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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 Table of n, a(n) for n=1..105. 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 Adjacent sequences: A143229 A143230 A143231 * A143233 A143234 A143235 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.)