A207433 G.f.: Sum_{n>=0} Product_{k=1..n} (q^k - 1) where q = (1-x^3)/(1-x).

1, 1, 3, 11, 56, 350, 2609, 22582, 222625, 2462969, 30219676, 407276420, 5981197376, 95073427910, 1626294895274, 29788176027819, 581704672430937, 12064521684969823, 264843222932272690, 6135057298705027024, 149559103545555671423, 3827360866024134614644

Offset: 0

Author

Paul D. Hanna, Feb 17 2012

Keywords

• nonn
• easy

Comments

Motivated by Peter Bala's identity described in A158690:

Sum_{n>=0} Product_{k=1..n} (q^k - 1) =

Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1),

here q = (1+x)/(1+x^2). See cross-references for other examples.

At present Bala's identity is conjectural and needs formal proof.

a(n) = number of upper triangular matrices with entries from {0,1,2} with no zero rows such that the sum of the entries is n, that is, row Fishburn matrices of size n with entries from {0,1,2}. Cf. A179525. - Peter Bala, Nov 05 2017

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 11*x^3 + 56*x^4 + 350*x^5 + 2609*x^6 +...
Let q = (1-x^3)/(1-x) = 1 + x + x^2, then
A(x) = 1 + (q-1) + (q-1)*(q^2-1) + (q-1)*(q^2-1)*(q^3-1) + (q-1)*(q^2-1)*(q^3-1)*(q^4-1) + (q-1)*(q^2-1)*(q^3-1)*(q^4-1)*(q^5-1) +...
Also, we have the identity:
A(x) = 1 + (q-1)/q + (q-1)*(q^3-1)/q^4 + (q-1)*(q^3-1)*(q^5-1)/q^9 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)/q^16 + (q-1)*(q^3-1)*(q^5-1)*(q^7-1)*(q^9-1)/q^25 +...
From _Peter Bala_, Nov 05 2017: (Start)
a(3) = 11: The eleven row Fishburn matrices of size 3 with entries in {0,1,2} are
/1 0\ /2 0\ /0 1\ /0 2\ /1 1\
\0 2/ \0 1/ \0 2/ \0 1/ \0 1/
and
/1 0 0\ /0 1 0\ /0 0 1\ /1 0 0\ /0 1 0\ /0 0 1\
|0 1 0| |0 1 0| |0 0 1| |0 0 1| |0 0 1| |0 0 1|.
\0 0 1/ \0 0 1/ \0 0 1/ \0 0 1/ \0 0 1/ \0 0 1/
(End)
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..260
• Hsien-Kuei Hwang, Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.

Crossrefs

Cf. A158690, A158691, A179525, A207386, A207397.

Programs

PARI

{a(n)=local(A=1+x,q=(1+x+x^2 +x*O(x^n))); A=sum(m=0,n,prod(k=1,m,(q^k-1)));polcoeff(A,n)}

PARI

{a(n)=local(A=1+x,q=(1+x+x^2 +x*O(x^n))); A=sum(m=0,n,q^(-m^2)*prod(k=1,m,(q^(2*k-1)-1)));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

Formula

G.f.: Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1) where q = (1-x^3)/(1-x). [Based on Peter Bala's conjecture in A158690]

a(n) ~ exp(Pi^2/24) * 2^(2*n+3/2) * 3^(n+1) * n! / Pi^(2*n+2). - Vaclav Kotesovec, Aug 22 2017