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.
Tony Bartoletti has authored 12 sequences. Here are the ten most recent ones:
For n = 2, the a(2)=2 solutions are
0 1
0 0 0 0
T(3,3) = 1 since, given any permutation of <1,2,3>, the third element will be the first to be adjacent to previous elements (modulo 3), and these 6 permutations are indistinguishable given rotations and reflection. Sample table (left-justified): .....1 .....1........2 .....2........6........4 .....6.......18.......28........8 ....24.......72......128......120.......16 ...120......360......672......840......496.......32 ...720.....2160.....4128.....5760.....5312.....2016.......64 ..5040....15120....29280....43200....47616....32928.....8128......128 .40320...120960...236160...360000...435264...387072...201728....32640......256 362880..1088640..2136960..3326400..4249920..4314240..3121152..1226880...130816......512
# Counting by exhaustive examination after a C program by Bartoletti. def A249796_row(n): def F(p, n): for k in range(2, n): a = mod(p[k] + 1, n) b = mod(p[k] - 1, n) fa, fb = false, false for i in range(k): if a == p[i] : fa = true if b == p[i] : fb = true if fa and fb: counts[k] += 1 return counts = [0]*n for p in Permutations(n): F(p, n) for k in range(2, n): counts[k] = counts[k] / (2*n) return counts for n in range(9): A249796_row(n) # Peter Luschny, Nov 11 2014
The diameter of S_5 is 10, given this set of generators, since there is no sequence shorter than 0010010121 (i.e., LLXLLXLXRX) that will transform (1,2,3,4,5) into (2,1,5,4,3), and there is no permutation of (1,2,3,4,5) that requires more than a length-10 composition of L, X and R. Thus a(5) = 10.
a[1] = 0; a[n_] := GraphDiameter[CayleyGraph[PermutationGroup[{Cycles[{{1, 2}}], Cycles[{Range[n]}], InversePermutation[Cycles[{Range[n]}]]}]]]; (* Ben Whitmore, Jan 09 2018 *)
def a(n): return PermutationGroup([[(1, 2)], [tuple(1..n)], PermutationGroupElement([tuple(1..n)])^(-1)]).cayley_graph().diameter() # Max Alekseyev, Sep 09 2011
def a186752(n): t = tuple(1..n); G = PermutationGroup([[(1, 2)], [t], PermutationGroupElement([t])^(-1)]); return G.cayley_graph().distance(G.one(),G(list(t)[::-1])) # Max Alekseyev, Sep 09 2011
f[n_] := If[OddQ@ n, 2^(n/2 + 1/2), Block[{k = IntegerExponent[n, 2] - 1}, 2^(n/2 - 1) + 2^((n/2 - 2^k)/(2^(k + 1)))]]; Array[f, 41] (* Robert G. Wilson v, Aug 08 2011 *)
def a(n):
if n%2: return 2**((n + 1)//2)
k=bin(n - 1)[2:].count('1') - bin(n)[2:].count('1')
return 2**(n//2 - 1) + 2**((n//2 - 2**k)//(2**(k + 1)))
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 29 2017
f[n_] := Block[{k = IntegerExponent[n, 2]}, 2^n/2 + 2^((n - 2^k)/(2^(k + 1)))]; Array[f, 32] (* Robert G. Wilson v, Aug 08 2011 *)
def a185333(n):
if n%2: return 2**((n + 1)//2)
k=bin(n - 1)[2:].count('1') - bin(n)[2:].count('1')
return 2**(n//2 - 1) + 2**((n//2 - 2**k)//(2**(k + 1)))
def a(n): return a185333(2*n)
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 29 2017, after the formula
a(3)=3 since (in postfix): 111** = 11*1* = 1, 111*+ = 11*1+ = 111+* = 11+1* = 2 and 111++ = 11+1+ = 3. Note that at n=7, the 11 possible values produced are the set {1,2,3,4,5,6,7,8,9,10,12}. This is the first n for which there are "skipped" values in the set.
b:= proc(n) option remember; `if`(n=1, {1}, {seq(seq(seq(
[f+g, f*g][], g=b(n-i)), f=b(i)), i=1..iquo(n, 2))})
end:
a:= n-> nops(b(n)):
seq(a(n), n=1..35); # Alois P. Heinz, May 05 2019
ReplaceListRepeated[forms_,rerules_]:=Union[Flatten[FixedPointList[Function[pre,Union[Flatten[ReplaceList[#,rerules]&/@pre,1]]],forms],1]];
Table[Length[Select[ReplaceListRepeated[{Array[1&,n]},{{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x+y]],{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x*y]]}],Length[#]==1&]],{n,10}] (* Gus Wiseman, Sep 29 2018 *)
from functools import cache
@cache
def f(m):
if m == 1: return {1}
out = set()
for j in range(1, m//2+1):
for x in f(j):
for y in f(m-j):
out.update([x + y, x * y])
return out
def a(n): return len(f(n))
print([a(n) for n in range(1, 40)]) # Michael S. Branicky, Aug 03 2022
a(2) = 2 since given two 1-ohm resistors, a series circuit yields 2 ohms, while a parallel circuit yields 1/2 ohms.
r:= proc(n) option remember; `if`(n=1, {1}, {seq(seq(seq(
[f+g, 1/(1/f+1/g)][], g in r(n-i)), f in r(i)), i=1..n/2)})
end:
a:= n-> nops(r(n)):
seq(a(n), n=1..15); # Alois P. Heinz, Apr 02 2015
r[n_] := r[n] = If[n == 1, {1}, Union @ Flatten @ {Table[ Table[ Table[ {f+g, 1/(1/f+1/g)}, {g, r[n-i]}], {f, r[i]}], {i, 1, n/2}]}]; a[n_] := Length[r[n]]; Table[a[n], {n, 1, 15}] (* Jean-François Alcover, May 28 2015, after Alois P. Heinz *)
\\ not efficient; just to show the method
N=10;
L=vector(N); L[1]=[1];
{ for (n=2, N,
my( T = Set( [] ) );
for (k=1, n\2,
for (j=1, #L[k],
my( r1 = L[k][j] );
for (i=1, #L[n-k],
my( r2 = L[n-k][i] );
T = setunion(T, Set([r1+r2, r1*r2/(r1+r2) ]) );
);
);
);
T = vecsort(Vec(T), , 8);
L[n] = T;
); }
for(n=1, N, print1(#L[n], ", ") );
\\ Joerg Arndt, Mar 07 2015
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