OFFSET
0,3
COMMENTS
The convolution of this sequence results in A073710 and is equal to the first differences of the unique terms of this sequence.
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
FORMULA
G.f. satisfies: A(x) = A(x^2)^2/(1-x).
G.f.: Product_{n>=0} 1/(1-x^(2^n))^(2^n). [Paul D. Hanna, May 01 2010]
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 3*x^3 + 10*x^4 + 10*x^5 + 22*x^6 + 22*x^7 +...
where A(x) = A(x^2)^2/(1-x) and thus
A(x) = 1 / [(1-x)*(1-x^2)^2*(1-x^4)^4*(1-x^8)^8*(1-x^16)^16*...].
Compare A(x)*(1-x) to A(x)^2:
A(x)*(1-x) = 1 + 2*x^2 + 7*x^4 + 12*x^6 + 35*x^8 + 58*x^10 + 133*x^12 +...
A(x)^2 = 1 + 2*x + 7*x^2 + 12*x^3 + 35*x^4 + 58*x^5 + 133*x^6 + 208*x^7 +...
Also note that
A(x)^2/(1-x) = 1 + 3*x + 10*x^2 + 22*x^3 + 57*x^4 + 115*x^5 + 248*x^6 + 456*x^7 +...
MATHEMATICA
terms = 42; For[m = 1; A = 1, m <= 2*terms, m = 2*m, A = ((1+x)*(Normal[A] /. x -> x^2))^2 + O[x]^m]; Join[{1}, Differences[CoefficientList[A, x] ]][[1 ;; terms]] (* Jean-François Alcover, Mar 06 2013, updated Apr 23 2016 *)
PROG
(PARI) {a(n)=polcoeff(prod(j=0, #binary(n), 1/(1-x^(2^j)+x*O(x^n))^(2^j)), n)} \\ Paul D. Hanna, May 01 2010
(Haskell)
a073709 n = a073709_list !! n
a073709_list = 1 : zipWith (-) (tail a073708_list) a073708_list
--- Reinhard Zumkeller, Jun 13 2013
CROSSREFS
KEYWORD
easy,nice,nonn
AUTHOR
Paul D. Hanna, Aug 05 2002
STATUS
approved