|
|
A350579
|
|
The first number in A350578 to appear n times.
|
|
2
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
|
|
LINKS
|
|
|
EXAMPLE
|
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.
|
|
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 *)
|
|
PROG
|
(Python)
from itertools import count
from collections import Counter
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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|