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.

A220498 Number of E-toothpicks (or tridents) added at n-th stage to the structure of the equilateral triangle of A220478.

Original entry on oeis.org

0, 2, 2, 2, 4, 2, 4, 4, 4, 6, 4, 6, 8, 2, 4, 4, 6, 10, 6, 14, 8, 10, 14, 8, 12, 14, 4, 8, 8, 10, 16, 12, 22, 16, 16, 18, 12, 14, 16, 16, 16, 10, 12, 20, 14, 22, 22, 18, 18, 24, 18, 28, 18, 20, 28, 22, 28, 20, 18, 18, 22, 32, 32, 26, 24, 22, 28, 28, 32, 34, 20, 20, 28
Offset: 0

Views

Author

Omar E. Pol, Feb 19 2013

Keywords

Comments

Essentially the first differences of A220478.

Crossrefs

Formula

a(n) = 1 + A161331(n+1)/6 = 2*A211976(n).

A231348 Number of triangles after n-th stage in a cellular automaton based in isosceles triangles of two sizes (see Comments lines for precise definition).

Original entry on oeis.org

0, 1, 3, 7, 11, 15, 23, 33, 41, 45, 53, 65, 81, 91, 111, 133, 149, 153, 161, 173, 189, 201, 225, 253, 285, 295, 315, 343, 383, 405, 449, 495, 527, 531, 539, 551, 567, 579, 603, 631, 663, 675, 699, 731, 779, 807, 863, 923, 987, 997, 1017, 1045, 1085, 1113, 1169, 1233, 1313, 1335, 1379, 1439, 1527, 1573, 1665, 1759, 1823
Offset: 0

Views

Author

Omar E. Pol, Dec 15 2013

Keywords

Comments

On the semi-infinite square grid the structure of this C.A. contains "black" triangles and "gray" triangles (see the Links section). Both types of triangles have two sides of length 5^(1/2). Every black triangle has a base of length 2 hence its height is 2 and its area is 2. Every gray triangle has a base of length 2^(1/2) hence its height is 3/(2^(1/2)) and its area is 3/2. Both types of triangles are arranged in the same way as the triangles of Sierpinski gasket (see A047999 and A006046). The black triangles are arranged in vertical direction. On the other hand the gray triangles are arranged in diagonal direction in the holes of the structure formed by the black triangles. Note that the vertices of all triangles coincide with the grid points.
The sequence gives the total number of triangles (black and gray) in the structure after n-th stage. A231349 (the first differences) gives the number of triangles added at n-th stage.
For a more complex structure see A233780.

Examples

			We start at stage 0 with no triangles, so a(0) = 0.
At stage 1 we add a black triangle, so a(1) = 1.
At stage 2 we add two black triangles, so a(2) = 1+2 = 3.
At stage 3 we add two black triangles and two gray triangles from the vertices of the master triangle, so a(3) = 3+2+2 = 7.
At stage 4 we add four black triangles, so a(4) = 7+4 = 11.
At stage 5 we add two black triangles and two gray triangles from the vertices of the master triangle, so a(5) = 11+2+2 = 15.
At stage 6 we add four black triangles and four gray triangles, so a(6) = 15+4+4 = 23.
At stage 7 we add four black triangles and six gray triangles, so a(7) = 23+4+6 = 33.
At stage 8 we add eight black triangles, so a(8) = 33+8 = 41.
And so on.
Note that always we add both black triangles and gray triangles except if n is a power of 2. In this case at stage 2^k we add only 2^k black triangles, for k >= 0.
		

Crossrefs

A211964 Right triangle from the snowflake (or E-toothpick) structure of A161330 (see Comments lines for definition).

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 8, 10, 12, 15, 17, 20, 24, 25, 27, 29, 32, 37, 40, 47, 51, 56, 63, 67, 73, 80, 82, 86, 90, 95, 103, 109, 120, 128, 136, 145, 151, 158, 166, 174, 182, 187, 193, 203, 210, 221, 232, 241, 250, 262, 271, 285, 294, 304, 318, 329, 343, 353, 362, 371
Offset: 0

Views

Author

Omar E. Pol, Dec 17 2012

Keywords

Comments

If n >> 1 the structure looks like a right triangle, which is essentially half of one of the six wedges of the E-toothpick (or snowflake) structure of A161330. The sequence gives the number of E-toothpicks in the structure after n stages. A211976 (the first differences) gives the number added at the n-th stage.

Crossrefs

Formula

a(n) = (((A161330(n+1) - 2)/6) + n)/2.
a(n) = A220478(n)/2. - Omar E. Pol, Feb 19 2012
Showing 1-3 of 3 results.