/((1-x)^2*(1+x+x^2)). (End)
a(n) = 2*(3*n + sqrt(3)*sin((2*Pi*n)/3)) / 9. - Colin Barker, Mar 08 2017

A274933 Maximal number of non-attacking queens on a quarter chessboard containing n^2 squares.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 57, 58, 59, 60, 61, 62
Offset: 1

Views

Author

N. J. A. Sloane, Jul 13 2016

Keywords

Comments

Take the quarter-board formed from a 2n-1 X 2n-1 chessboard by joining the center square to the top two corners. There are n^2 squares. If n = 11, 2n-1 = 21 and the board looks like this, with 11^2 = 121 squares (the top row is the top of the chessboard, the single cell at the bottom is at the center of the board):
OOOOOOOOOOOOOOOOOOOOO
-OOOOOOOOOOOOOOOOOOO-
--OOOOOOOOOOOOOOOOO--
---OOOOOOOOOOOOOOO---
----OOOOOOOOOOOOO----
-----OOOOOOOOOOO-----
------OOOOOOOOO------
-------OOOOOOO-------
--------OOOOO--------
---------OOO---------
----------O----------
The main question is, how does a(n) behave when n is large? (See A287866.)
This is a bisection of A287864. - Rob Pratt, Jun 04 2017

Examples

			For n=6 the maximal number is 5:
OOXOOOOOOOO
-OOOOOOXOO-
--OXOOOOO--
---OOOXO---
----OOO----
-----X-----
Examples from _Rob Pratt_, Jul 13 2016:
(i) For n=15 the maximal number is 13:
OOOOOOXOOOOOOOOOOOOOOOOOOOOOO
-OOOOOOOOOOOOOOOOOOXOOOOOOOO-
--OOOOOXOOOOOOOOOOOOOOOOOOO--
---OOOOOOOOOOOOOOOXOOOOOOO---
----OOOOOOOOOOOXOOOOOOOOO----
-----OOOOOOOXOOOOOOOOOOO-----
------OOOXOOOOOOOOOOOOO------
-------OOOOOOOOOXOOOOO-------
--------OOXOOOOOOOOOO--------
---------OOOOOOOOXOO---------
----------OOOOXOOOO----------
-----------XOOOOOO-----------
------------OXOOO------------
-------------OOO-------------
--------------O--------------
(ii) For n=31 the maximal number is 28:
OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOXOOOOOOOOOOOOOOOOO
-OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOXOOOOOOOOOOOOOOOOOO-
--OOOOOOOOOOOOOOXOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO--
---OOOOOOOOOOOOOOOXOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO---
----OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOXOOOOOOOOOOOOOO----
-----OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOXOOOOOOOOOOOOOOO-----
------OOOOOOOOOOOXOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO------
-------OOOOOOOOOOOOXOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO-------
--------OOOOOOOOOOOOOOOOOOOOXOOOOOOOOOOOOOOOOOOOOOOOO--------
---------OOOOOOOOOOOOOOOOOOOOOOOOOOOOOXOOOOOOOOOOOOO---------
----------OOOOOOOOOOOOOOOOOXOOOOOOOOOOOOOOOOOOOOOOO----------
-----------OOOOOOOOOOOOOOOOOOOOOOOOOOXOOOOOOOOOOOO-----------
------------OOOOOOOOOOOOOOOOOOOOOOOOOOOXOOOOOOOOO------------
-------------OOOOOOOOOOOOOOOOOOXOOOOOOOOOOOOOOOO-------------
--------------OOOOOOOOOOXOOOOOOOOOOOOOOOOOOOOOO--------------
---------------OOOOOXOOOOOOOOOOOOOOOOOOOOOOOOO---------------
----------------OOOOOOOXOOOOOOOOOOOOOOOOOOOOO----------------
-----------------OOOOOOOOOOOOOOOOOOOXOOOOOOO-----------------
------------------OOOOXOOOOOOOOOOOOOOOOOOOO------------------
-------------------OOOOOOOOOOOOOOOOXOOOOOO-------------------
--------------------OXOOOOOOOOOOOOOOOOOOO--------------------
---------------------OOOOOOOOOOOOOXOOOOO---------------------
----------------------OOOOOOOOXOOOOOOOO----------------------
-----------------------OOOXOOOOOOOOOOO-----------------------
------------------------OOOOOOOOOXOOO------------------------
-------------------------XOOOOOOOOOO-------------------------
--------------------------OOOOOOXOO--------------------------
---------------------------OOXOOOO---------------------------
----------------------------OOOOO----------------------------
-----------------------------OOO-----------------------------
------------------------------O------------------------------

Links

Crossrefs

Cf. A274616, A287864, A287866.

Formula

Since there can be at most one queen per row, a(n) <= n. In fact, since there cannot be a queen on both rows 1 and 2, a(n) <= n-1 for n>1. - N. J. A. Sloane, Jun 04 2017

Extensions

Terms a(n) with n >= 15 corrected and extended by Rob Pratt, Jul 13 2016
a(46)-a(67) from Andy Huchala, Mar 27 2024

A287866 a(n) = n - A274933(n).

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Jun 04 2017

Keywords

Comments

It would be nice to know how fast this grows.

Crossrefs

Cf. A274933, A287864, A287867.

Extensions

a(46)-a(99) from Andy Huchala, Mar 28 2024
Showing 1-6 of 6 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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