|
| |
| |
|
|
|
1, 1, 1, 2, 2, 3, 4, 6, 7, 11, 13, 20, 24, 37, 44, 68, 81, 125, 149, 230, 274, 423, 504, 778, 927, 1431, 1705, 2632, 3136, 4841, 5768, 8904, 10609, 16377, 19513, 30122, 35890, 55403, 66012, 101902, 121415, 187427, 223317, 344732, 410744, 634061, 755476, 1166220
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,4
|
|
|
COMMENTS
|
Bruce (see link) formulated the sequence using the following two equations:
a(2n) = a(2n-1)+a(2n-3),
a(2n+1) = a(2n-1)+a(2n-2),
with n>1 and initial conditions a(1)=a(2)=a(3)= 1.
These equations lead to a pair of tribonacci-type recurrence equations, for n>2:
a(2n+1) = a(2n-1)+a(2n-3)+a(2n-5),
a(2n+2) = a(2n)+a(2n-2)+a(2n-4).
It could be more appropriate to consider the sequence as a kind of two-dimensional tribonacci sequence (a(2n-1),(a(2n)), i.e. as (1, 1), (1, 2), (2, 3), (4, 6), (7, 11), (13, 20), (24, 37), (44, 68), (81, 125), (149, 230), (274, 423), (504, 778), (927, 1431), (1705, 2632), (3136, 4841),... since after the first three initial pairs, the next pair can be obtained by adding three previous pairs component-wise. However, the first three initial pairs (1, 1), (1, 2), (2, 3) are redundant in comparison with the original integer sequence that needs only three initial integers 1, 1 and 1.
One method to construct the two-dimensional sequence is by using the well-known tribonacci-related morphism f with f(a) = ab, f(b) = ac, f(c) = a on the monoid of strings over the alphabet {a, b, c}. Using component-wise map, the following sequence of pairs is obtained: (c,b), (a, ac), (ab, aba), (abac, abacab), (abacaba, abacabaabac), (abacabaabacab, abacabaabacababacaba), ...; which is initialized by the pair (c,b) and any pair (x,y) is followed by (f(x),f(y)). The length of every string in every component consitutes the two-dimensional sequence.
|
|
|
LINKS
|
|
|
|
FORMULA
|
G.f.: x*(1+x+x^3)/(1-x^2-x^4-x^6). [corrected by G. C. Greubel, Nov 03 2018]
a(1) = a(2) = a(3) = 1; for n>1:
a(2n) = a(2n-1) + a(2n-3),
a(2n+1) = a(2n-1) + a(2n-2).
|
|
|
EXAMPLE
|
The first 14 pairs of string and its length are
(c, 1);
(b, 1);
(a, 1);
(ac, 2);
(ab, 2);
(aba, 3);
(abac, 4);
(abacab, 6);
(abacaba, 7);
(abacabaabac, 11);
(abacabaabacab, 13);
(abacabaabacababacaba, 20);
(abacabaabacababacabaabac, 24);
(abacabaabacababacabaabacabacabaabacab, 37); ...
|
|
|
MAPLE
|
with(StringTools):
# The following procedure defines the morphism f
Morphf := proc (x::string) local Start, L, Init, i;
Init := x;
L := length(Init);
Start := 1;
for i from Start to 2*L do
if Init[i] = "c" then
Init := Insert(Init, i, "a"); i := i+1; L := L+1;
Init := Delete(Init, i-1 .. i-1); i := i-1; L := L-1;
elif Init[i] = "b" then
Init := Insert(Init, i, "ac"); i := i+2; L := L+2;
Init := Delete(Init, i-2 .. i-2); i := i-1; L := L-1;
elif Init[i] = "a" then
Init := Insert(Init, i, "b"); i := i+1; L := L+1;
end if;
end do;
eval(Init);
end proc:
#The following procedure is intended to create sequence of
#strings c, b, a, ac, ab, aba, abac, ..., etc, obtained by
#iterating the morphism f n times but it starts from the third
#string "a", i.e. leaving the first two strings "c" and "b"
#behind:
TribWord := proc (x1, x2::string, n) local A, B, C, i;
A := x1; B := x2;
for i to n do
if type(i, odd) = true then
A := Morphf(A);
C := A;
else
B := Morphf(B); C := B
end if;
end do;
eval(C);
end proc;
#The following command will print a(1), a(2), ..., a(30).
for i to 30 do
printf("%d%s", length(TribWord("c", "b", i-2)), `, `);
end do
|
|
|
MATHEMATICA
|
LinearRecurrence[{0, 1, 0, 1, 0, 1}, {1, 1, 1, 2, 2, 3}, 48] (* Bruno Berselli, Jun 25 2012 *)
|
|
|
PROG
|
(PARI) x='x+O('x^50); Vec(x*(1+x+x^3)/(1-x^2-x^4-x^6)) \\ G. C. Greubel, Nov 03 2018
(Magma) I:=[1, 1, 1, 2, 2, 3]; [n le 6 select I[n] else Self(n-2) + Self(n-4) + Self(n-6): n in [1..50]]; // G. C. Greubel, Nov 03 2018
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
