[go: up one dir, main page]

login
A255978
a(n) = a(n-1) + a(n-2) + (1 + (-1)^(a(n-1) + a(n-2))) with a(0)=0, a(1)=1.
3
0, 1, 1, 4, 5, 9, 16, 25, 41, 68, 109, 177, 288, 465, 753, 1220, 1973, 3193, 5168, 8361, 13529, 21892, 35421, 57313, 92736, 150049, 242785, 392836, 635621, 1028457, 1664080, 2692537, 4356617, 7049156, 11405773, 18454929, 29860704, 48315633, 78176337, 126491972, 204668309, 331160281, 535828592
OFFSET
0,4
COMMENTS
This is a minimally modified Fibonacci sequence (A000045) in that it preserves characteristic properties of the original sequence: a(n) is a function of the sum of the preceding two terms, the ratio of two consecutive terms tends to the Golden Mean, and the initial two terms are the same as in the Fibonacci sequence. See A253197 and A253198 for other members of this family.
FORMULA
a(n) = a(n-1) + a(n-2) + (1 + (-1)^(a(n-1) + a(n-2))), a(0)=0, a(1)=1.
G.f.: x*(1+2*x^2-x^3)/((1-x)*(1+x+x^2)*(1-x-x^2)). - Joerg Arndt, Mar 16 2015
a(n) = a(n-1) + a(n-2) + a(n-3) - a(n-4) - a(n-5) for n>4. - Colin Barker, Mar 28 2015
a(n) = 2*Fibonacci(n) - (1 if n != 0 (mod 3)). - Nicolas Bělohoubek, Sep 29 2021
EXAMPLE
For n = 2, a(2) = 0 + 1 + (1 + (-1)^(0 + 1)) = 1.
For n = 3, a(3) = 1 + 1 + (1 + (-1)^(1 + 1)) = 4.
For n = 4, a(4) = 1 + 4 + (1 + (-1)^(1 + 4)) = 5.
For n = 5, a(5) = 4 + 5 + (1 + (-1)^(4 + 5)) = 9.
MATHEMATICA
RecurrenceTable[{a[n]==a[n-1]+a[n-2] +(1+(-1)^(a[n-1]+a[n-2])), a[0]==0, a[1]==1}, a, {n, 0, 50}]
CoefficientList[Series[x (1 + 2 x^2 - x^3) / ((1 - x) (1 + x + x^2) (1 - x - x^2)), {x, 0, 70}], x] (* Vincenzo Librandi, Mar 24 2015 *)
LinearRecurrence[{1, 1, 1, -1, -1}, {0, 1, 1, 4, 5}, 50] (* Harvey P. Dale, Mar 26 2019 *)
PROG
(PARI) concat(0, Vec(x*(1+2*x^2-x^3)/((1-x)*(1+x+x^2)*(1-x-x^2)) + O(x^30))) \\ Michel Marcus, Mar 23 2015
(Magma) [n le 2 select (n-1) else Self(n-1)+Self(n-2)+(1+(-1)^(Self(n-1)+Self(n-2))): n in [1..45] ]; // Vincenzo Librandi, Mar 24 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Waldemar Puszkarz, Mar 12 2015
STATUS
approved