A350579 The first number in A350578 to appear n times.

0, 42, 153, 504, 921, 921, 17729, 17729, 17729, 17729, 60610, 60610, 109617, 109617, 109617, 109617, 109617, 109617, 109599, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 2184074, 2184074, 2184074

Offset: 1

Author

Scott R. Shannon, Jan 07 2022

Keywords

• nonn

Comments

See A350578 for further details.

Examples

			a(1) = 0 as A350578(0) = 0, thus 0 is the first number to appear one time.
a(2) = 42 as A350578(20) = A350578(24) = 42, thus 42 is the first number to appear two times. This is also true for A005132.
a(3) = 153 as A350578(74) = A350578(78) = A350578(114) = 153, thus 153 is the first number to appear three times.
		

Links

• Index entries for sequences related to Recamán's sequence

Crossrefs

Cf. A350578, A005132, A335299, A064389, A261573, A064387.

Programs

Mathematica

a[0]=0;a[n_]:=a[n]=If[a[n-1]-n>=0&&Count[Array[a,n-1,0],a[n-1]-n]<=Count[Array[a,n-1,0],a[n-1]+n],a[n-1]-n,a[n-1]+n];
Table[k=0;While[Max[Last/@(c=Tally@Array[a,++k,0])]!=i];a[k-1],{i,6}] (* Giorgos Kalogeropoulos, Jan 07 2022 *)

Python

from itertools import count
from collections import Counter
def A350579(n):
    b, bcounter = 0, Counter({0})
    for m in count(1):
        if bcounter[b] == n: return b
        b += -m if b-m >= 0 and bcounter[b-m] <= bcounter[b+m] else m
        bcounter[b] += 1 # Chai Wah Wu, Jan 08 2022