A280521 From the "Fibonachos" game: Number of phases of the following routine: during each phase, the player removes objects from a pile of n objects as the Fibonacci sequence until the pile contains fewer objects than the next Fibonacci number. The phases continue until the pile is empty.

1, 1, 2, 1, 2, 2, 1, 2, 2, 3, 2, 1, 2, 2, 3, 2, 3, 3, 2, 1, 2, 2, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 1, 2, 2, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 4, 4, 3, 2, 1, 2, 2, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 4, 4, 3, 2, 3, 3, 4, 3, 4, 4, 3, 4, 4, 5, 4, 3, 2

Offset: 1

Author

Peter Kagey, Jan 04 2017

Keywords

• nonn

Comments

From the Fibonachos link: "Two people are sharing a plate of nachos. They take turns dividing the nachos, each taking the n-th Fibonacci number of nachos on the n-th turn. When the number of nachos left is less than the next Fibonacci number, they start the sequence over. What number of nachos (less than 500) requires the most number of restarts? How would you generate numbers of nachos with a high number of restarts?"

Ones appear at indices in A000071.

Also the number of iterations of A066628(n + 1) required to reach 0.

Examples

			a(1) = 1 via [[1]];
a(2) = 1 via [[1, 1]];
a(3) = 2 via [[1, 1], [1]];
a(4) = 1 via [[1, 1, 2]];
a(5) = 2 via [[1, 1, 2], [1]];
a(6) = 2 via [[1, 1, 2], [1, 1]];
a(7) = 1 via [[1, 1, 2, 3]];
a(8) = 2 via [[1, 1, 2, 3], [1]];
a(9) = 2 via [[1, 1, 2, 3], [1, 1]];
a(10) = 3 via [[1, 1, 2, 3], [1, 1], [1]];
An example of counting iterations of A066628(n + 1) to reach zero:
a(10) = 3 because A066628(A066628(A066628(10 + 1) + 1) + 1) = 0. Fewer iterations fails to reach zero.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
• N. J. A. Sloane, Winter Fruits: New Problems from OEIS. (Sequence mentioned from 12:50-19:50.)
• Reddit user Teblefer, Fibonachos

Crossrefs

Cf. A000045, A000071, A066628, A280523 (records).

See A280053 for other sequences based on this construction. - N. J. A. Sloane, Jan 07 2017

Programs

Maple

A280521 := proc(n)
    local a,nres,i ;
    a := 0 ;
    nres := n;
    while nres > 0 do
        for i from 1 do
            if A000071(i) > nres then
                break;
            end if;
        end do:
        nres := nres-A000071(i-1) ;
        a := a+1 ;
    end do:
    a ;
end proc:
seq(A280521(n),n=1..80) ; # R. J. Mathar, Mar 05 2017

PARI

a(n)=my(s); while(n, my(k,t); while((t=fibonacci(k++))<=n, n-=t); s++); s \\ Charles R Greathouse IV, Jan 04 2017