A147646 If A139251 is written as a triangle with rows of lengths 1, 2, 4, 8, 16, ..., the n-th row begins with 2^n followed by the first 2^n-1 terms of the present sequence.
4, 8, 12, 12, 16, 28, 32, 20, 16, 28, 36, 40, 60, 88, 80, 36, 16, 28, 36, 40, 60, 88, 84, 56, 60, 92, 112, 140, 208, 256, 192, 68, 16, 28, 36, 40, 60, 88, 84, 56, 60, 92, 112, 140, 208, 256, 196, 88, 60, 92, 112, 140, 208, 260, 224, 172, 212, 296, 364, 488, 672, 704, 448, 132
Offset: 1
Keywords
Examples
Further comments: A139251 as a triangle is: . 1 . 2 4 . 4 4 8 12 . 8 4 8 12 12 16 28 32 . 16 4 8 12 12 16 28 32 20 16 28 36 40 60 88 80 . 32 4 8 12 12 16 28 32 20 16 28 36 40 60 88 80 36 16 28 36 40 60 88 84 56 ... leading to the present sequence: . 4 8 12 12 16 28 32 20 16 28 36 40 60 88 80 36 16 28 36 40 60 88 84 56 ... Note that this can also be written as a triangle: . 4 8 . 12 12 16 28 . 32 20 16 28 36 40 60 88 . 80 36 16 28 36 40 60 88 84 56 60 92 112 140 208 256 . 192 68 16 28 36 40 60 88 84 56 60 92 112 140 208 256 196 88 60 92 112 140 ... The first column is (n+1)2^n (where n is the row number), the second column is 2^(n+1)+4, and the rest exhibits the same constant column behavior, where the rows converge to: . 16 28 36 40 60 88 84 56 60 92 112 140 208 256 196 88 60 92 112 140 ... Once again this can be written as a triangle: . 16 . 28 36 40 60 . 88 84 56 60 92 112 140 208 . 256 196 88 60 92 112 140 208 260 224 172 212 296 364 488 672 . 704 452 152 60 92 112 140 208 260 224 172 212 296 364 488 672 708 480 236 ... and this behavior continues ad infinitum.
Links
- David Applegate, Table of n, a(n) for n = 1..2047
- David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
- N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
- Index entries for sequences related to toothpick sequences
Crossrefs
Programs
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Maple
S:=proc(n) option remember; local i,j; if n <= 0 then RETURN(0); fi; if n <= 2 then RETURN(2^(n+1)); fi; i:=floor(log(n)/log(2)); j:=n-2^i; if j=0 then RETURN(2*n+4); fi; if j<2^i-1 then RETURN(2*S(j)+S(j+1)); fi; if j=2^i-1 then RETURN(2*S(j)+S(j+1)-4); fi; -1; end; # N. J. A. Sloane, May 18 2009
Formula
Letting n = 2^i + j for 0 <= j < 2^i, we have the recurrence (see A139251 for proof):
a(1) = 4
a(2) = 8
a(n) = 2n+4 = 2*a(n/2) - 4 if j = 0
a(n) = 2*a(j) + a(j+1) - 4 if j = 2^i-1
a(n) = 2*a(j) + a(j+1) if 1 <= j < 2^i-1
Comments