The OEIS is supported by the many generous donors to the OEIS Foundation. Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 60, we have over 367,000 sequences, and we’ve crossed 11,000 citations (which often say “discovered thanks to the OEIS”). Other ways to Give
 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A171922 For definition see Comments lines. 10
 1, 2, 2, 4, 2, 4, 6, 4, 2, 11, 6, 11, 2, 4, 12, 4, 16, 11, 12, 14, 16, 6, 24, 2, 29, 9, 29, 4, 24, 12, 32, 14, 24, 11, 16, 29, 23, 6, 38, 8, 41, 26, 32, 40, 38, 16, 24, 2, 41, 43, 41, 29, 42, 12, 9, 71, 4, 11, 35, 53, 6, 11, 24, 14, 71, 23, 9, 11, 32, 35, 47, 2, 58, 24, 58 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Constructed in an attempt to find the lexicographically earliest sequence of positive integers, not all 1's, with the property that for n >= 2, if a(n-1) = k, then min(m : a(n+m) = a(n), m > 0) = k. However, the sequence is well-defined, even if should fail to satisfy that property. The sequence is constructed as follows: 1) Given a(n-1) = k, we require min(m : a(n+m) = a(n), m > 0) = k. 2) If a(1) = a(2) = 1, we find by induction that a(n) = 1 for all n, so this is forbidden. 3) For n > 1, if a(n) = k then a(n+b_n(i)) = k for all i, with b_n(0) = 0 and b_n(i+1) = b_n(i) + a(n+b_n(i)-1). Hence every such k appears infinitely often. 4) Hence any a(n) not forced to be equal to a previous a(m) must have some new, never-seen-before value (or violate (1)). Whether such force exists is completely determined by the a(m): 1 <= m < n. 5) We can fully characterize a valid sequence by C = [ c_i ], the distinct values that it takes in order of first appearance. We can then generate the original sequence using (1) and (4). The desired sequence is that generated by the lexically earliest C. 6) Given a(n) = k, we must avoid a(n+m) = k-m for all m > 0, else we would have a(n+1) = a(n+1+k) = a(n+m+1), violating (1). 7) Given known a(n-1) = x, a(n+1) = y and trying to find a(n) = k, we have a(n+1) = a(n+k+1) = y. So by (6) and (3) we must avoid b_n(i) = y+1 for all i. We make the (unproved) assumption that defending against both (6) and (7) is sufficient to avoid backtracking. That appears to work, and produces the current sequence. The associated sequence C is A171921. The sequence has the property that its forwards van Eck transform (see A171898) is the same sequence prefixed with 0. - N. J. A. Sloane, Oct 23 2010 LINKS Hugo van der Sanden, Table of n, a(n) for n = 1..15137 Hugo van der Sanden, Perl program for this and related sequences N. J. A. Sloane, Transforms CROSSREFS Cf. A171921, A171898, A171925, A171926, A171927, A171930, A171939, A171940, A309681. Sequence in context: A260723 A059214 A091820 * A306743 A140821 A063789 Adjacent sequences: A171919 A171920 A171921 * A171923 A171924 A171925 KEYWORD nonn AUTHOR Hugo van der Sanden and N. J. A. Sloane, Oct 23 2010, Oct 24 2010 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified December 2 01:36 EST 2023. Contains 367505 sequences. (Running on oeis4.)