A266535 -id:A266535 - cp's OEIS frontend

A266540 Partial sums of A266539.

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, 92, 103, 114, 127, 140, 155, 170, 171, 172, 175, 178, 183, 188, 195, 202, 211, 220, 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

Offset: 1

Omar E. Pol, Jan 02 2016

nonn

Also A266535 and twice the terms of A256249 interleaved, or in other words A266535 and A266538 interleaved.
It appears that this sequence has a fractal (or fractal-like) behavior.
First differs from both A266510 and A266530 at a(25), with which it shares infinitely many terms.
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.
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)).

Robert Israel, Table of n, a(n) for n = 1..10000
Yuri Nikolayevsky and Ioannis Tsartsaflis, Cohomology of N-graded Lie algebras of maximal class over Z_2, arXiv:1512.87676 [math.RA], (2016); see pp. 2 and 6.
Index entries for sequences related to the Josephus Problem

Cf. A006257 (Josephus problem), A256249, A266535, A266510, A266530, A266538, A266539.

A006257[0]:=0: for n from 1 to 100 do A006257[n]:=(A006257[n-1]+1) mod n +1: end do:
ListTools:-PartialSums([seq(A006257[i]$2,i=0..100)]); # Robert Israel, Jan 13 2016

Join[{0, 0}, Table[{k, k}, {n, 1, 6}, {k, 1, 2^n-1, 2}] // Flatten] // Accumulate (* Jean-François Alcover, Sep 19 2018 *)

a(2n-1) = A266535(n).
a(2n) = 2 * A256249(n-1) = A266538(n-1).
a(n) = (a(n-1) + a(n+1))/2, if n is an odd number greater than 1.
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

A266539 Terms of A006257 (Josephus problem) repeated.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 3, 3, 1, 1, 3, 3, 5, 5, 7, 7, 1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 13, 13, 15, 15, 1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 13, 13, 15, 15, 17, 17, 19, 19, 21, 21, 23, 23, 25, 25, 27, 27, 29, 29, 31, 31, 1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 13, 13, 15, 15, 17, 17, 19, 19, 21, 21, 23, 23, 25, 25

Offset: 1

Omar E. Pol, Jan 02 2016

First differs from both A266509 and A266529 at a(25), and shares with them infinitely many terms.

			Written as an irregular triangle in which the row lengths are twice the terms of A011782 the sequence begins:
   0, 0;
   1, 1;
   1, 1, 3, 3;
   1, 1, 3, 3, 5, 5, 7, 7;
   1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 13, 13, 15, 15;
   ...
Row sums give 0 together with A004171.

Partial sums give A266540.

Cf. A004171, A006257, A011782, A256249, A266509, A266529, A266535.

A006257[0]:=0: for n from 1 to 100 do A006257[n]:=(A006257[n-1]+1) mod n +1: end do:
seq(A006257[i]$2,i=0..100); # Robert Israel, Jan 13 2016

Join[{0, 0}, Table[SeriesCoefficient[(1 + x^2)/((-1 + x)^2 (1 + x)), {x, 0, m}], {n, 6}, {m, 0, 2^n - 1}]] // Flatten (* Michael De Vlieger, Jan 05 2016 *)

G.f.: (x^2 + x^4)/(1 - x - x^2 + x^3) - (1 - x)^(-1)*Sum_{k>=1} 2^k*x^(2^(k+1)). - Robert Israel, Jan 13 2016

Offset changed to 1 by Ivan Neretin, Feb 09 2017

A266538 Twice the partial sums of A006257 (Josephus problem).

Original entry on oeis.org

0, 2, 4, 10, 12, 18, 28, 42, 44, 50, 60, 74, 92, 114, 140, 170, 172, 178, 188, 202, 220, 242, 268, 298, 332, 370, 412, 458, 508, 562, 620, 682, 684, 690, 700, 714, 732, 754, 780, 810, 844, 882, 924, 970, 1020, 1074, 1132, 1194, 1260, 1330, 1404, 1482, 1564, 1650, 1740, 1834, 1932, 2034, 2140, 2250, 2364, 2482, 2604

Offset: 0

Omar E. Pol, Jan 12 2016

Index entries for sequences related to the Josephus Problem.

Bisection of A266540.

Cf. A006257, A256249, A266535, A266540.

2*Accumulate[Flatten[{0, Table[Range[1, 2^n - 1, 2], {n, 0, 7}]}]] (* Jake L Lande, Aug 05 2024 *)

a(n) = 2 * A256249(n).

cp's OEIS Frontend

A266540 Partial sums of A266539.

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A266539 Terms of A006257 (Josephus problem) repeated.

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A266538 Twice the partial sums of A006257 (Josephus problem).

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