A111317 - OEIS

%I #37 Nov 18 2022 08:56:24

%S 1,1,0,0,1,0,-1,1,1,-1,0,1,-1,0,2,-1,-1,2,-1,-2,3,1,-3,2,1,-4,2,3,-4,

%T 1,4,-5,0,6,-5,-2,7,-5,-4,10,-3,-7,10,-2,-10,11,1,-13,11,4,-16,11,9,

%U -19,8,12,-22,7,19,-24,2,24,-26,-3,32,-25,-10,37,-25,-18,45,-21,-29,49,-17,-39,56,-8,-51,58,0,-65,61,14,-78,59,27,-92

%N Let f(a,q) = Product_{j>=0} (1 - a*q^j); g.f. is f(q^2,q^3) / f(q,q^3).

%C Convolution inverse of A111165.

%H Alois P. Heinz, <a href="/A111317/b111317.txt">Table of n, a(n) for n = 0..10000</a>

%H G. E. Andrews and B. C. Berndt, <a href="http://www.ams.org/notices/200801/tx080100018p.pdf">Your Hit Parade: The Top Ten Most Fascinating Formulas in Ramanujan's Lost Notebook</a>, Notices Amer. Math. Soc., 55 (No. 1, 2008), 18-30. See p. 25, Equation (39).

%F Euler transform of period 3 sequence [ 1, -1, 0, ...]. - _Michael Somos_, Dec 23 2007

%F G.f.: Product_{k>=0} (1 - x^(3*k+2)) / (1 - x^(3*k+1)).

%F G.f.: exp( Sum_{n>=1} 1/(1 + x^n + x^(2n)) * x^n/n ). - _Paul D. Hanna_, Jan 23 2010

%F From _Peter Bala_, Dec 2012: (Start)

%F Let F(x) denote the o.g.f. of this sequence. For positive integer n >= 2, the real number F(1/n) has the simple continued fraction expansion 1 + 1/(n-1 + 1/(1 + 1/(n^2-1 + 1/(1 + 1/(n^3-1 + 1/(1 + ...)))))).

%F For n >= 2, F(-1/n) has the simple continued fraction expansion

%F 1/(1 + 1/(n-1 + 1/(n^2+1 + 1/(n^3-1 + ...)))). Examples are given below. Cf. A005169 and A143951.

%F (End)

%e From _Peter Bala_, Dec 2012: (Start)

%e F(1/10) = Sum_{n>=0} a(n)/10^n has the simple continued fraction expansion 1 + 1/(9 + 1/(1 + 1/(99 + 1/(1 + 1/(999 + 1/(1 + ...)))))).

%e F(-1/10) = Sum_{n>=0} (-1)^n*a(n)/10^n has the simple continued fraction expansion 1/(1 + 1/(9 + 1/(101 + 1/(999 + 1/(1001 + ...))))).

%e (End)

%p a:= proc(n) option remember; `if`(n=0, 1,

%p       add(add(d*[0, 1, -1][irem(d, 3)+1],

%p       d=numtheory[divisors](j))*a(n-j), j=1..n)/n)

%p     end:

%p seq(a(n), n=0..80);  # _Alois P. Heinz_, Apr 01 2014

%t a[n_] := a[n] = If[n == 0, 1, Sum[Sum[d*{0, 1, -1}[[Mod[d, 3]+1]], {d, Divisors[j]}]*a[n-j], {j, 1, n}]/n]; Table[a[n], {n, 0, 80}] (* _Jean-François Alcover_, Apr 09 2014, after _Alois P. Heinz_ *)

%o (PARI) {a(n) = if( n<0, 0, polcoeff( prod(k=0, n\3, (1 - x^(3*k+2)) / (1 - x^(3*k+1)), 1 + x * O(x^n)), n))} /* _Michael Somos_, Dec 23 2007 */

%o (PARI) {a(n)=polcoeff(exp(sum(m=1,n+1,1/(1+x^m+x^(2*m)+x*O(x^n))*x^m/m)),n)} \\ _Paul D. Hanna_, Jan 23 2010

%o (Sage) # uses[EulerTransform from A166861]

%o b = BinaryRecurrenceSequence(-1, -1)

%o a = EulerTransform(b)

%o print([a(n) for n in range(88)]) # _Peter Luschny_, Nov 17 2022

%Y Cf. A005169, A143951.

%K sign,look

%O 0,15

%A _N. J. A. Sloane_, Nov 09 2005