A078649 -id:A078649 - cp's OEIS frontend

A156351 a(n) = Sum_{k=1..n} (-1)^K(k+1)*(K(k+1)-K(k)) where K(k) = A000002(k).

1, 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8, 8, 9, 10, 10, 11, 11, 12, 13, 14, 14, 15, 16, 17, 17, 18, 18, 19, 20, 20, 21, 22, 23, 23, 24, 25, 25, 26, 26, 27, 28, 29, 29, 30, 31, 32, 32, 33, 34, 34, 35, 35, 36, 37, 37, 38, 38, 39, 40, 41, 41, 42, 43, 43, 44, 44, 45, 46, 46, 47, 48, 49, 49

Offset: 1

Benoit Cloitre, Feb 08 2009

nonn

a(n)=1 plus the number of symbol changes in the first n terms of A078880. - Jean-Marc Fedou and Gabriele Fici, Mar 18 2010

Jon Maiga, Table of n, a(n) for n = 1..10000
J.M. Fedou and G. Fici, Some remarks on differentiable sequences and recursivity, Journal of Integer Sequences 13(3): Article 10.3.2 (2010). [From Jean-Marc Fedou and _Gabriele Fici_, Mar 18 2010]

Partial sums of A156728.

a2 = {1, 2, 2}; Do[ a2 = Join[a2, {1 + Mod[n - 1, 2]}], {n, 3, 80}, {i, 1, a2[[n]]}]; a[n_] := Sum[(-1)^a2[[k + 1]]*(a2[[k + 1]] - a2[[k]]), {k, 1, n}]; Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Jun 18 2013 *)

n - A054353(a(n)) = 1 if n is in A078649, n - A054353(a(n)) = 0 otherwise. A078649(n + 1 - a(n)) - n takes values among {0,1,2,3}.

a(n) = gcd(a(a(n-1)),2) + a(n-2) (conjectured). - Jon Maiga, Dec 07 2021

A157687 a(n) = n - A054353(A156351(n)).

Original entry on oeis.org

0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0

Offset: 1

Benoit Cloitre, Mar 04 2009

nonn

Cf. A156351, A156728, A157684, A157685, A157686

a(n) = 1 iff n is in A078649, a(n) = 0 iff n is in A054353.
a(n) = n-A054353(A157684(n)+A157685(n)).
a(n) = 1 - A156728(n). - Alan Michael Gómez Calderón, Dec 19 2024

A156256 Number of 1's separating successive 2's in the Kolakoski sequence A000002.

Original entry on oeis.org

0, 2, 1, 0, 1, 0, 2, 2, 0, 1, 2, 1, 0, 2, 2, 1, 0, 1, 0, 2, 1, 0, 1, 2, 2, 0, 1, 0, 2, 1, 0, 1, 0, 2, 2, 1, 0, 1, 2, 0, 1, 0, 2, 1, 0, 1, 0, 2, 2, 0, 1, 2, 1, 0, 1, 0, 2, 2, 1, 0, 2, 2, 0, 1, 2, 2, 0, 1, 0, 2, 1, 0, 1, 2, 0, 1, 0, 1, 2, 2, 0, 1, 0, 2, 1, 0, 1, 2, 2, 0, 1, 2, 1, 0, 2, 2, 1, 0, 1, 2

Offset: 1

Benoit Cloitre, Feb 07 2009

nonn

After deleting 0's in this sequence it remains the bisection of Kolakoski sequence A000002(2n+1) n>=1 given by A100428.
This is because A100428 gives the lengths of runs of 1's in Kolakoski sequence. - Jean-Christophe Hervé, Oct 14 2014
The Kolakovski sequence can be obtained back (except the initial 1) by the following substitution rules: insert 2 between two successive nonzero values and 0 -> 22, 1 -> 1, 2 -> 11. - Jean-Christophe Hervé, Oct 14 2014

			The Kolakoski sequence begins with 122112122122, thus this one begins 0, 2, 1, 0, 1, 0. - _Jean-Christophe Hervé_, Oct 14 2014

Cf. A000002, A078649, A100428, A248806.

a(n) = A078649(n+1)-A078649(n)-2.

Better name from Jean-Christophe Hervé, Oct 15 2014

cp's OEIS Frontend

A156351 a(n) = Sum_{k=1..n} (-1)^K(k+1)*(K(k+1)-K(k)) where K(k) = A000002(k).

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A157687 a(n) = n - A054353(A156351(n)).

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A156256 Number of 1's separating successive 2's in the Kolakoski sequence A000002.

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