A246559 List of one-sided polyominoes in binary coding, ordered by number of bits, then value of the binary code. Can be read as irregular table with row lengths A000988.

0, 1, 3, 7, 11, 15, 23, 27, 30, 39, 54, 75, 31, 47, 55, 62, 79, 91, 94, 143, 181, 182, 188, 203, 286, 314, 406, 551, 566, 1099, 63, 95, 111, 126, 159, 175, 183, 189, 190, 207, 219, 221, 222, 252, 287, 303, 315, 318, 347, 350, 378, 407, 413, 476, 504

Offset: 1

Author

M. F. Hasler, Aug 29 2014

Keywords

• nonn

Comments

The binary coding (as suggested in a post to the SeqFan list by F. T. Adams-Watters) is obtained by summing the powers of 2 corresponding to the numbers covered by the polyomino, when the points of the quarter-plane are numbered by antidiagonals, and the animal is pushed to both borders as to obtain the smallest possible value. See example for further details.

The smallest value for an n-omino is the sum 2^0+...+2^(n-1) = 2^n-1 = A000225(n), and the largest value, obtained for the straight n-omino (in x direction), is 2^0+2^1+2^3+...+2^A000217(n-1) = A181388(n-1).

Examples

			Number the points of the first quadrant as follows:
...
9 ...
5 8 ...
2 4 7 ...
0 1 3 6 10 ...
The "empty" 0-omino is represented by the empty sum equal to 0 = a(1).
The monomino is represented by a square on 0, and the binary code 2^0 = 1 = a(2).
The dominos ".." and ":" would be represented by 2^0+2^1 = 3 and 2^0+2^2 = 5. Since they are equivalent up to rotation, only 3 = a(3) is listed.
The A000988(3) = 2 one-sided trominoes are represented by 2^0+2^1+2^3 = 11 (...) and 2^0+2^1+2^2 = 7 (:.). Again these values are listed in increasing size as a(4) and a(5).
		

Links

• John Mason, Table of n, a(n) for n = 1..46575
• F. T. Adams-Watters, Re: Sequence proposal by John Mason, SeqFan list, Aug 24 2014

Crossrefs

See A246521 and A246533 for enumeration of free and fixed polyominoes.

Programs

PARI

rot(P,T=[0,1;-1,0])=P=Set(apply(x->x*T,P));apply(x->x-[P[1][1],0],P)
onesided(L,N=apply(p2n,L))={ local(L=L, R=apply(P->setsearch(L,rot(P)),L), cleanup(i)=my(m=N[i]); while(m!=N[i=R[i]], if( m>N[i], m=N[i], L[i]=0))); for(i=1,#L, L[i] && cleanup(i));if(#L>1,select(P->P,L),L)}
for(i=0,5,print(Set(apply(p2n,onesided(L=if(i,grow(L),[[]])))))) \\ see A246533 for grow() and p2n()