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A136348
a(1)=1. Let m be the number of terms in the longest run of consecutive equal terms from among the first n terms of the sequence. a(n+1) = the number of terms, from among the first n terms of the sequence, that divide m.
2
1, 1, 2, 3, 3, 3, 5, 5, 5, 5, 3, 3, 3, 3, 3, 6, 6, 6, 6, 6, 6, 17, 17, 17, 17, 17, 17, 17, 2, 2, 2, 2, 2, 2, 2, 2, 11, 11, 11, 11, 11, 11, 11, 11, 11, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 25, 11, 12, 12, 12, 12, 12
OFFSET
1,3
LINKS
EXAMPLE
Among the first 20 terms of the sequence the run of five 3's (from a(11) to a(15)) is the longest run of consecutive equal terms. (So m = 5.) Among the first 20 terms of the sequence, there are six which divide 5. (a(1),a(2),a(7),a(8),a(9),a(10)) So a(21) = 6.
PROG
(PARI) See Links section.
CROSSREFS
Cf. A136347.
Sequence in context: A361159 A184995 A372546 * A353710 A284397 A284374
KEYWORD
nonn,look
AUTHOR
Leroy Quet, Dec 25 2007
EXTENSIONS
More terms from Rémy Sigrist, Aug 02 2018
STATUS
approved