A151550 Expansion of g.f. Product_{n >= 1} (1 + x^(2^n-1) + 2*x^(2^n)).
1, 1, 2, 1, 3, 4, 4, 1, 3, 4, 5, 5, 10, 12, 8, 1, 3, 4, 5, 5, 10, 12, 9, 5, 10, 13, 15, 20, 32, 32, 16, 1, 3, 4, 5, 5, 10, 12, 9, 5, 10, 13, 15, 20, 32, 32, 17, 5, 10, 13, 15, 20, 32, 33, 23, 20, 33, 41, 50, 72, 96, 80, 32, 1, 3, 4, 5, 5, 10, 12, 9, 5, 10, 13, 15, 20, 32, 32, 17, 5, 10, 13
Offset: 0
Keywords
Examples
From _Omar E. Pol_, Jun 09 2009, edited by _N. J. A. Sloane_, Jun 17 2009: May be written as a triangle: 0; 1; 1,2; 1,3,4,4; 1,3,4,5,5,10,12,8; 1,3,4,5,5,10,12,9,5,10,13,15,20,32,32,16; 1,3,4,5,5,10,12,9,5,10,13,15,20,32,32,17,5,10,13,15,20,32,33,23,20,33,41,... The rows of the triangle converge to A151555.
References
- D. 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
Links
- N. J. A. Sloane, Table of n, a(n) for n = 0..16383
- 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.], which is also available at arXiv:1004.3036v2, [math.CO], 2010.
- N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Crossrefs
For generating functions of the form Product_{k>=c} (1+a*x^(2^k-1)+b*x^2^k) for the following values of (a,b,c) see: (1,1,0) A160573, (1,1,1) A151552, (1,1,2) A151692, (2,1,0) A151685, (2,1,1) A151691, (1,2,0) A151688 and A152980, (1,2,1) A151550, (2,2,0) A151693, (2,2,1) A151694.
Cf. A000079. - Omar E. Pol, Jun 09 2009
Programs
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Mathematica
terms = 100; CoefficientList[Product[(1+x^(2^n-1) + 2 x^(2^n)), {n, 1, Log[2, terms] // Ceiling}] + O[x]^terms, x] (* Jean-François Alcover, Aug 05 2018 *)
Formula
To get a nice recurrence, change the offset to 0 and multiply the g.f. by x as in the triangle in the example lines. Then we have: a(0)=0; a(2^i)=1; a(2^i-1)=2^(i-1) for i >= 1; otherwise write n = 2^i+j with 1 <= j <= 2^i-2, then a(n) = a(2^i+j) = 2*a(j) + a(j+1).
Comments