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.
1.3017...
Table starts .1..1....1.....1......1......1......1......1......1......1......1......1......1 .1..1....1.....1......1......1......1......1......1......1......1......1......1 .1..2....2.....2......2......2......2......2......2......2......2......2......2 .0..3....4.....4......4......4......4......4......4......4......4......4......4 .0..5...11....12.....12.....12.....12.....12.....12.....12.....12.....12.....12 .0..7...29....39.....40.....40.....40.....40.....40.....40.....40.....40.....40 .0.10...77...138....153....154....154....154....154....154....154....154....154 .0.13..202...503....638....659....660....660....660....660....660....660....660 .0.18..532..1864...2825...3085...3113...3114...3114...3114...3114...3114...3114 .0.24.1395..6936..12938..15438..15893..15929..15930..15930..15930..15930..15930 .0.34.3664.25868..60458..81200..86857..87599..87644..87645..87645..87645..87645 .0.44.9605.96512.285664.442206.502092.513649.514795.514850.514851.514851.514851 ... Some solutions for n=6 k=4 ..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0 ..1....1....1....1....1....1....1....1....1....1....1....1....1....1....1....1 ..2....2....2....2....2....2....2....2....2....2....0....2....2....2....0....2 ..3....3....0....0....1....3....0....3....1....0....2....3....3....3....2....1 ..1....2....3....3....0....0....3....1....3....2....1....4....1....4....0....0 ..0....0....1....0....3....3....2....3....1....1....2....1....2....0....1....2
Some solutions for n=6 k=4 ..0....1....1....0....4....4....4....0....2....2....1....2....1....4....1....1 ..2....0....4....4....3....0....0....4....1....3....4....0....0....2....0....3 ..1....4....2....1....2....3....2....1....0....4....3....2....2....1....2....1 ..4....3....4....2....3....1....4....2....4....1....2....4....4....3....4....4 ..1....0....3....0....0....4....2....3....2....0....1....3....0....4....2....3 ..0....2....1....3....1....0....3....1....4....4....0....0....1....3....0....1
from itertools import product
def T(n, k):
if n == 1: return k+1
symbols = "".join(chr(48+i) for i in range(k+1))
squares = ["".join(u)*2 for r in range(1, n//2 + 1)
for u in product(symbols, repeat = r)]
words = ("0" + "".join(w) for w in product(symbols, repeat=n-1))
return (k+1)*sum(all(s not in w for s in squares) for w in words)
def atodiag(maxd): # maxd antidiagonals
return [T(n, d+1-n) for d in range(1, maxd+1) for n in range(1, d+1)]
print(atodiag(11)) # Michael S. Branicky, May 20 2021
There are 96 quaternary squarefree words of length 4: each of the words 0102, 0120, 0121, 0123 has 4!=24 images under the permutations of the set {0,1,2,3}. - _Arseny Shur_, Apr 26 2015
G.f. = 1 + 4*x + 12*x^2 + 36*x^3 + 96*x^4 + 264*x^5 + 696*x^6 + 1848*x^7 + ....
def isf(s): # incrementally squarefree (check factors ending in last letter)
for l in range(1, len(s)//2 + 1):
if s[-2*l:-l] == s[-l:]: return False
return True
def aupton(nn, verbose=False):
alst, sfs = [1], set("1")
for n in range(1, nn+1):
an = 4*len(sfs)
sfsnew = set(s+i for s in sfs for i in "0123" if isf(s+i))
alst, sfs = alst+[an], sfsnew
if verbose: print(n, an)
return alst
print(aupton(14)) # Michael S. Branicky, Jun 20 2022
ab~ac (cycle b,c), ab~bc~ca and ac~ba~cb (cycle a,b,c) => a(1) = 6/6 = 1.
(* This program is not convenient for a large number of terms *) a[n_] := a[n] = (1/6)*Length[ DeleteCases[ Tuples[ Range[3], n + 1], {a___, b__, b__, c___}]]; Reap[ Do[ Print["a[", n, "] = ", a[n]]; Sow[a[n]], {n, 1, 12}]][[2, 1]] (* Jean-François Alcover, Jul 24 2013 *)
a:=[1,2,3,2,1]; b:=[2,3,1,3,2]; c:=[3,1,2,1,3]; S:=[1];
for m from 1 to 6 do S:=subs({1=a[],2=b[],3=c[]},S); od: S;
my(table=[0,1,2,1,0]); a(n) = my(v=digits(n,5)); sum(i=1,#v,table[v[i]+1]) %3+1; \\ Kevin Ryde, Jul 31 2020
0.267786840217889112376671403584302552555...
digits = 100; k0 = 5; dk = 5; Clear[r]; r[k_] := r[k] = Sum[(-1)^(n-1)*2/(2^(2^(n+1)-1)-1) * Product[2^(2^m-1)/(2^(2^m-1)-1), {m, 2, n}], {n, 1, k}] // N[#, digits+10]&; r[k0]; r[k = k0 + dk]; While[RealDigits[r[k], 10, digits+10] != RealDigits[r[k - dk], 10, digits+10], Print["k = ", k]; k = k + dk]; RealDigits[r[k], 10, digits] // First
For n = 7 the maximal squarefree words are 0102010 and the 5 others induced by permuting the symbols.
extend:= proc(w) local res,t,wt,i;
res:= NULL;
for t from "0" to "2" do
wt:= cat(w,t);
if andmap(i -> wt[-2*i..-i-1] <> wt[-i..-1], [$1..floor(length(wt)/2)]) then res:= wt,res fi
od:
res
end proc:
L[1]:= ["0"]:
for n from 1 to 43 do
A[n]:= 0: L[n+1]:= NULL;
for t in L[n] do
r:= extend(t);
if [r] = [] then A[n]:= A[n]+3
else L[n+1]:= L[n+1],r
fi
od;
L[n+1]:= [L[n+1]];
od:
seq(A[n],n=1..43); # Robert Israel, Feb 09 2017
def isf(s): # incrementally squarefree
for l in range(1, len(s)//2 + 1):
if s[-2*l:-l] == s[-l:]: return False
return True
def aupton(nn, verbose=False):
alst, sfs = [], set("0")
for n in range(1, nn+1):
an, sfsnew = 0, set()
for s in sfs:
se = [s+i for i in "012" if isf(s+i)]
if len(se) > 0: sfsnew.update(se)
else: an += 3
alst, sfs = alst+[an], sfsnew
if verbose: print(n, an)
return alst
print(aupton(40)) # Michael S. Branicky, Jul 21 2021
a(1) = 1 because a,b,c and d are similar. a(2) = 1 because aa is not squarefree; so ab is the only valid case. a(3) = 2 counting aba and abc. a(4) = 4 counting abac, abca, abcb and abcd. a(5) = 11 counting abaca,abacb,abcab,abcac,abcba,abacd,abcad,abcbd,abcda,abcdb and abcdc.
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