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A108229
n occurs Lucas number L(n) times (A000204).
0
1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8
OFFSET
1,2
COMMENTS
This is the Lucas number equivalent of "n occurs A000045(n) times" (A072649), which is one of an infinite number of sequences derived from the Self-Counting Sequence [1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ... (A002024)] which consists of 1 copy of 1, 2 copies of 2, 3 copies of 3 and so on. These include Golomb's sequence, also known as Silverman's sequence (A001462) and the like. As with these others, the challenge is to give a surprisingly simple closed-form formula for a(n).
FORMULA
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(3*sqrt(3)) (A073010). - Amiram Eldar, Feb 18 2024
EXAMPLE
Because the first few Lucas numbers L(n), for n = 1, 2, 3, ... are 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, the current sequence consists of 1 one, 3 twos, 4 threes, 7 fours, 11 fives, 29 sixes, 47 sevens, 76 eights, 123 nines and so on.
MATHEMATICA
Flatten[Table[Table[n, {LucasL[n]}], {n, 8}]] (* Harvey P. Dale, Feb 04 2015 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Jul 23 2005
STATUS
approved