1,2

The nachos sequence based on a sequence of positive numbers S starting with 1 is defined as follows: To find a(n) we start with a pile of n nachos.

During each phase, we successively remove S(1), then S(2), then S(3), ..., then S(i) nachos from the pile until fewer than S(i+1) remain. Then we start a new phase, successively removing S(1), then S(2), ..., then S(j) nachos from the pile until fewer than S(j+1) remain. Repeat. a(n) is the number of phases required to empty the pile.

Suggested by the Fibonachos sequence A280521, which is the case when S is 1,1,2,3,5,8,13,... (A000045).

If S = 1,2,3,4,5,... we get A057945.

If S = 1,2,3,5,7,11,... (A008578) we get A280055.

If S = triangular numbers we get A281367.

If S = squares we get the present sequence.

If S = powers of 2 we get A100661.

Needs a more professional Maple program.

Comment from Matthew C. Russell, Jan 30 2017 (Start):

Theorem: Any nachos sequence based on a sequence S = {1=s1 < s2 < s3 < ...} is unbounded.

Proof: S is the (infinite) set of numbers that we are allowed to subtract. (In the case of Fibonachos, this is the set of Fibonaccis themselves, not the partial sums.)

Suppose that n is a positive integer, with the number of stages of the process denoted by a(n).

Let s_m be the smallest element of S that is greater than n.

Then, if you start the process at N = n + s1 + s2 + s3 + ... + s_(m-1), you will get stuck when you hit n, and will have to start the process over again. Thus you will take a(n) + 1 stages of the process here, so a(N) = a(n) + 1.

(End)

Lars Blomberg, Table of n, a(n) for n = 1..10000

Reddit user Teblefer, Fibonachos

If n = 10, in the first phase we successively remove 1, then 4 nachos, leaving 5 in the pile. The next square is 9, which is bigger than 5, so we start a new phase. We remove 1, then 4 nachos, and now the pile is empty. There were two phases, so a(10)=2.

S:=[seq(i^2, i=1..1000)];

phases := proc(n) global S; local a, h, i, j, ipass;

a:=1; h:=n;

for ipass from 1 to 100 do

for i from 1 to 100 do

j:=S[i];

if j>h then a:=a+1; break; fi;

h:=h-j;

if h=0 then return(a); fi;

od;

od;

return(-1);

end;

t1:=[seq(phases(i), i=1..1000)];

# Alternative:

A280053 := proc(n)

local a, nres, i ;

a := 0 ;

nres := n;

while nres > 0 do

for i from 1 do

if A000330(i) > nres then

break;

end if;

end do:

nres := nres-A000330(i-1) ;

a := a+1 ;

end do:

a ;

end proc:

seq(A280053(n), n=1..80) ; # R. J. Mathar, Mar 05 2017

A280053[n_] := Module[{a, nres, i}, a = 0; nres = n; While[nres > 0, For[i = 1, True, i++, If[i(i+1)(2i+1)/6 > nres, Break[]]]; nres = nres - i(i-1)(2i-1)/6; a++]; a];

Table[A280053[n], {n, 1, 90}] (* Jean-François Alcover, Mar 16 2023, after R. J. Mathar *)

Cf. A000045, A008578, A280521, A057945, A281367, A100661, A280055.

For indices of first occurrences of 1,2,3,4,... see A280054.

Sequence in context: A145172 A338484 A338493 * A283365 A381423 A053824

Adjacent sequences: A280050 A280051 A280052 * A280054 A280055 A280056

N. J. A. Sloane, Jan 07 2017

approved