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 A049802 a(n) = n mod 2 + n mod 4 + ... + n mod 2^k, where 2^k <= n < 2^(k+1). 4
 0, 0, 1, 0, 2, 2, 4, 0, 3, 4, 7, 4, 7, 8, 11, 0, 4, 6, 10, 8, 12, 14, 18, 8, 12, 14, 18, 16, 20, 22, 26, 0, 5, 8, 13, 12, 17, 20, 25, 16, 21, 24, 29, 28, 33, 36, 41, 16, 21, 24, 29, 28, 33, 36, 41, 32, 37, 40, 45, 44, 49, 52, 57, 0, 6, 10, 16, 16 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,5 COMMENTS There is the following connection between this sequence and A080277: A080277(n) = n + n*floor(log_2(n)) - a(n). Since A080277(n) is the solution to a prototypical recurrence in the analysis of the algorithm Merge Sort, that is, T(0) := 0, T(n) := 2*T(floor(n/2)) + n, the sequence a(n) seems to be the major obstacle when trying to find a simple, sum-free solution to this recurrence. It seems hard to get rid of the sum. - Peter C. Heinig (algorithms(AT)gmx.de), Oct 21 2006 When n = 2^k with k > 0 then a(n+1) = k. For this reason, when n-1 is a Mersenne prime then n - 1 = M(p) = 2^p - 1 = 2^a(n+1) - 1 and p = a(n+1) is prime. - David Morales Marciel, Oct 23 2015 LINKS Metin Sariyar, Table of n, a(n) for n = 1..32000 (terms 1..1000 from Paolo P. Lava) B. Dearden, J. Iiams, and J. Metzger, A Function Related to the Rumor Sequence Conjecture, J. Int. Seq. 14 (2011), #11.2.3, Example 7. FORMULA From Robert Israel, Oct 23 2015: (Start) a(2*n) = 2*a(n). a(2*n+1) = 2*a(n) + A070939(n) for n >= 1. G.f. A(x) satisfies A(x) = 2*(1+x)*A(x^2) + (x/(1-x^2))*Sum_{i>=1} x^(2^i). (End) MAPLE with(numtheory); P:=proc(q) local a, b, c, n; a:=0; for n from 1 to q do a:=convert(n, binary, decimal); b:=1; c:=0; while (a mod 10^b)

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Last modified August 18 12:29 EDT 2022. Contains 356212 sequences. (Running on oeis4.)