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A160552 a(0)=0, a(1)=1; a(2^i+j) = 2*a(j) + a(j+1) for 0 <= j < 2^i. 38
0, 1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 5, 11, 17, 15, 1, 3, 5, 7, 5, 11, 17, 15, 5, 11, 17, 19, 21, 39, 49, 31, 1, 3, 5, 7, 5, 11, 17, 15, 5, 11, 17, 19, 21, 39, 49, 31, 5, 11, 17, 19, 21, 39, 49, 35, 21, 39, 53, 59, 81, 127, 129, 63, 1, 3, 5, 7, 5, 11, 17, 15, 5, 11, 17, 19, 21, 39, 49, 31 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,4
COMMENTS
This recurrence is patterned after the one for A152980, but without the special cases.
Sequence viewed as triangle:
0,
1,
1, 3,
1, 3, 5, 7,
1, 3, 5, 7, 5, 11, 17, 15,
1, 3, 5, 7, 5, 11, 17, 15, 5, 11, 17, 19, 21, 39, 49, 31.
The rows converge to A151548.
Also the sum of the terms in the k-th row (k >= 1) is 4^(k-1). Proof by induction. - N. J. A. Sloane, Jan 21 2010
If this sequence [1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 5, 11, 17, 15, ...] is convolved with [1, 2, 2, 2, 2, 2, 2, ...] we obtain A139250, the toothpick sequence. Example: A139250(5) = 15 = (1, 2, 2, 2, 2) * (3, 1, 3, 1, 1). - Gary W. Adamson, May 19 2009
Starting with 1 and convolved with [1, 2, 0, 0, 0, ...] = A151548. - Gary W. Adamson, Jun 04 2009
Refer to A162956 for the analogous triangle using N=3. - Gary W. Adamson, Jul 20 2009
It appears that the sums of two successive terms give the positive terms of A139251. - Omar E. Pol, Feb 18 2015
LINKS
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.]
FORMULA
G.f.: x*(1+2*x)/(1+x) + (4*x^2/(1+2*x))*(-1 + Product_{k>=1} (1 + x^(2^k-1) + 2*x^(2^k))). - N. J. A. Sloane, May 23 2009, based on Gary W. Adamson's comment above and the known g.f. for A139250.
It appears that a(n) = A169708(n)/4, n >= 1. - Omar E. Pol, Feb 15 2015
It appears that a(n) = A139251(n) - a(n-1), n >= 1. - Omar E. Pol, Feb 18 2015
EXAMPLE
a(2) = a(2^1+0) = 2*a(0) + a(1) = 1, a(3) = a(2^1+1) = 2*a(1) + a(2) = 3*a(2^i) = 2*a(0) + a(1) = 1.
MAPLE
S:=proc(n) option remember; local i, j; if n <= 1 then RETURN(n); fi; i:=floor(log(n)/log(2)); j:=n-2^i; 2*S(j)+S(j+1); end; # N. J. A. Sloane, May 18 2009
H := x*(1+2*x)/(1+x) + (4*x^2/(1+2*x))*(mul(1+x^(2^k-1)+2*x^(2^k), k=1..20)-1); series(H, x, 120); # N. J. A. Sloane, May 23 2009
MATHEMATICA
Nest[Join[#, 2 # + Append[Rest@#, 1]] &, {0}, 7] (* Ivan Neretin, Feb 09 2017 *)
CROSSREFS
For the recurrence a(2^i+j) = C*a(j) + D*a(j+1), a(0) = A, a(1) = B for following values of (A B C D) see: (0 1 1 1) A118977, (1 0 1 1) A151702, (1 1 1 1) A151570, (1 2 1 1) A151571, (0 1 1 2) A151572, (1 0 1 2) A151703, (1 1 1 2) A151573, (1 2 1 2) A151574, (0 1 2 1) A160552, (1 0 2 1) A151704, (1 1 2 1) A151568, (1 2 2 1) A151569, (0 1 2 2) A151705, (1 0 2 2) A151706, (1 1 2 2) A151707, (1 2 2 2) A151708.
Sequence in context: A371928 A179760 A334852 * A256263 A006257 A323555
KEYWORD
nonn
AUTHOR
David Applegate, May 18 2009
STATUS
approved

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Last modified April 24 19:52 EDT 2024. Contains 371963 sequences. (Running on oeis4.)