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A141323
Number of groups of order L(n).
2
1, 1, 1, 2, 1, 1, 5, 1, 1, 4, 1, 1, 4, 2, 1, 9, 1, 1, 30, 1, 1, 11, 5, 1, 4, 1, 2, 9, 2, 1
OFFSET
0,4
COMMENTS
This is to A140987 as Lucas numbers A000032 are to Fibonacci numbers A000045.
FORMULA
a(n) = A000001(A000032(n)).
EXAMPLE
a(6) = 5 because the Lucas number L(6) (starting at L(0) = 2) is 18 and there are 5 groups of order 18.
CROSSREFS
KEYWORD
more,nonn
AUTHOR
Jonathan Vos Post, Aug 02 2008
EXTENSIONS
a(16)-a(29) from Eric M. Schmidt, Jun 20 2014
STATUS
approved