

A280053


"Nachos" sequence based on squares.


6



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
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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_(m1), 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:=hj;
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 := nresA000330(i1) ;
a := a+1 ;
end do:
a ;
end proc:
seq(A280053(n), n=1..80) ; # R. J. Mathar, Mar 05 2017


CROSSREFS

Cf. A000045, A008578, A280521, A057945, A281367, A100661, A280055.
For indices of first occurrences of 1,2,3,4,... see A280054.
Sequence in context: A092196 A100878 A145172 * A283365 A053824 A033925
Adjacent sequences: A280050 A280051 A280052 * A280054 A280055 A280056


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Jan 07 2017


STATUS

approved



