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.
109 is in the sequence because its binary expansion is 1101101.
import Data.Set (singleton, deleteFindMin, insert)
a247648 n = a247648_list !! (n-1)
a247648_list = f $ singleton 1 where
f s = x : f (insert (4 * x + 1) $ insert (2 * x + 1) s')
where (x, s') = deleteFindMin s
-- Reinhard Zumkeller, Sep 25 2014
vitopart := proc (n) local L, i, j, N, p, t: N := 2*n: L := ListTools:-Reverse(convert(N, base, 2)): j := 0: for i to nops(L) do if L[i] = 0 then j := j+1: p[j] := numboccur(L[1 .. i], 1) end if end do: sort([seq(p[t], t = 1 .. j)], `>=`) end proc: a := proc (n) if n = 1 then 1 elif `mod`(n, 2) = 0 then a((1/2)*n) elif `mod`(n, 2) = 1 and `mod`((1/2)*n-1/2, 2) = 0 then a((1/2)*n-1/2)+1 else a((1/2)*n-1/2) end if end proc: A := {}: for n to 254 do if a(n) = nops(vitopart(n)) then A := `union`(A, {n}) else end if end do: A; # program is based on my comment; the command vitopart(n) yields the integer partition having viabin number n. # Emeric Deutsch, Aug 29 2017
Select[Range@ 256, And[First@ # == Last@ # == 1, NoneTrue[Map[Length, Select[Split[#], First@ # == 0 &]], # > 1 &]] &@ IntegerDigits[#, 2] &] (* Michael De Vlieger, Aug 29 2017 *)
isok(k) = if (k%2, my(b=binary(k)); #select(x->(x==0), vector(#b-1, k, b[k]+b[k+1])) == 0); \\ Michel Marcus, Jun 15 2024
A247648_list = [n for n in range(1,10**5) if n % 2 and not '00' in bin(n)] # Chai Wah Wu, Sep 25 2014
The first few generations are: ..........X.......... ........XXXXX........ ......X.X.X.X.X...... ....XX..X.X.X..XX.... (f^3) ..X...X...X...X...X.. XXXX.XXX.XXX.XXX.XXXX ... f^3 mod 2 = x^6 + x^5 + x^2 + 1/x^2 + 1/x^5 + 1/x^6 + 1 has 7 terms, so a(3) = 7. From _Omar E. Pol_, Mar 02 2015: (Start) Also, written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins: 1; 5; 5, 7; 5,17, 7,19; 5,25,17,19, 7,31,19,25; 5,25,25,35,17,61,19,71, 7,35,31,41,19,71,25,77; 5,25,25,35,25,85,35,95,17,85,61,71,19,91,71,77,7,35,35,49,31,107,41,121,19, ... (End) It follows from the Generalized Run Length Transform result mentioned in the comments that in each row the first quarter of the terms (and no more) are equal to 5 times the beginning of the sequence itself. It cannot be said that the rows converge (in any meaningful sense) to five times the sequence. - _N. J. A. Sloane_, Mar 03 2015
import sympy
from functools import reduce
from operator import mul
x = sympy.symbols('x')
f = 1/x**2+1/x+1+x+x**2
A247649_list, g = [1], 1
for n in range(1,1001):
s = [int(d,2) for d in bin(n)[2:].split('00') if d != '']
g = (g*f).expand(modulus=2)
if len(s) == 1:
A247649_list.append(g.subs(x,1))
else:
A247649_list.append(reduce(mul,(A247649_list[d] for d in s)))
# Chai Wah Wu, Sep 25 2014
For an example of calculation by reversing Fibonacci binary digits, see reference in link, p. 144: On the basis (1,1,2,3,5,8) n=13 is written as 110101, Reversing all but the most AND least significant digits gives 101011, which evaluates to 16, so a(13)=16. On the basis (1,1,2,3,5,8) n=14 is written as 101101, Reversing all but the most AND least significant digits gives 101101, which evaluates to 14, so a(14)=14.
(*Produce indices of maximal Fibonacci expansion (recursively)*)
MaxFibInd[n_] := Module[{t = Floor[Log[GoldenRatio, Sqrt[5]*n + 1]] - 1}, Piecewise[{{{1}, n == 1}, {Append[MaxFibInd[n - Fibonacci[t]], t], n > 1}},]];
(*Define a(n)*)
a[n_] := Module[{MFI = MaxFibInd[n]}, Apply[Plus, Fibonacci[Last[MFI] - MFI + 1]]];
(*Generate DATA*)
Array[a, 67]
Here is the neighborhood: [0, 0, X, 0, 0] [0, 0, X, 0, 0] [X, X, X, X, X] [0, 0, X, 0, 0] [0, 0, X, 0, 0] which contains a(1) = 9 ON cells. The second and third generations are: [0, 0, 0, 0, X, 0, 0, 0, 0] [0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 0, 0, 0, X, 0, 0, 0, 0] [0, 0, 0, 0, 0, 0, 0, 0, 0] [X, 0, X, 0, X, 0, X, 0, X] (again with 9 ON cells) [0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 0, 0, 0, X, 0, 0, 0, 0] [0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 0, 0, 0, X, 0, 0, 0, 0] [0, 0, 0, 0, 0, 0, X, 0, 0, 0, 0, 0, 0] [0, 0, 0, 0, 0, 0, X, 0, 0, 0, 0, 0, 0] [0, 0, 0, 0, X, X, 0, X, X, 0, 0, 0, 0] [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 0, X, 0, 0, X, X, X, 0, 0, X, 0, 0] [0, 0, X, 0, X, 0, 0, 0, X, 0, X, 0, 0] [X, X, 0, 0, X, 0, X, 0, X, 0, 0, X, X] (with 37 ON cells) [0, 0, X, 0, X, 0, 0, 0, X, 0, X, 0, 0] [0, 0, X, 0, 0, X, X, X, 0, 0, X, 0, 0] [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] [0, 0, 0, 0, X, X, 0, X, X, 0, 0, 0, 0] [0, 0, 0, 0, 0, 0, X, 0, 0, 0, 0, 0, 0] [0, 0, 0, 0, 0, 0, X, 0, 0, 0, 0, 0, 0] The terms can be arranged into blocks of sizes 1,1,2,4,8,16,32,...: 1, 9, 9, 37, 9, 65, 37, 157, 9, 81, 65, 237, 37, 293, 157, 713, 9, 81, 81, 333, 65, 473, 237, 1077, 37, 333, 293, 1129, 157, 1285, 713, 2737, 9, 81, 81, 333, 81, 585, 333, 1413, 65, 585, 473, 1733, 237, 1933, 1077, 4337, 37, 333, 333, 1369, 293, 2125, 1129, 4969, 157, 1413, 1285, 5041, 713, 5561, 2737, 11421, ... The final terms in the rows are A246315.
C:=f->subs({x=1, y=1}, f);
# Find number of ON cells in CA for generations 0 thru M defined by rule
# that cell is ON iff number of ON cells in nbd at time n-1 was odd
# where nbd is defined by a polynomial or Laurent series f(x, y).
OddCA:=proc(f, M) global C; local n, a, i, f2, p;
f2:=simplify(expand(f)) mod 2;
a:=[]; p:=1;
for n from 0 to M do a:=[op(a), C(p)]; p:=expand(p*f2) mod 2; od:
lprint([seq(a[i], i=1..nops(a))]);
end;
f:=1/x^2+1/x+1+x+x^2+1/y^2+1/y+y+y^2;
OddCA(f, 70);
c[f_] := f /. {x -> 1, y -> 1};
OddCA[f_, M_] := Module[{a = {}, f2, p = 1}, f2 = PolynomialMod[f, 2]; Do[ AppendTo[a, c[p]]; Print[a]; p = PolynomialMod[p f2, 2], {n, 0, M}]; a];
f = 1/x^2 + 1/x + 1 + x + x^2 + 1/y^2 + 1/y + y + y^2;
OddCA[f, 70] (* Jean-François Alcover, May 24 2020, after Maple *)
import sympy
from operator import mul
from functools import reduce
x, y = sympy.symbols('x y')
f = ((1/x**2+1/x+1+x+x**2)*(1/y**2+1/y+1+y+y**2)).expand(modulus=2)
A247650_list, g = [1], 1
for n in range(1, 101):
s = [int(d, 2) for d in bin(n)[2:].split('00') if d != '']
g = (g*f).expand(modulus=2)
if len(s) == 1:
A247650_list.append(g.subs([(x, 1), (y, 1)]))
else:
A247650_list.append(reduce(mul, (A247650_list[d] for d in s)))
# Chai Wah Wu, Sep 25 2014
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