A123916 Number of binary words whose (unique) decreasing Lyndon decomposition is into Lyndon words each with an odd number of 1's; EULER transform of A000048.
1, 1, 2, 3, 6, 10, 19, 34, 65, 120, 229, 432, 829, 1583, 3051, 5874, 11370, 22012, 42756, 83113, 161917, 315723, 616588, 1205232, 2358604, 4619485, 9055960, 17766086, 34880215, 68524486, 134707150, 264960828, 521449025
Offset: 1
Keywords
Examples
The binary words 1111, 1101, 1001, 0101, 0111, 0001 of length 4 decompose as 1*1*1*1, 1*1*01, 1*001, 01*01, 0111, 0001 and each subword has an odd number of 1's, therefore a(4)=6. G.f. A(x) = x + x^2 + 2*x^3 + 3*x^4 + 6*x^5 + 10*x^6 + 19*x^7 + 34*x^8 + ... such that A(x)^2 * (1 - 2*x) = A(x^2).
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..700
Programs
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PARI
/* G.f. A(x) satisfies: A(x)^2 = A(x^2)/(1 - 2*x) */ {a(n) = my(A=x); for(i=1,n, A = sqrt( subst(A,x, x^2)/(1 - 2*x +x*O(x^n)))); polcoeff(A,n)} for(n=1,50, print1(a(n),", ")) \\ Paul D. Hanna, Apr 17 2016
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PARI
/* As the EULER transform of A000048 */ {A000048(n) = sumdiv(n, d, (d%2)*(moebius(d)*2^(n/d)))/(2*n)} \\ Michael B. Porter {a(n) = polcoeff( prod(k=1,n, 1/(1 - x^k +x*O(x^n))^A000048(k)), n-1)} for(n=1,50, print1(a(n),", ")) \\ Paul D. Hanna, Apr 17 2016
Formula
Prod_{n>=1} 1/(1-q^n)^A000048(n) = 1 + sum_{n>=1} a(n) q^n.
G.f. A(x) satisfies: A(x)^2 = A(x^2) / (1 - 2*x). - Paul D. Hanna, Apr 17 2016
a(n) ~ c * 2^n / sqrt(n), where c = 0.3412831644583761326654... . - Vaclav Kotesovec, Apr 18 2016
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = v^2 * (v^2 - 2*u^2*v - u^4) + 2*w*u^4. - Michael Somos, Jun 27 2017
Comments