A280523 -id:A280523 - cp's OEIS frontend

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

Peter Kagey, Jan 04 2017

nonn

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.

			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.

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

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

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

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

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

A215004 a(0) = a(1) = 1; for n>1, a(n) = a(n-1) + a(n-2) + floor(n/2).

Original entry on oeis.org

1, 1, 3, 5, 10, 17, 30, 50, 84, 138, 227, 370, 603, 979, 1589, 2575, 4172, 6755, 10936, 17700, 28646, 46356, 75013, 121380, 196405, 317797, 514215, 832025, 1346254, 2178293, 3524562, 5702870, 9227448, 14930334, 24157799, 39088150, 63245967, 102334135, 165580121

Offset: 0

Alex Ratushnyak, Jul 31 2012

If the first two terms are {0,1}, we get A020956 except for the first term.

If the first two terms are {1,2}, we get A281362.

Colin Barker, Table of n, a(n) for n = 0..1000
Jean-Luc Baril, Sergey Kirgizov, and Armen Petrossian, Dyck Paths with catastrophes modulo the positions of a given pattern, Australasian J. Comb. (2022) Vol. 84, No. 2, 398-418.
Nathan Fox, Proof of formula for a(n).
Index entries for linear recurrences with constant coefficients, signature (2,1,-3,0,1).

Cf. A020956, except for first term: same formula, seed {0,1}.

Cf. A000045, A281362.

[Fibonacci(n+3)-(2*n+5-(-1)^n)/4: n in [0..40]]; // _G. C. Greubel, Feb 01 2018

Table[((-1)^n - 2 n + 8 Fibonacci[n] + 4 LucasL[n] - 5)/4, {n, 0, 20}] (* Vladimir Reshetnikov, May 18 2016 *)
RecurrenceTable[{a[0]==a[1]==1,a[n]==a[n-1]+a[n-2]+Floor[n/2]},a,{n,40}] (* or *) LinearRecurrence[{2,1,-3,0,1},{1,1,3,5,10},40] (* Harvey P. Dale, Jul 11 2020 *)

Vec(-(x^3-x+1)/((x-1)^2*(x+1)*(x^2+x-1)) + O(x^100)) \\ Colin Barker, Sep 16 2015

a(n)=([0,1,0,0,0;0,0,1,0,0;0,0,0,1,0;0,0,0,0,1;1,0,-3,1,2]^n* [1;1;3;5;10])[1,1] \\ Charles R Greathouse IV, Jan 16 2017

prpr = prev = 1
for n in range(2,100):
    print(prpr, end=', ')
    curr = prpr+prev + n//2
    prpr = prev
    prev = curr

[fibonacci(n+3) -(n+2+(n%2))//2 for n in range(41)] # G. C. Greubel, Apr 05 2024

From Colin Barker, Sep 16 2015: (Start)
a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) + a(n-5) for n>4.
G.f.: (1-x+x^3) / ((1-x)^2*(1+x)*(1-x-x^2)). (End)
a(n) = Fibonacci(n+3) - floor((n+3)/2). - Nathan Fox, Jan 27 2017
a(n) = (-3/4 + (-1)^n/4 + (2^(-n)*((1-t)^n*(-2+t) + (1+t)^n*(2+t)))/t + (-1-n)/2) where t=sqrt(5). - Colin Barker, Feb 09 2017
From G. C. Greubel, Apr 05 2024: (Start)
a(n) =  Fibonacci(n+3) - (1/4)*(2*n + 5 - (-1)^n).
E.g.f.: 2*exp(x/2)*( cosh(sqrt(5)*x/2) + (2/sqrt(5))*sinh(sqrt(5)*x/2) ) - (1/2)*( (x+2)*cosh(x) + (x+3)*sinh(x) ). (End)

A280522 The number of restarts for the routine described by A280521.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 2, 1, 2, 2, 1, 0, 1, 1, 2, 1, 2, 2, 1, 2, 2, 3, 2, 1, 0, 1, 1, 2, 1, 2, 2, 1, 2, 2, 3, 2, 1, 2, 2, 3, 2, 3, 3, 2, 1, 0, 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

Offset: 1

Peter Kagey, Jan 04 2017

nonn

This is simply one less than the number of stages, but is seems like an equally good measure, so it gets its own entry. - N. J. A. Sloane, Jan 27 2017

Peter Kagey, Table of n, a(n) for n = 1..10000

Cf. A280521, A280523.

a(n) = A280521(n) - 1.

cp's OEIS Frontend

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.

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A215004 a(0) = a(1) = 1; for n>1, a(n) = a(n-1) + a(n-2) + floor(n/2).

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A280522 The number of restarts for the routine described by A280521.

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