A280053 "Nachos" sequence based on squares.

1, 2, 3, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 2, 3, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 1, 2, 3, 4, 5, 2, 3, 4, 5, 6, 3, 4, 5, 6, 2, 3, 4, 5, 6, 3, 4, 5, 6, 7, 4, 5, 6, 7, 3, 4, 2, 3, 4, 5, 6, 3

Offset

1,2

Comments

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)

Links

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

Reddit user Teblefer, Fibonachos

Example

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.

Maple

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)];
# 2nd program
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

Mathematica

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 *)

Crossrefs

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 A053824 A033925

Adjacent sequences: A280050 A280051 A280052 * A280054 A280055 A280056

Keyword

nonn

Author

N. J. A. Sloane, Jan 07 2017

Status

approved