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.

A209721 1/4 the number of (n+1) X 3 0..2 arrays with every 2 X 2 subblock having distinct clockwise edge differences.

Original entry on oeis.org

3, 4, 5, 7, 9, 13, 17, 25, 33, 49, 65, 97, 129, 193, 257, 385, 513, 769, 1025, 1537, 2049, 3073, 4097, 6145, 8193, 12289, 16385, 24577, 32769, 49153, 65537, 98305, 131073, 196609, 262145, 393217, 524289, 786433, 1048577, 1572865, 2097153, 3145729
Offset: 1

Views

Author

R. H. Hardin, Mar 12 2012

Keywords

Comments

Column 2 of A209727.
From Richard Locke Peterson, Apr 26 2020: (Start)
The formula a(n) = 2*a(n-2)-1 also fits empirically. With the given initial numbers a(1)=3, a(2)=4, a(3)=5, this new formula implies the old empirical formula. (But it is not established that the old empirical formula is true, so it is not established that the new formula is true either.) Furthermore, if the initial numbers had somehow, for example, been 3,4,6 instead, the new formula no longer implies the old formula.
If the new formula actually is true, it follows that a(n) is the number of distinct integer triangles that can be formed with sides of length a(n-1) and a(n-2), since the greatest length the third side can have is a(n-1)+a(n-2)-1, and the least length would be a(n-1)-a(n-2)+1. (End)
Conjectures: a(n) = A029744(n+1)+1. Also, a(n) = positions of the zeros in A309019(n+2) - A002487(n+2). - George Beck, Mar 26 2022

Examples

			Some solutions for n=4
..2..1..2....1..2..1....0..2..1....2..0..1....1..2..0....2..1..2....0..1..0
..0..2..0....2..0..2....1..0..2....1..2..0....2..0..1....0..2..0....2..0..2
..1..0..1....0..1..0....0..2..1....2..0..1....1..2..0....1..0..1....1..2..1
..0..2..0....2..0..2....1..0..2....1..2..0....2..0..1....0..2..0....2..0..2
..1..0..1....0..1..0....0..2..1....2..0..1....1..2..0....2..1..2....1..2..1
		

Crossrefs

The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Formula

Empirical: a(n) = a(n-1) +2*a(n-2) -2*a(n-3).
Empirical g.f.: x*(3+x-5*x^2)/((1-x)*(1-2*x^2)). [Colin Barker, Mar 23 2012]

A209722 1/4 the number of (n+1) X 4 0..2 arrays with every 2 X 2 subblock having distinct clockwise edge differences.

Original entry on oeis.org

4, 5, 6, 8, 10, 14, 18, 26, 34, 50, 66, 98, 130, 194, 258, 386, 514, 770, 1026, 1538, 2050, 3074, 4098, 6146, 8194, 12290, 16386, 24578, 32770, 49154, 65538, 98306, 131074, 196610, 262146, 393218, 524290, 786434, 1048578, 1572866, 2097154, 3145730
Offset: 1

Views

Author

R. H. Hardin, Mar 12 2012

Keywords

Comments

Column 3 of A209727.

Examples

			Some solutions for n=4:
..2..1..2..1....2..1..2..1....1..2..1..2....1..0..2..0....2..1..2..1
..0..2..0..2....0..2..0..2....2..0..2..0....0..2..1..2....0..2..0..2
..2..1..2..1....1..0..1..0....0..1..0..1....1..0..2..0....1..0..1..0
..0..2..0..2....0..2..0..2....2..0..2..0....0..2..1..2....0..2..0..2
..2..1..2..1....2..1..2..1....0..1..0..1....1..0..2..0....1..0..1..0
		

Crossrefs

Cf. A209727.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Formula

Empirical: a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3).
Conjectures from Colin Barker, Jul 12 2018: (Start)
G.f.: x*(4 + x - 7*x^2) / ((1 - x)*(1 - 2*x^2)).
a(n) = 3*2^(n/2 - 1) + 2 for n even.
a(n) = 2^((n + 1)/2) + 2 for n odd.
(End)

A209723 1/4 the number of (n+1) X 5 0..2 arrays with every 2 X 2 subblock having distinct clockwise edge differences.

Original entry on oeis.org

6, 7, 8, 10, 12, 16, 20, 28, 36, 52, 68, 100, 132, 196, 260, 388, 516, 772, 1028, 1540, 2052, 3076, 4100, 6148, 8196, 12292, 16388, 24580, 32772, 49156, 65540, 98308, 131076, 196612, 262148, 393220, 524292, 786436, 1048580, 1572868, 2097156
Offset: 1

Views

Author

R. H. Hardin, Mar 12 2012

Keywords

Comments

Column 4 of A209727.

Examples

			Some solutions for n=4:
..2..1..2..0..2....0..2..0..1..0....0..1..0..1..0....0..1..0..1..0
..0..2..0..1..0....2..1..2..0..2....2..0..2..0..2....2..0..2..0..2
..2..1..2..0..2....0..2..0..1..0....0..1..0..1..0....0..1..0..1..0
..0..2..0..1..0....2..1..2..0..2....2..0..2..0..2....2..0..2..0..2
..2..1..2..0..2....0..2..0..1..0....1..2..1..2..1....0..1..0..1..0
		

Crossrefs

Cf. A209727.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A029744 = {s(n), n>=1}, the numbers 2^k and 3*2^k, as the parent: A029744 (s(n)); A052955 (s(n)-1), A027383 (s(n)-2), A354788 (s(n)-3), A347789 (s(n)-4), A209721 (s(n)+1), A209722 (s(n)+2), A343177 (s(n)+3), A209723 (s(n)+4); A060482, A136252 (minor differences from A354788 at the start); A354785 (3*s(n)), A354789 (3*s(n)-7). The first differences of A029744 are 1,1,1,2,2,4,4,8,8,... which essentially matches eight sequences: A016116, A060546, A117575, A131572, A152166, A158780, A163403, A320770. The bisections of A029744 are A000079 and A007283. - N. J. A. Sloane, Jul 14 2022

Formula

Empirical: a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3).
Conjectures from Colin Barker, Jul 12 2018: (Start)
G.f.: x*(6 + x - 11*x^2) / ((1 - x)*(1 - 2*x^2)).
a(n) = 3*2^(n/2 - 1) + 4 for n even.
a(n) = 2^((n + 1)/2) + 4 for n odd.
(End)

A209724 1/4 the number of (n+1) X 6 0..2 arrays with every 2 X 2 subblock having distinct clockwise edge differences.

Original entry on oeis.org

8, 9, 10, 12, 14, 18, 22, 30, 38, 54, 70, 102, 134, 198, 262, 390, 518, 774, 1030, 1542, 2054, 3078, 4102, 6150, 8198, 12294, 16390, 24582, 32774, 49158, 65542, 98310, 131078, 196614, 262150, 393222, 524294, 786438, 1048582, 1572870, 2097158
Offset: 1

Views

Author

R. H. Hardin, Mar 12 2012

Keywords

Comments

Column 5 of A209727.
Conjecture: a(1) = 8; for n > 1, a(n) is the smallest integer m such that m = ((2x * a(n-1)) /(x+1)) - x , with x a positive nontrivial divisor of m. (This is true at least for a(1) to a(100).) - Enric Reverter i Bigas, Oct 11 2020

Examples

			Some solutions for n=4:
..2..1..2..1..2..1....2..0..2..0..1..0....2..1..2..1..2..1....0..1..0..1..0..2
..0..2..0..2..0..2....1..2..1..2..0..2....0..2..0..2..0..2....2..0..2..0..2..1
..1..0..1..0..1..0....2..0..2..0..1..0....2..1..2..1..2..1....0..1..0..1..0..2
..0..2..0..2..0..2....1..2..1..2..0..2....0..2..0..2..0..2....2..0..2..0..2..1
..1..0..1..0..1..0....2..0..2..0..1..0....1..0..1..0..1..0....0..1..0..1..0..2
		

Crossrefs

Formula

Empirical: a(n) = a(n-1) +2*a(n-2) -2*a(n-3).
Conjectures from Colin Barker, Mar 07 2018: (Start)
G.f.: x*(8 + x - 15*x^2) / ((1 - x)*(1 - 2*x^2)).
a(n) = 3*2^(n/2-1) + 6 for n even.
a(n) = 2^((n+1)/2) + 6 for n odd.
(End)

A209725 1/4 the number of (n+1) X 7 0..2 arrays with every 2 X 2 subblock having distinct clockwise edge differences.

Original entry on oeis.org

12, 13, 14, 16, 18, 22, 26, 34, 42, 58, 74, 106, 138, 202, 266, 394, 522, 778, 1034, 1546, 2058, 3082, 4106, 6154, 8202, 12298, 16394, 24586, 32778, 49162, 65546, 98314, 131082, 196618, 262154, 393226, 524298, 786442, 1048586, 1572874, 2097162
Offset: 1

Views

Author

R. H. Hardin, Mar 12 2012

Keywords

Comments

Column 6 of A209727.

Examples

			Some solutions for n=4:
..1..0..1..0..1..0..1....2..0..2..0..1..0..2....0..1..0..1..0..1..0
..0..2..0..2..0..2..0....1..2..1..2..0..2..1....2..0..2..0..2..0..2
..1..0..1..0..1..0..1....2..0..2..0..1..0..2....0..1..0..1..0..1..0
..0..2..0..2..0..2..0....1..2..1..2..0..2..1....2..0..2..0..2..0..2
..1..0..1..0..1..0..1....2..0..2..0..1..0..2....0..1..0..1..0..1..0
		

Crossrefs

Cf. A209727.

Formula

Empirical: a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3).
Conjectures from Colin Barker, Jul 12 2018: (Start)
G.f.: x*(12 + x - 23*x^2) / ((1 - x)*(1 - 2*x^2)).
a(n) = 3*2^(n/2 - 1) + 10 for n even.
a(n) = 2^((n + 1)/2) + 10 for n odd.
(End)

A209726 1/4 the number of (n+1) X 8 0..2 arrays with every 2 X 2 subblock having distinct clockwise edge differences.

Original entry on oeis.org

16, 17, 18, 20, 22, 26, 30, 38, 46, 62, 78, 110, 142, 206, 270, 398, 526, 782, 1038, 1550, 2062, 3086, 4110, 6158, 8206, 12302, 16398, 24590, 32782, 49166, 65550, 98318, 131086, 196622, 262158, 393230, 524302, 786446, 1048590, 1572878, 2097166
Offset: 1

Views

Author

R. H. Hardin, Mar 12 2012

Keywords

Comments

Column 7 of A209727.

Examples

			Some solutions for n=4:
..1..2..0..2..1..2..0..2....0..1..0..2..0..1..0..2....1..2..0..2..0..2..1..2
..2..0..1..0..2..0..1..0....2..0..2..1..2..0..2..1....2..0..1..0..1..0..2..0
..1..2..0..2..1..2..0..2....0..1..0..2..0..1..0..2....1..2..0..2..0..2..1..2
..2..0..1..0..2..0..1..0....2..0..2..1..2..0..2..1....2..0..1..0..1..0..2..0
..1..2..0..2..1..2..0..2....0..1..0..2..0..1..0..2....1..2..0..2..0..2..1..2
		

Crossrefs

Cf. A209727.

Formula

Empirical: a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3).
Conjectures from Colin Barker, Jul 12 2018: (Start)
G.f.: x*(16 + x - 31*x^2) / ((1 - x)*(1 - 2*x^2)).
a(n) = 3*2^(n/2 - 1) + 14 for n even.
a(n) = 2^((n + 1)/2) + 14 for n odd.
(End)
Showing 1-6 of 6 results.