A281367 "Nachos" sequence based on triangular numbers.

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

Offset: 1

Author

N. J. A. Sloane, Jan 30 2017

Keywords

• nonn

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 the present sequence.

If S = squares we get A280053.

If S = powers of 2 we get A100661.

More than the usual number of terms are shown in order to distinguish this sequence from A104246.

Examples

			If n = 14, in the first phase we successively remove 1, then 3, then 6 nachos, leaving 4 in the pile. The next triangular number is 10, which is bigger than 4, so we start a new phase. We remove 1, then 3 nachos, and now the pile is empty. There were two phases, so a(14)=2.
		

Links

• Lars Blomberg, Table of n, a(n) for n = 1..10000
• Reddit user Teblefer, Fibonachos

Crossrefs

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

For indices of first occurrences of 1,2,3,4,... see A281368.

Different from A104246.

Programs

Maple

S:=[seq(i*(i+1)/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
A281367 := proc(n)
    local a,nres,i ;
    a := 0 ;
    nres := n;
    while nres > 0 do
        for i from 1 do
            if A000292(i) > nres then
                break;
            end if;
        end do:
        nres := nres-A000292(i-1) ;
        a := a+1 ;
    end do:
    a ;
end proc:
seq(A281367(n),n=1..80) ; # R. J. Mathar, Mar 05 2017

Mathematica

tri[n_] := n (n + 1) (n + 2)/6;
A281367[n_] := Module[{a = 0, nres = n, i}, While[nres > 0, For[i = 1, True, i++, If[tri[i] > nres, Break[]]]; nres -= tri[i-1]; a++]; a];
Table[A281367[n], {n, 1, 99}] (* Jean-François Alcover, Apr 11 2024, after R. J. Mathar *)