A079948 First differences of A079000.

3, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Author

N. J. A. Sloane, Feb 22 2003

Keywords

• nonn

Comments

Alternate description of sequence: start with a(1)=3; apply 1->2, 2->11, 3->21; iterate. - Matthew Vandermast, Mar 08 2003

References

• N. J. A. Sloane, Seven Staggering Sequences, in Homage to a Pied Puzzler, E. Pegg Jr., A. H. Schoen and T. Rodgers (editors), A. K. Peters, Wellesley, MA, 2009, pp. 93-110.

Links

• B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
• B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.
• N. J. A. Sloane, Seven Staggering Sequences.

Programs

Mathematica

b[1] = 1; b[n_] := (k = Floor[Log[2, (n+3)/6]]; j = n - (9*2^k-3); 12*2^k - 3 + 3*j/2 + Abs[j]/2); Array[b, 106] // Differences (* Jean-François Alcover, Sep 02 2018 *)

Formula

After first two terms, a run of length 3*2^k 1's followed by a run of length 3*2^k 2's, for k = 0, 1, ...

a(n) = floor(log_2(8*(floor((n+3)/3))/3)) - floor(log_2(floor((n+3)/3))) for n>2; with a(1)=3 and a(2)=2. - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 22 2003

Also a(n) = A079882(A002264(n+3)) for n>2, where A002264=floor(n/3). - Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Feb 22 2003