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User: Loeky Haryanto

Loeky Haryanto has authored 1 sequences.

A213816 Tribonacci sequences A000073 and A001590 interleaved.

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

Offset: 1

Loeky Haryanto, Jun 22 2012

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.

			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); ...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Ian Bruce, A Modified Tribonacci Sequence, The Fibonacci Quarterly 22, no.3 (1984), 244-246.
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1,0,1).

Cf. A000073.

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

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

LinearRecurrence[{0, 1, 0, 1, 0, 1}, {1, 1, 1, 2, 2, 3}, 48] (* Bruno Berselli, Jun 25 2012 *)

x='x+O('x^50); Vec(x*(1+x+x^3)/(1-x^2-x^4-x^6)) \\ G. C. Greubel, Nov 03 2018

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).

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User: Loeky Haryanto

A213816 Tribonacci sequences A000073 and A001590 interleaved.

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