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.
LinearRecurrence[{0,3,0,4}, {1,5,7,19}, 40] (* Georg Fischer, Aug 18 2021 *)
from itertools import islice, count import sympy def A225654gen(): # generator of terms from sympy.abc import x f, g, blist, c = 1/x**2+1/x+1+x+x**2, 1, [1], 1 yield c for n in count(1): s = [int(d,2) for d in bin(n)[2:].split('00') if d != ''] g = (g*f).expand(modulus=2) if len(s) == 1: blist.append(g.subs(x,1)) else: blist.append(prod(blist[d] for d in s)) c += blist[-1] yield c A225654_list = list(islice(A225654gen(),26)) # Chai Wah Wu, Dec 23 2021
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
a247647 = a007088 . a247648 -- Reinhard Zumkeller, Sep 25 2014
With[{upto=500},Map[FromDigits,Select[IntegerString[Range[1,upto,2],2],StringFreeQ[#,"00"]&]]] (* Paolo Xausa, Dec 06 2023 *)
A247647_list = [int(bin(n)[2:]) for n in range(1,10**5) if n % 2 and not '00' in bin(n)] # Chai Wah Wu, Sep 25 2014
From _Omar E. Pol_, Mar 01 2015: (Start) Written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins: 1; 6; 6,12; 6,16,12,24; 6,36,16,32,12,36,24,48; 6,36,36,72,16,56,32,64,12,72,36,72,24,68,48,96; 6,36,36,72,36,96,72,144,16,96,56,112,32,100,64,128,12,72,72,144,36,120,72,144,24,144,68,136... ... In each row the first quarter of the terms (and no more) are equal to 6 times the beginning of the sequence itself (corrected after Sloane's comment in A247649, Mar 03 2015). (End)
r1:=proc(f) local g,n; g:=n->nops(expand(f^n) mod 2); [seq(g(n),n=0..90)]; end; r1(1+x+x^2+x^3);
a[n_] := Count[CoefficientList[(1 + x + x^2 + x^3 + x^4 + x^5)^n, x], _?OddQ];
Table[a[n], {n, 0, 90}] (* Jean-François Alcover, Apr 06 2017 *)
a(n) = {my(pol=(1+x+x^2+x^3+x^4+x^5)*Mod(1,2)); subst(lift(pol^n), x, 1);} \\ Michel Marcus, Mar 01 2015
From _Omar E. Pol_, Mar 01 2015: (Start) Written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins: 1; 3; 3,9; 3,7,9,17; 3,9,7,21,9,17,17,33; 3,9,9,27,7,17,21,43,9,27,17,51,17,35,33,67; 3,9,9,27,9,21,27,51,7,21,17,51,21,41,43,83,9,27,27,81,17,43,51,113,17,51,35,105,33,67,67,137; Thanks to _Michel Marcus_ we can see the first few terms of the next four rows as shown below: 3,9,9,27,9,21,27,51,9,27,21,63,27,51,51,99,7,21,... 3,9,9,27,9,21,27,51,9,27,21,63,27,51,51,99,9,27,27,... 3,9,9,27,9,21,27,51,9,27,21,63,27,51,51,99,9,27,27,81,... 3,9,9,27,9,21,27,51,9,27,21,63,27,51,51,99,9,27,27,81,21,... ... Apparently in each row the first quarter of the terms (and no more) are equal to 3 times the beginning of the sequence itself (comment corrected after Sloane's comment in A247649, Mar 03 2015). (End)
Table[PolynomialMod[(1+x+x^4)^n,2]/.x->1,{n,0,80}]
Table[Count[CoefficientList[Expand[(1+x+x^4)^n],x],?OddQ],{n,0,80}] (* _Harvey P. Dale, Apr 15 2012 *)
a(n) = {my(pol = (xx^4 + xx + 1)*Mod(1,2)); subst(lift(pol^n), xx, 1);} \\ Michel Marcus, Mar 01 2015
tabf(nn, k=16) = {nbpt = 0; for (n=0, nn, if (n==0, nbt = 1, nbt = 2^(n-1)); for (m=nbpt, nbpt+nbt-1, if (m-nbpt >k, k++; break); print1(nbopd(m), ",");); print(); nbpt += nbt;);} \\ Michel Marcus, Mar 03 2015
r1:=proc(f) local g,n; g:=n->nops(expand(f^n) mod 2); [seq(g(n),n=0..90)]; end; r1(1+x+x^2+x^3);
a[n_] := Count[(List @@ Expand[(1+x+x^3+x^4)^n]) /. x -> 1, _?OddQ]; a[0] = 1;
Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Apr 04 2017 *)
a(n) = {my(pol=(1+x+x^3+x^4)*Mod(1,2)); subst(lift(pol^n), x, 1);} \\ Michel Marcus, Mar 01 2015
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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