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 A143232 Sum of denominators of Egyptian fraction expansion of A004001(n) - n/2. 1

%I #8 May 05 2023 07:48:02

%S 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,

%T 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,

%U 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

%N Sum of denominators of Egyptian fraction expansion of A004001(n) - n/2.

%C A004001 is the Hofstadter-Conway \$10,000 Sequence and A004001(n) - n/2 is increasingly larger versions of the batrachion Blancmange function.

%H H. Rich, <a href="http://www.jsoftware.com/pipermail/programming/2008-August/011551.html">Hofstadter-Conway \$10,000 Sequence</a>, Programming Archives, 2008.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Hofstadter-Conway10000-DollarSequence.html">Hofstadter-Conway \$10,000 Sequence</a>.

%F a(n) = Sum of denominators of Egyptian fraction expansion of A004001(n) - n/2 .

%F 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)) .

%e 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.

%o (J) a004001 =: ((] +&:\$: -) \$:@:<:)@.(2&<) M.

%o a143232 =: (+ 3 * 1&|)@:(a004001 - -:)"0

%Y Cf. A004001.

%K nonn,easy

%O 1,1

%A Dan Bron (j (at) bron.us), Jul 31 2008

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Last modified August 15 19:07 EDT 2024. Contains 375173 sequences. (Running on oeis4.)