A343878 a(n) is the least k such that A342585(k) = n.
1, 2, 5, 9, 11, 21, 25, 30, 47, 39, 59, 71, 96, 100, 126, 115, 160, 178, 197, 217, 221, 261, 243, 265, 336, 322, 374, 419, 397, 479, 425, 485, 551, 583, 649, 618, 723, 653, 801, 690, 727, 887, 930, 974, 889, 932, 1115, 976, 1260, 1310, 1023, 1414, 1070, 1522
Offset: 0
Examples
We have: n: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21 A342585: 0, 1, 1, 0, 2, 2, 2, 0, 3, 2, 4, 1, 1, 0, 4, 4, 4, 1, 4, 0, 5 So: - a(0) = 1, a(1) = 2, a(2) = 5, a(3) = 9, a(4) = 11, a(5) = 21.
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..10000
- Rémy Sigrist, PARI program for A343878
Programs
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Mathematica
Block[{a, c, k, m, nn = 54}, c[0] = 1; a = {0}~Join~Reap[Do[k = 0; While[IntegerQ[c[k]], Set[m, c[k]]; Sow[m]; If[IntegerQ@ c[m], c[m]++, c[m] = 1]; k++]; Sow[0]; c[0]++, nn]][[-1, -1]]; TakeWhile[Array[FirstPosition[a, #][[1]] &, nn, 0], IntegerQ]] (* Michael De Vlieger, Oct 12 2021 *)
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PARI
See Links section.
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Python
def A343878(n): k, c = 0, dict() while True: m, r = 0, 1 while r > 0: k += 1 r = c.get(m,0) if n == r: return k c[r] = c.get(r,0)+1 m += 1 # Chai Wah Wu, Aug 31 2021
Formula
For n >= 1, a(n) <= A343880(n) + 1. - Peter Munn, May 08 2021
Extensions
Name shortened by Peter Munn, May 08 2021
Comments