A215006 a(0)=0, a(n+1) is the least k>a(n) such that k+a(n)+n+1 is a Fibonacci number.

0, 1, 2, 3, 6, 10, 18, 30, 51, 84, 139, 227, 371, 603, 980, 1589, 2576, 4172, 6756, 10936, 17701, 28646, 46357, 75013, 121381, 196405, 317798, 514215, 832026, 1346254, 2178294, 3524562, 5702871, 9227448, 14930335, 24157799, 39088151, 63245967, 102334136, 165580121

Offset: 0

Author

Alex Ratushnyak, Jul 31 2012

Keywords

• nonn
• easy

Comments

Same definition for k:

k+b(n)+n is a square for each term b(n) of A097063 except the first;

k+b(n)+n+1 is a square for each term b(n) of A007590 except the first;

k+b(n)+n is a cube for each term b(n) of the sequence 0, 7, 18, 43, 78, 133, 204, 301, 420, 571, 750, 967, 1218, 1513, 1848, 2233, 2664, 3151, 3690, 4291, 4950, 5677, 6468, 7333, ... (last digit repeats with period 10);

k+b(n)+n is a triangular number for each term b(n) of A002378 (oblong numbers);

k+b(n)+n is an oblong number for each term b(n) of A000217 (triangular numbers);

k+b(n)+n is a prime for each term b(n) of the sequence 0, 1, 2, 6, 7, 11, 12, 18, 21, 23, 26, 30, 31, 35, 40, 42, 43, 47, 48, 60, 69, 73, 78, 80, 87, 99, 102, 104, 107, 115, 118, 120, 125, 135, ...

Examples

			For n + 1 = 7, a(n + 1) = 30 is the least k > a(n) = a(6) = 18 such that k + a(n) + n + 1 = 30 + 18 + 6 + 1 = 55 is a Fibonacci number. - _David A. Corneth_, Sep 03 2016
		

Links

• Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).

Crossrefs

Cf. A000045, A000217, A002378, A007590, A097063.

Programs

Magma

[n le 3 select n else Self(n)+Self(n-1)+Floor(n/2)-1: n in [0..40]]; // Bruno Berselli, Jul 31 2012

Mathematica

Join[{0}, LinearRecurrence[{2, 1, -3, 0, 1}, {1, 2, 3, 6, 10}, 39]] (* Jean-François Alcover, Oct 05 2017 *)

Python

prpr = 0
prev = 1
fib = [0]*100
for n in range(100):
    fib[n] = prpr
    curr = prpr+prev
    prpr = prev
    prev = curr
a = 0
for n in range(1,55):
    print(a, end=',')
    b = c = 0
    while c <= a:
        c = fib[b] - a - n
        b += 1
    a=c

Python

print(0, end=',')
prpr = 1
prev = 2
for n in range(3,56):
    print(prpr, end=',')
    curr = prpr+prev + n//2 - 1
    prpr = prev
    prev = curr

Formula

a(n) = a(n-1) +a(n-2) +floor(n/2) -1 with n>1, a(0)=0, a(1)=1.

From Bruno Berselli, Jul 31 2012: (Start)

G.f.: x*(1-2*x^2+x^3+x^4)/((1+x)*(1-x)^2*(1-x-x^2)).

a(n) = Fibonacci(n+2)-A004526(n+1) with n>0, a(0)=0.

a(n) = A129696(n-1)+1 with n>1, a(0)=0, a(1)=1. (End)

Extensions

Definition corrected by David A. Corneth, Sep 03 2016