A295559 Same as A161645 except that triangles must always grow outwards.
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
Keywords
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.]
Links
- Lars Blomberg, Table of n, a(n) for n = 0..10000
- R. Reed, The Lemming Simulation Problem, Mathematics in School, 3 (#6, Nov. 1974), front cover and pp. 5-6. [Scanned photocopy of pages 5, 6 only, with annotations by R. K. Guy and N. J. A. Sloane]
- N. J. A. Sloane, Illustration of first 7 generations of A161644 and A295560 (edge-to-edge version)
- N. J. A. Sloane, Illustration of first 11 generations of A161644 and A295560 (vertex-to-vertex version) [Include the 6 cells marked x to get A161644(11), exclude them to get A295560(11).]
- N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Programs
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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
Comments