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
Keywords
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.
Programs
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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
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