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A349002 The number of Lyndon words of size n from an alphabet of 4 letters and 1st, 2nd and 3rd letter of the alphabet with equal frequency in the words. 1
1, 1, 0, 2, 6, 12, 34, 120, 354, 1082, 3636, 12270, 40708, 139062, 484866, 1692268, 5944470, 21134808, 75625330, 271720146, 982116648, 3569558058, 13025614962, 47714385708, 175470892468, 647508620070, 2396613522804, 8896422981608, 33114570409896, 123566641829256 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,4
COMMENTS
Counts a subset of the Lyndon words in A027377. Here there is no requirement of how often the 4th letter of the alphabet occurs in the admitted word, only on the frequency of the 1st to 3rd letter of the alphabet.
LINKS
FORMULA
n*a(n) = Sum_{d|n} mu(d)*A344560(n/d) where mu = A008683.
EXAMPLE
Examples for the alphabet {0,1,2,3}:
a(0)=1 counts (), the empty word.
a(3)=2 counts (021) (012).
a(4)=6 counts (0321) (0231) (0312) (0132) (0213) (0123).
a(5)=12 counts (03321) (03231) (02331) (03312) (03132) (01332) (03213) (02313) (03123) (01323) (02133) (01233).
a(6)=34 counts (020211) (002211) (012021) (002121) (010221) (001221) (033321) (033231) (032331) (023331) (012102) (011202) (002112) (010212) (001212) (033312) (011022) (010122) (001122) (033132) (031332) (013332) (033213) (032313) (023313) (033123) (031323) (013323) (032133) (023133) (031233) (013233) (021333) (012333).
PROG
(PARI) a(n) = if(n>0, sumdiv(n, d, moebius(n/d)*sum(k=0, d\3, d!/(k!^3*(d-3*k)!)))/n, n==0) \\ Andrew Howroyd, Jan 14 2023
CROSSREFS
Sequence in context: A088808 A076278 A221989 * A204512 A099576 A303479
KEYWORD
nonn
AUTHOR
R. J. Mathar, Nov 05 2021
EXTENSIONS
Terms corrected and extended by Andrew Howroyd, Jan 14 2023
STATUS
approved

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Last modified March 29 07:27 EDT 2024. Contains 371265 sequences. (Running on oeis4.)