%I #60 Feb 22 2023 18:27:22
%S 0,0,1,2,3,4,7,10,11,12,15,18,23,28,35,42,43,44,47,50,55,60,67,74,83,
%T 92,103,114,127,140,155,170,171,172,175,178,183,188,195,202,211,220,
%U 231,242,255,268,283,298,315,332,351,370,391,412,435,458,483,508,535,562,591,620,651,682,683,684,687,690,695,700
%N Partial sums of A266539.
%C Also A266535 and twice the terms of A256249 interleaved, or in other words A266535 and A266538 interleaved.
%C It appears that this sequence has a fractal (or fractal-like) behavior.
%C First differs from both A266510 and A266530 at a(25), with which it shares infinitely many terms.
%C For an illustration of initial terms consider the diagram of A256249 in the fourth quadrant of the square grid together with a reflected copy in the second quadrant.
%C Also the third sequence of Betti numbers of the Lie algebra m_0(n) over Z_2. See the Nikolayevsky-Tsartsaflis paper, pages 2 and 6. Note that a(n) is denoted by b_3(m_0(n)).
%H Robert Israel, <a href="/A266540/b266540.txt">Table of n, a(n) for n = 1..10000</a>
%H Yuri Nikolayevsky and Ioannis Tsartsaflis, <a href="http://arxiv.org/abs/1512.07676">Cohomology of N-graded Lie algebras of maximal class over Z_2</a>, arXiv:1512.87676 [math.RA], (2016); see pp. 2 and 6.
%H <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>
%F a(2n-1) = A266535(n).
%F a(2n) = 2 * A256249(n-1) = A266538(n-1).
%F a(n) = (a(n-1) + a(n+1))/2, if n is an odd number greater than 1.
%F G.f.: (x^3+x^5)/(1-2*x+2*x^3-x^4) - x*(1-x)^(-2)*Sum_{k>=1} 2^k*x^(2^(1+k)). - _Robert Israel_, Jan 13 2016
%p A006257[0]:=0: for n from 1 to 100 do A006257[n]:=(A006257[n-1]+1) mod n +1: end do:
%p ListTools:-PartialSums([seq(A006257[i]$2,i=0..100)]); # _Robert Israel_, Jan 13 2016
%t Join[{0, 0}, Table[{k, k}, {n, 1, 6}, {k, 1, 2^n-1, 2}] // Flatten] // Accumulate (* _Jean-François Alcover_, Sep 19 2018 *)
%Y Cf. A006257 (Josephus problem), A256249, A266535, A266510, A266530, A266538, A266539.
%K nonn
%O 1,4
%A _Omar E. Pol_, Jan 02 2016