%I #35 Jul 21 2018 17:35:36
%S 1,-1,1,-2,3,-4,6,-10,15,-22,34,-52,78,-118,180,-274,415,-630,958,
%T -1454,2206,-3350,5088,-7724,11726,-17806,27036,-41046,62320,-94624,
%U 143668,-218130,331191,-502854,763486,-1159206,1760038,-2672286,4057356,-6160326,9353294,-14201206,21561836
%N Defining sequence for an inverse Fredholm-Rueppel triangle.
%C A104975 is the sequence array for this sequence.
%C abs(a(n)) is the numbers of compositions of n into parts of the form 2^k-1, see example. [_Joerg Arndt_, Jan 06 2013]
%H Robert Israel, <a href="/A104977/b104977.txt">Table of n, a(n) for n = 0..5509</a>
%H Johann Cigler, <a href="https://mathoverflow.net/questions/292231">Relation between a continued fraction and partitions</a>, mathoverflow question 292231, Feb 05 2018.
%H Johann Cigler, <a href="https://arxiv.org/abs/1803.05164">A curious class of Hankel determinants</a>, arXiv:1803.05164 [math.CO], 2018.
%F G.f.: x / (sum_{k>=0} x^(2^k)). (corrected by _Joerg Arndt_, Jan 06 2013)
%F G.f.: 1 / (1 + sum(n>=1, x^(2^n-1) ) ), replace the '+' by '-' to obtain the g.f. for compositions into parts 2^k-1. [_Joerg Arndt_, Jan 06 2013]
%F G.f.: 1 - x / (1 + x / (1 + x / (1 - x / (1 + x / (1 - x / ...))))) = 1 + b(1) * x / (1 + b(2) * x / (1 + b(3) * x / ...)) where b(n) = (-1)^ A090678(n+1) [Conjecture]. - _Michael Somos_, Jan 03 2013
%F Convolution inverse is A209229 with 0 preprended. - _Michael Somos_, Jan 03 2013
%F a(n) = sum(m=0..n, T(n,m)*(-1)^(m)), where T(n,m)=(1+(-1)^(n-m))/2 *sum(k=1..(n-m)/2, binomial(m,k)*T((n-m)/2,k)), T(n,n)=1. - _Vladimir Kruchinin_, Mar 18 2015
%e G.f. = 1 - x + x^2 - 2*x^3 + 3*x^4 - 4*x^5 + 6*x^6 - 10*x^7 + 15*x^8 - 22*x^9 + 34*x^10 + ...
%e From _Joerg Arndt_, Jan 06 2013: (Start)
%e There are abs(a(8)) = 15 compositions of 8 into parts 2^k-1:
%e [ 1] [ 1 1 1 1 1 1 1 1 ]
%e [ 2] [ 1 1 1 1 1 3 ]
%e [ 3] [ 1 1 1 1 3 1 ]
%e [ 4] [ 1 1 1 3 1 1 ]
%e [ 5] [ 1 1 3 1 1 1 ]
%e [ 6] [ 1 1 3 3 ]
%e [ 7] [ 1 3 1 1 1 1 ]
%e [ 8] [ 1 3 1 3 ]
%e [ 9] [ 1 3 3 1 ]
%e [10] [ 1 7 ]
%e [11] [ 3 1 1 1 1 1 ]
%e [12] [ 3 1 1 3 ]
%e [13] [ 3 1 3 1 ]
%e [14] [ 3 3 1 1 ]
%e [15] [ 7 1 ]
%e (End)
%p N:= 100: # to get a(0)..a(N)
%p S:= series(x/add(x^(2^k),k=0..ilog2(N+1)),x,N+2):
%p [seq](coeff(S, x, j), j = 0 .. N); # _Robert Israel_, Feb 07 2018
%t T[n_, n_] = 1; T[n_, m_] := T[n, m] =(1 + (-1)^(n-m))/2 Sum[Binomial[m, k]* T[(n-m)/2, k], {k, 1, (n-m)/2}];
%t a[n_] := Sum[T[n, m] (-1)^m, {m, 0, n}];
%t Array[a, 50, 0] (* _Jean-François Alcover_, Jul 21 2018, after _Vladimir Kruchinin_ *)
%o (PARI)
%o N = 66; q = 'q + O('q^N); L = 2+ceil(log(N)/log(2));
%o gf = q / sum(n=0, L, q^(2^n) );
%o /* gf = 1 / (1 + sum(n=1, L, q^(2^n-1) ) ); */ /* same */
%o v = Vec(gf)
%o /* _Joerg Arndt_, Jan 06 2013 */
%o (Maxima)
%o T(n,m):=if n=m then 1 else (1+(-1)^(n-m))/2*sum(binomial(m,k)*T((n-m)/2,k),k,1,(n-m)/2); makelist(sum(T(n,m)*(-1)^(m),m,0,n),n,0,20); /* _Vladimir Kruchinin_, Mar 18 2015 */
%Y Cf. A090678, A209229.
%K easy,sign
%O 0,4
%A _Paul Barry_, Mar 30 2005
%E Added more terms, _Joerg Arndt_, Jan 06 2013
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