A104977 Defining sequence for an inverse Fredholm-Rueppel triangle.

1, -1, 1, -2, 3, -4, 6, -10, 15, -22, 34, -52, 78, -118, 180, -274, 415, -630, 958, -1454, 2206, -3350, 5088, -7724, 11726, -17806, 27036, -41046, 62320, -94624, 143668, -218130, 331191, -502854, 763486, -1159206, 1760038, -2672286, 4057356, -6160326, 9353294, -14201206, 21561836

Offset: 0

Author

Paul Barry, Mar 30 2005

Keywords

• easy
• sign

Comments

A104975 is the sequence array for this sequence.

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]

Examples

			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 + ...
From _Joerg Arndt_, Jan 06 2013: (Start)
There are abs(a(8)) = 15 compositions of 8 into parts 2^k-1:
[ 1]  [ 1 1 1 1 1 1 1 1 ]
[ 2]  [ 1 1 1 1 1 3 ]
[ 3]  [ 1 1 1 1 3 1 ]
[ 4]  [ 1 1 1 3 1 1 ]
[ 5]  [ 1 1 3 1 1 1 ]
[ 6]  [ 1 1 3 3 ]
[ 7]  [ 1 3 1 1 1 1 ]
[ 8]  [ 1 3 1 3 ]
[ 9]  [ 1 3 3 1 ]
[10]  [ 1 7 ]
[11]  [ 3 1 1 1 1 1 ]
[12]  [ 3 1 1 3 ]
[13]  [ 3 1 3 1 ]
[14]  [ 3 3 1 1 ]
[15]  [ 7 1 ]
(End)
		

Links

• Robert Israel, Table of n, a(n) for n = 0..5509
• Johann Cigler, Relation between a continued fraction and partitions, mathoverflow question 292231, Feb 05 2018.
• Johann Cigler, A curious class of Hankel determinants, arXiv:1803.05164 [math.CO], 2018.

Crossrefs

Cf. A090678, A209229.

Programs

Maple

N:= 100: # to get a(0)..a(N)
S:= series(x/add(x^(2^k),k=0..ilog2(N+1)),x,N+2):
[seq](coeff(S, x, j), j = 0 .. N); # Robert Israel, Feb 07 2018

Mathematica

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}];
a[n_] := Sum[T[n, m] (-1)^m, {m, 0, n}];
Array[a, 50, 0] (* Jean-François Alcover, Jul 21 2018, after Vladimir Kruchinin *)

Maxima

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 */

PARI

N = 66;  q = 'q + O('q^N);  L = 2+ceil(log(N)/log(2));
gf = q / sum(n=0, L, q^(2^n) );
/* gf = 1 / (1 + sum(n=1, L, q^(2^n-1) ) ); */  /* same */
v = Vec(gf)
/* Joerg Arndt, Jan 06 2013 */

Formula

G.f.: x / (sum_{k>=0} x^(2^k)). (corrected by Joerg Arndt, Jan 06 2013)

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]

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

Convolution inverse is A209229 with 0 preprended. - Michael Somos, Jan 03 2013

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

Extensions

Added more terms, Joerg Arndt, Jan 06 2013