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A064064
n-th step is to add a(n) to each previous number a(k) (including itself, i.e., k <= n) to produce n+1 more terms of the sequence, starting with a(0)=1.
7
1, 2, 3, 4, 4, 5, 6, 5, 6, 7, 8, 5, 6, 7, 8, 8, 6, 7, 8, 9, 9, 10, 7, 8, 9, 10, 10, 11, 12, 6, 7, 8, 9, 9, 10, 11, 10, 7, 8, 9, 10, 10, 11, 12, 11, 12, 8, 9, 10, 11, 11, 12, 13, 12, 13, 14, 9, 10, 11, 12, 12, 13, 14, 13, 14, 15, 16, 6, 7, 8, 9, 9, 10, 11, 10, 11, 12, 13, 10, 7, 8, 9, 10, 10
OFFSET
0,2
COMMENTS
Each number eventually appears A001190 times (binary rooted trees can be constructed by combining earlier trees in a similar manner with the n-th tree having a(n) endpoints).
The number of leaves in the tree of rank n+1 according to the Colijn-Plazzotta ranking scheme for unlabeled binary rooted trees. - Noah A Rosenberg, Jun 14 2022
LINKS
C. Colijn and G. Plazzotta, A Metric on Phylogenetic Tree Shapes, Systematic Biology, volume 67, number 1, January 2018, pages 113-126, with section 2.3 number of tips v_a(1) = a(n) for tree R_a = n+1.
N. A. Rosenberg, On the Colijn-Plazzotta numbering scheme for unlabeled binary rooted trees, Discr. Appl. Math. 291 (2021), 88-98.
FORMULA
a(0) = 1 and a(n+1) = a(A002262(n)) + a(A003056(n)) for any n >= 0. - Rémy Sigrist, Aug 07 2017
a(n) = (A064002(n+1) + 1)/2. - Kevin Ryde, Jul 22 2022
EXAMPLE
Start with (1). So after initial step we have (*1*, 1+1 = 2), then (1, *2*, 1+2 = 3, 2+2 = 4), then (1, 2, *3*, 4, 1+3 = 4, 2+3 = 5, 3+3 = 6), then (1, 2, 3, *4*, 4, 5, 6, 1+4 = 5, 2+4 = 6, 3+4 = 7, 4+4 = 8), then (1, 2, 3, 4, *4*, 5, 6, 5, 6, 7, 8, 1+4 = 5, 2+4 = 6, 3+4 = 7, 4+4 = 8, 4+4 = 8), etc.
MATHEMATICA
a[0]=1; a[n_]:=With[{s = Floor[(Sqrt[8*n - 7] - 1)/2]}, a[s] + a[n - s*(s + 1)/2 - 1]]; Array[a, 84, 0] (* Harry Richman, Feb 24 2024 *)
PROG
(PARI) a(n) = if (n==0, return (1), return (a(A002262(n-1))+a(A003056(n-1)))) \\ Rémy Sigrist, Aug 07 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Henry Bottomley, Aug 31 2001
STATUS
approved