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.

A161644 Number of ON states after n generations of cellular automaton based on triangles.

Original entry on oeis.org

0, 1, 4, 10, 16, 22, 34, 52, 64, 70, 82, 106, 136, 160, 190, 232, 256, 262, 274, 298, 328, 358, 400, 466, 532, 568, 598, 658, 742, 814, 892, 988, 1036, 1042, 1054, 1078, 1108, 1138, 1180, 1246, 1312, 1354, 1396, 1474, 1588, 1702, 1816, 1966, 2104, 2164, 2194
Offset: 0

Views

Author

David Applegate and N. J. A. Sloane, Jun 15 2009

Keywords

Comments

Analog of A151723 and A151725, but here we are working on the hexagonal net where each triangular cell has three neighbors (meeting along its edges). A cell is turned ON if exactly one of its three neighbors is ON. An ON cell remains ON forever.
We start with a single ON cell.
There is a dual version where the triangular cells meet vertex-to-vertex. The counts are the same: the two versions are isomorphic. Reed (1974) uses the vertex-to-vertex version. See the two Sloane "Illustration" links below to compare the two versions.
It appears that a(n) is also the number of polytoothpicks added in a toothpick structure formed by V-toothpicks but starting with a Y-toothpick: a(n) = a(n-1)+(A182632(n)-A182632(n-1))/2. (Checked up to n=39.) - Omar E. Pol, Dec 07 2010 and R. J. Mathar, Dec 17 2010
It appears that the behavior is similar to A161206. - Omar E. Pol, Jan 15 2016
It would be nice to have a formula or recurrence.
If new triangles are required to always move outwards we get A295559 and A295560.
From Paul Cousin, May 23 2025: (Start)
This is ETA rule 242 (11110010 in binary):
-----------------------------------------------
|state of the cell |1|1|1|1|0|0|0|0|
|sum of the neighbors' states |3|2|1|0|3|2|1|0|
|cell's next state |1|1|1|1|0|0|1|0|
----------------------------------------------- (End)

References

  • R. Reed, The Lemming Simulation Problem, Mathematics in School, 3 (#6, Nov. 1974), front cover and pp. 5-6. [Describes the dual structure where new triangles are joined at vertices rather than edges.]
  • S. Ulam, On some mathematical problems connected with patterns of growth of figures, pp. 215-224 of R. E. Bellman, ed., Mathematical Problems in the Biological Sciences, Proc. Sympos. Applied Math., Vol. 14, Amer. Math. Soc., 1962. See Example 3.

Crossrefs

Programs

  • PARI
    \\ See Links section.

Formula

a(n) = (A182632(n) - 1)/2, n >= 1. - Omar E. Pol, Mar 07 2013

Extensions

Edited by N. J. A. Sloane, Jan 10 2010 and Nov 27 2017

A161645 First differences of A161644: number of new ON cells at generation n of the triangular cellular automaton described in A161644.

Original entry on oeis.org

0, 1, 3, 6, 6, 6, 12, 18, 12, 6, 12, 24, 30, 24, 30, 42, 24, 6, 12, 24, 30, 30, 42, 66, 66, 36, 30, 60, 84, 72, 78, 96, 48, 6, 12, 24, 30, 30, 42, 66, 66, 42, 42, 78, 114, 114, 114, 150, 138, 60, 30, 60, 84, 90, 114, 174, 198, 132, 90, 144, 210, 192, 192, 210, 96, 6, 12, 24
Offset: 0

Views

Author

David Applegate and N. J. A. Sloane, Jun 15 2009

Keywords

Comments

See the comments in A161644.
It appears that a(n) is also the number of V-toothpicks or Y-toothpicks added at the n-th stage in a toothpick structure on hexagonal net, starting with a single Y-toothpick in stage 1 and adding only V-toothpicks in stages >=2 (see A161206, A160120, A182633). - Omar E. Pol, Dec 07 2010

Examples

			From _Omar E. Pol_, Apr 08 2015: (Start)
The positive terms written as an irregular triangle in which the row lengths are the terms of A011782:
1;
3;
6,6;
6,12,18,12;
6,12,24,30,24,30,42,24;
6,12,24,30,30,42,66,66,36,30,60,84,72,78,96,48;
6,12,24,30,30,42,66,66,42,42,78,114,114,114,150,138,60,30,60,84,90,114,174,198,132,90,144,210,192,192,210,96;
...
It appears that the right border gives A003945.
(End)
		

References

  • R. Reed, The Lemming Simulation Problem, Mathematics in School, 3 (#6, Nov. 1974), front cover and pp. 5-6. [Describes the dual structure where new triangles are joined at vertices rather than edges.]

Crossrefs

A295559 Same as A161645 except that triangles must always grow outwards.

Original entry on oeis.org

0, 1, 3, 6, 6, 6, 12, 18, 12, 6, 12, 18, 18, 18, 30, 42, 24, 6, 12, 18, 18, 18, 30, 42, 30, 18, 30, 42, 42, 42, 66, 90, 48, 6, 12, 18, 18, 18, 30, 42, 30, 18, 30, 42, 42, 42, 66, 90, 54, 18, 30, 42, 42, 42, 66, 90, 66, 42, 66, 90, 90, 90, 138, 186, 96, 6, 12
Offset: 0

Views

Author

N. J. A. Sloane, Nov 27 2017

Keywords

Comments

Note that Reed's version has errors (see A295558).

References

  • R. Reed, The Lemming Simulation Problem, Mathematics in School, 3 (#6, Nov. 1974), front cover and pp. 5-6. [Describes the dual structure where new triangles are joined at vertices rather than edges.]

Crossrefs

Programs

  • PARI
    \\ Empirically discovered algorithm.
    AVal(k)=6*(2^(k+1)-1)
    BVal(k, kk)={ local v; v = 3 * 2^k; for (j=0,kk-1,v += 6 * 2^j);v}
    A295559(n)={ local (len,zeros,ones,r);
    if(n==0, return(0));
    if(n==1, return(1));
    if(n==2, return(3));
    n++; len=length(binary(n));
    zeros=ones=0; i=bittest(n,0);  \\ Skip trailing 1
    while(bittest(n,i)==0,zeros++;i++);
    for(j=i+1,len-2,ones+=bittest(n,j));
    if (bittest(n,0)==1,
    if (len==zeros+2, r=BVal(1, zeros-1), if (zeros==0, r=BVal(ones+1, ones+1), r=BVal(ones+2, ones+zeros))),
    if (len==zeros+1, r=AVal(zeros-2), r=AVal(ones+zeros-1)));
    r;}
    vector(200,i,A295559(i-1))
    \\ Lars Blomberg, Dec 20 2017

Formula

From Lars Blomberg, Dec 20 2017: (Start)
Empirically (correct to 3*10^6 terms):
Convert n+1 to binary and view it as 1a1b or 1a1b1,
where a is zero or more digits, let "ones" be the number of 1's in a,
and b is zero or more 0's, let "zeros" be the number of 0's.
Let "len" be the total number of binary digits.
Then r=A295559(n) is determined by ones, zeros, len, and the parity of n+1, as follows:
if (n==0,1,2) r=0,1,3
else if (n+1 is odd)
if (len==zeros+2) r=BVal(1, zeros-1) else if (zeros==0) r=BVal(ones+1, ones+1) else r=BVal(ones+2, ones+zeros)
else
if (len==zeros+1) r=AVal(zeros-2) else r=AVal(ones+zeros-1)
and
AVal(k)=6*(2^(k+1)-1)
BVal(k, kk)=3*2^k + sum(j=0,kk-1, 6 * 2^j) (End)

Extensions

Terms a(18) and beyond from Lars Blomberg, Dec 20 2017
Showing 1-3 of 3 results.