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 1-4 of 4 results.

A246521 List of free polyominoes in binary coding, ordered by number of bits, then value of the binary code. Can be read as irregular table with row lengths A000105 (in which case the offset is 0).

Original entry on oeis.org

0, 1, 3, 7, 11, 15, 23, 27, 30, 75, 31, 47, 62, 79, 91, 94, 143, 181, 182, 188, 406, 1099, 63, 95, 111, 126, 159, 175, 183, 189, 190, 207, 219, 221, 222, 252, 347, 350, 378, 407, 413, 476, 504, 1103, 1115, 1118, 1227, 1244, 2127, 2229, 2230, 2236, 2292, 2451, 2454, 2460, 33867, 127
Offset: 1

Views

Author

M. F. Hasler, Aug 28 2014

Keywords

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 placed (and flipped/rotated) as to obtain the smallest possible value, which in particular implies pushing it to both borders. 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, is 2^0 + 2^1 + 2^3 + ... + 2^A000217(n-1) = A181388(n-1).
See A246533 for the variant that lists fixed polyominoes.

Examples

			Number the points of the first quadrant as follows:
   ...   ...   ...
    9 13 18 24 31 ...
    5  8 12 17 23 ...
    2  4  7 11 16 ...
    0  1  3  6 10 ...
An animal occupying squares numbered k1, ..., kN will be represented by a term a(n) = 2^k1 + ... + 2^kN, the position and orientation being chosen as to minimize this value:
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 free domino is rotated to the ".." configuration represented by 2^0 + 2^1 (since this is smaller than the ":" configuration with value 2^0 + 2^2).
The A000105(3) = 2 free triominoes are represented by 2^0 + 2^1 + 2^3 = [...] and 2^0 + 2^1 + 2^2 = [:.]. The latter value is smaller, therefore the L-shaped triomino is listed before the straight one.
From _M. F. Hasler_, Jan 25 2021: (Start)
Writing all N-ominoes on row N, the table begins:
  N | a(m .. m+k), m = 1 + Sum_{j<N} A000105(j), k = A000105(N) - 1
----+--------------------------------------------------------------
  0 | a(1) = 0 = []
  1 | a(2) = 1 = 2^0 = [.]
  2 | a(3) = 3 = 2^0 + 2^1 = [..]
  3 | a(4) = 7 = [:.], a(5) = 11 = [...]
  4 | 15 = [:..], 23 = [::], 27 = [.:.], 30 = [':.], 75 = [....]
... | ...
(End)
		

Crossrefs

See A246533 and A246559 for lists of fixed and one-sided polyominoes.

Extensions

More terms from John Mason, Aug 29 2014

A246534 a(n) = Sum_{k=1..n} 2^(T(k)-1), where T(k)=k(k+1)/2 = A000217(k).

Original entry on oeis.org

0, 1, 5, 37, 549, 16933, 1065509, 135283237, 34495021605, 17626681066021, 18032025190548005, 36911520172609651237, 151152638972001256489509, 1238091191924352276155613733, 20283647694843594776223406899749, 664634281540152780046679753547072037
Offset: 0

Views

Author

M. F. Hasler, Aug 28 2014

Keywords

Comments

Similar to A181388, this occurs as binary encoding of a straight n-omino lying on the y-axis, when the grid points of the first quadrant (N x N, N={0,1,2,...}) are given the weight 2^k, with k=0, 1,2, 3,4,5, ... filled in by antidiagonals.
Numbers k such that the k-th composition in standard order (row k of A066099) is a reversed initial interval. - Gus Wiseman, Apr 02 2020

Examples

			Label the cells of an infinite square matrix with 0,1,2,3,... along antidiagonals:
  0 1 3 6 10 ...
  2 4 7 ...
  5 8 ...
  9 ...
  ....
Now any subset of these cells can be represented by the sum of 2 raised to the power written in the given cells. In particular, the subset consisting of the first cell in the first 1, 2, 3, ... rows is represented by 2^0, 2^0+2^2, 2^0+2^2+2^5, ...
		

Crossrefs

The version for prime (rather than binary) indices is A002110.
The non-strict generalization is A114994.
The non-reversed version is A164894.
Intersection of A333256 and A333217.
Partial sums of A036442.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    normQ[m_]:=Or[m=={},Union[m]==Range[Max[m]]];
    Select[Range[0,1000],normQ[stc[#]]&&Greater@@stc[#]&] (* Gus Wiseman, Apr 02 2020 *)
  • PARI
    t=0;vector(20,n,t+=2^(n*(n+1)/2-1)) \\ yields the vector starting with a[1]=1
    
  • PARI
    t=0;vector(20,n,if(n>1,t+=2^(n*(n-1)/2-1))) \\ yields the vector starting with 0
    
  • Python
    a = 0
    for n in range(1,17): print(a, end =', '); a += 1<<(n-1)*(n+2)//2 # Ya-Ping Lu, Jan 23 2024

A175317 a(n) = Sum_{d|n} A007955(d) where A007955(m) = product of divisors of m.

Original entry on oeis.org

1, 3, 4, 11, 6, 42, 8, 75, 31, 108, 12, 1778, 14, 206, 234, 1099, 18, 5901, 20, 8116, 452, 498, 24, 333618, 131, 692, 760, 22166, 30, 810372, 32, 33867, 1104, 1176, 1238, 10085333, 38, 1466, 1538, 2568180, 42, 3112382, 44, 85690, 91386, 2142, 48, 255138610
Offset: 1

Views

Author

Jaroslav Krizek, Apr 01 2010

Keywords

Examples

			For n = 4, with b(n) = A007955(n), a(4) = b(1) + b(2) + b(4) = 1 + 2 + 8 = 11.
		

Crossrefs

Subsequences: A008864, A181388 \ {0}.

Programs

  • Mathematica
    a[n_] := DivisorSum[n, #^(DivisorSigma[0, #]/2) &]; Array[a, 50] (* Amiram Eldar, Oct 23 2021 *)
  • PARI
    a(n) = sumdiv(n, d, vecprod(divisors(d))); \\ Michel Marcus, Dec 09 2014 and Oct 23 2021
    
  • Python
    from math import isqrt
    from sympy import divisor_count, divisors
    def A175317(n): return sum(isqrt(d)**c if (c:=divisor_count(d)) & 1 else d**(c//2) for d in divisors(n,generator=True)) # Chai Wah Wu, Jun 24 2022

Formula

From Bernard Schott, Oct 26 2021: (Start)
a(1) = 1 (the only fixed point).
a(p) = p+1 for prime p only.
a(2^k) = A181388(k+1). (End)

Extensions

Corrected by Jaroslav Krizek, Apr 02 2010
Edited and more terms from Michel Marcus, Dec 09 2014

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.

Original entry on oeis.org

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

Views

Author

M. F. Hasler, Aug 29 2014

Keywords

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).
		

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()
Showing 1-4 of 4 results.