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A143232 Sum of denominators of Egyptian fraction expansion of A004001(n) - n/2. 1
2, 0, 2, 0, 2, 1, 2, 0, 2, 1, 3, 1, 3, 1, 2, 0, 2, 1, 3, 2, 3, 2, 4, 2, 4, 2, 3, 2, 3, 1, 2, 0, 2, 1, 3, 2, 4, 2, 4, 3, 5, 3, 5, 4, 5, 4, 5, 3, 5, 4, 5, 4, 5, 3, 5, 3, 4, 2, 4, 2, 3, 1, 2, 0, 2, 1, 3, 2, 4, 3, 4, 3, 5, 4, 6, 4, 6, 5, 7, 5, 7, 6, 7, 6, 7, 5, 7, 6, 8, 6, 8, 7, 8, 7, 8, 6, 8, 7, 8, 7, 8, 6, 8, 6, 7 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
A004001 is the Hofstadter-Conway $10,000 Sequence and A004001(n) - n/2 is increasingly larger versions of the batrachion Blancmange function.
LINKS
H. Rich, Hofstadter-Conway $10,000 Sequence, Programming Archives, 2008.
Eric Weisstein's World of Mathematics, Hofstadter-Conway $10,000 Sequence.
FORMULA
a(n) = Sum of denominators of Egyptian fraction expansion of A004001(n) - n/2 .
For practical purposes, a full Egyptian fraction algorithm is not necessary. Since the elements of A004001(n) - n/2 are either whole or their fractional part is .5, the sequence can be effected by a(n) = sefd(A004001(n) - n/2) with sefd(x) = x + 3 * (x - floor(x)) .
EXAMPLE
a(43) = 5 because A004001(43) = 25, so (A004001(43) - (43/2)) = 3.5 and the Egyptian fraction expansion of 3.5 is (1/1)+(1/1)+(1/1)+(1/2), so the denominators are 1,1,1,2 which sums to 5.
PROG
(J) a004001 =: ((] +&:$: -) $:@:<:)@.(2&<) M.
a143232 =: (+ 3 * 1&|)@:(a004001 - -:)"0
CROSSREFS
Cf. A004001.
Sequence in context: A305575 A247977 A359239 * A329981 A096030 A025815
KEYWORD
nonn,easy
AUTHOR
Dan Bron (j (at) bron.us), Jul 31 2008
STATUS
approved

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Last modified July 16 08:10 EDT 2024. Contains 374345 sequences. (Running on oeis4.)