

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, sumfree 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 n1 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



FORMULA

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/(1x^2))*Sum_{i>=1} x^(2^i). (End)


MAPLE

f:= proc(n) option remember; local m;
if n::even then 2*procname(n/2)
else m:= (n1)/2; 2*procname(m) + ilog2(m) + 1
fi
end proc:
f(1):= 0:


MATHEMATICA

Table[n * Floor@Log[2, n]  Sum[Floor[n*2^k]*2^k, {k, Log[2, n]}], {n, 100}] (* Federico Provvedi, Aug 17 2013 *)


PROG

(PARI) a(n) = sum(k=1, logint(n, 2), n % 2^k); \\ Michel Marcus, Dec 12 2019


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



