cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-9 of 9 results.

A158690 Expansion of the basic hypergeometric series 1 + (1 - exp(-t)) + (1 - exp(-t))*(1 - exp(-3*t)) + (1 - exp(-t))*(1 - exp(-3*t))*(1 - exp(-5*t)) + ... as a series in t.

Original entry on oeis.org

1, 1, 5, 55, 1073, 32671, 1431665, 85363615, 6646603073, 654896692351, 79656194515025, 11722538113191775, 2052949879753739873, 421931472111868912831, 100568330857984368195185
Offset: 0

Views

Author

Peter Bala, Mar 24 2009

Keywords

Comments

We appear to get the same sequence by expanding 1 - (1 - exp(t)) + (1 - exp(t))*(1 - exp(2*t)) - (1 - exp(t))*(1 - exp(2*t))*(1 - exp(3*t)) + ... as a series in t. Compare with A079144. For other sequences with generating functions of a similar type see A000364, A000464, A002105 and A002439.
From Peter Bala, Mar 13 2017: (Start)
It appears that the g.f. has two other forms: either F(exp(-t)) where F(q) = Sum_{n >= 0} q^(n+1)*Product_{k = 1..n} 1 - q^(2*k) = q + q^2 + q^3 - q^7 - q^8 - q^10 - q^11 - ... is a g.f. for A003475 or 1/2*G(exp(t)) where G(q) = 1 + Sum_{n >= 0} (-1)^n*q^(n+1)*Product_{k = 1..n} 1 - q^k = 1 + q - q^2 + 2*q^3 - 2*q^4 + q^5 + q^7 - 2*q^8 + ... is a g.f. for A003406. See Zagier, Example 1. (End)
From Peter Bala, Dec 18 2021: (Start)
Conjectures:
1) Taking the sequence modulo an integer k gives an eventually periodic sequence with period dividing phi(k). For example, the sequence taken modulo 16 begins [1, 1, 5, 7, 1, 15, 1, 15, 1, 15, 1, 15, ...] with an apparent pre-period of length 4 and a period of length 2.
2) Let i >= 0 and define a_i(n) = a(n+i). Then for each i the Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
If true, then for each i the expansion of exp( Sum_{n >= 1} a_i(n)*x^n/n ) has integer coefficients. For example, the expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 21*x^3 + 291*x^4 + 6861*x^5 + 246171*x^6 + 12458901*x^7 + 843915891*x^8 + 73640674461*x^9 + 8041227405771*x^10 + ... appears to have integer coefficients. (End)

Examples

			G.f. A(x) = 1 + x + 5*x^2 + 55*x^3 + 1073*x^4 + 32671*x^5 + 1431665*x^6 + ...
		

Crossrefs

Programs

  • Mathematica
    max = 14; se = Series[1 + Sum[ Product[1 - E^(-(2*k - 1)*t), {k, 1, n}], {n, 1, max}], {t, 0, max}]; CoefficientList[se, t]*Range[0, max]! (* Jean-François Alcover, Mar 06 2013 *)
  • PARI
    {a(n)=n!*polcoeff(sum(m=0, n, prod(k=1, m, 1-exp(-(2*k-1)*x+x*O(x^n)))), n)} \\ Paul D. Hanna, Aug 01 2012
    
  • PARI
    {a(n)=n!*polcoeff(sum(m=0, n, prod(k=1, m, exp(k*x+x*O(x^n))-1)), n)} \\ Paul D. Hanna, Aug 01 2012

Formula

Basic hypergeometric generating function: 1 + Sum_{n >= 0} Product_{k = 1..n} (1 - exp(2*k-1)*t) = 1 + t + 5*t^2/2! + 55*t^3/3! + ....
a(n) ~ 6*sqrt(2) * 12^n * (n!)^2 / Pi^(2*n+2). - Vaclav Kotesovec, May 04 2014
Conjectural g.f.: G(exp(t)) as a formal power series in t, where G(q) := Sum_{n >= 0} q^(2*n+1) * Product_{k = 1..2*n} (1 - q^k). - Peter Bala, May 16 2017
E.g.f.: Sum_{n>=0} exp(n*(n+1)/2*x) / Product_{k=0..n} (1 + exp(k*x)). - Paul D. Hanna, Oct 14 2020

A179525 G.f.: A(x) = Sum_{n>=0} Product_{k=1..n} ((1+x)^k - 1).

Original entry on oeis.org

1, 1, 2, 7, 33, 197, 1419, 11966, 115575, 1257718, 15223822, 202860828, 2950665011, 46516215168, 790009447590, 14379745626739, 279256447482090, 5763290215111558, 125960271446527241, 2906289188751628643, 70594767279197608011, 1800695322331687800336, 48122711251655255426539, 1344617808976210991187090, 39206731897407002624384182, 1190905492485213830900901986
Offset: 0

Views

Author

Paul D. Hanna, Jul 17 2010

Keywords

Comments

From Vít Jelínek, Feb 12 2012: (Start)
a(n) has the following combinatorial interpretations:
(1) the number of upper-triangular matrices over {0,1} having at least one '1'-entry in each row and having n '1'-entries in total. E.g., for n=2, this corresponds to these two matrices (with zeros represented as dots):
1. .1
.1 .1
(2) the number of upper-triangular matrices over {0,1} that are symmetric with respect to the northeast diagonal, have at least one '1'-entry in each row and column, have no '1'-entry on the northeast diagonal, and have 2n '1'-entries in total. For n=2, those are the two matrices
11. 1...
..1 .1..
..1 ..1.
...1
(3) the number of upper-triangular matrices over {0,1} that are symmetric with respect to the northeast diagonal, have at least one '1'-entry in each row and column, have at least one '1'-entry on the northeast diagonal, and have n '1'-entries on or above the northeast diagonal. For n=2, this corresponds to
11 1..
.1 .1.
..1
(End)
This is an example of Peter Bala's identity (cf. 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) at q = 1 + x. See cross-references for other examples.

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 33*x^4 + 197*x^5 + 1419*x^6 +...
A(x) = 1 + ((1+x)-1) + ((1+x)-1)*((1+x)^2-1) + ((1+x)-1)*((1+x)^2-1)*((1+x)^3-1) +...
Let q = 1+x, then g.f. also equals:
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 +...
		

Crossrefs

Cf. A207434 (log).

Programs

  • Mathematica
    a[ n_] := SeriesCoefficient[ Sum[ Product[ (1 + x)^j - 1, {j, k}], {k, 0, n}], {x, 0, n}]; (* Michael Somos, Jun 27 2017 *)
  • PARI
    {a(n) = polcoeff(sum(i=0, n, prod(j=1, i, (1+x)^j-1, 1+x*O(x^n))), n)};
    for(n=0, 30, print1(a(n), ", "))
    
  • PARI
    /* G.f. as a continued fraction: */
    {a(n) = local(CF=1+x*O(x)); for(k=0, n, CF=1/((1+x)^(n-k+1)-((1+x)^(n-k+2)-1)*CF)); polcoeff(1/(1-x*CF), n, x)}
    for(n=0, 30, print1(a(n), ", "))
    
  • PARI
    {a(n) = local(A=1+x, q=(1+x +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, 30, print1(a(n), ", "))
    
  • PARI
    /* Sum_{n>=0} n!*Product_{k=0..n-1} [Integral (1+x)^k dx] */
    {a(n) = my(A=1); A = sum(m=0,n, m! * prod(k=0,m-1, intformal((1+x)^k) +x*O(x^n)) );polcoeff(A,n)}
    for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Aug 16 2016

Formula

G.f.: 1/(1 - ((1+x)-1)/((1+x) - ((1+x)^2-1)/((1+x)^2 - ((1+x)^3-1)/((1+x)^3 - ((1+x)^4-1)/((1+x)^4 - ((1+x)^5-1)/((1+x)^5 -...)))))), (continued fraction) [Paul D. Hanna, Jan 29 2012]
G.f.: Sum_{n>=0} q^(-n^2) * Product_{k=1..n} (q^(2*k-1) - 1) where q = 1+x. [Based on Peter Bala's identity in comments]
Conjecturally, a(n) is asymptotically c*n!*(12/Pi^2)^n, where c=6*sqrt(2)*exp(-Pi^2/24)/Pi^2. - Vít Jelínek, Feb 12 2012 [This is correct: see Hwang and Jin, Table 3, p. 26. - Peter Bala, Jan 31 2021]
G.f.: Q(0), where Q(k)= 1 - (1-(1+x)^(2*k+1))/(1 - (1-(1+x)^(2*k+2))/(1 - (1+x)^(2*k+2) - 1/Q(k+1))); (continued fraction). Conjecture. - Sergei N. Gladkovskii, May 13 2013
From Peter Bala, May 16 2017: (Start)
G.f.: A(x) = 1/2*( 1 + Sum_{n >= 0} (1 + x)^(n+1)*Product_{k = 1..n} ((1 + x)^k - 1) ).
Conjectural g.f.: Sum_{n >= 0} 1/(1 + x)^(n+1)*Product_{k = 1..n} (1 - 1/(1 + x)^(2*k)).
Conjectural g.f.: Sum_{n >= 0} (1 + x)^(2*n+1)*Product_{k = 1..2*n} (1 - (1 + x)^k). Cf. A158690, which has e.g.f. A(exp(x) - 1). (End)

A067619 Total number of parts in all self-conjugate partitions of n. Also, sum of largest parts of all self-conjugate partitions of n.

Original entry on oeis.org

0, 1, 0, 2, 2, 3, 3, 4, 7, 8, 9, 10, 15, 16, 18, 23, 30, 32, 35, 42, 51, 59, 63, 73, 89, 100, 106, 125, 145, 160, 174, 198, 229, 255, 274, 310, 355, 388, 420, 472, 534, 582, 631, 701, 784, 859, 928, 1021, 1144, 1243, 1338, 1475, 1630, 1767, 1909, 2089, 2299
Offset: 0

Views

Author

Naohiro Nomoto, Feb 01 2002

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Sum[n*q^(2n-1)*Product[1+q^k, {k, 1, 2n-3, 2}], {n, 1, 30}], {q, 0, 60}], q]

Formula

G.f.: A(q) = Sum_{n >= 1} n*q^(2*n-1)*(1+q)*(1+q^3)*...*(1+q^(2*n-3)).
From Peter Bala, Aug 20 2017: (Start)
Let F(q) = Product_{i >= 1} (1 + q^(2*i-1)). Then A(q) = Sum_{n >= 0} ( F(q) - Product_{i = 1..n} (1 + q^(2*i-1)) ).
It follows that the above sum A(q) satisfies -A(q-1) = 1 + q + 3*q^2 + 12*q^3 + 61*q^4 + ..., the g.f. for A158691, row-Fishburn matrices of size n. (End)

Extensions

Edited by Dean Hickerson, Feb 11 2002

A207386 G.f.: Sum_{n>=0} Product_{k=1..n} (q^k - 1) where q = (1+x)/(1+x^3).

Original entry on oeis.org

1, 1, 2, 6, 28, 172, 1269, 10879, 106343, 1167970, 14241792, 190919195, 2790920003, 44184957237, 753152722642, 13752229833566, 267809474619299, 5540559819166056, 121355678158129804, 2805498395990301867, 68265999939081386947, 1744058001878302097109
Offset: 0

Views

Author

Paul D. Hanna, Feb 17 2012

Keywords

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^3). See cross-references for other examples.
At present Bala's identity is conjectural and needs formal proof.

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 28*x^4 + 172*x^5 + 1269*x^6 +...
Let q = (1+x)/(1+x^3) = 1/(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) +...
which also is proposed to equal:
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 +...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+x,q=(1+x)/(1+x^3 +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)/(1+x^3 +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=0,21,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)/(1+x^3). [Based on Peter Bala's conjecture in A158690]
a(n) ~ n! * 2^(2*n+3/2) * 3^(n+1) / (exp(Pi^2/24) * Pi^(2*n+2)). - Vaclav Kotesovec, Aug 22 2017

A207569 G.f.: Sum_{n>=0} Product_{k=1..n} ((1+x)^(2*k-1) - 1).

Original entry on oeis.org

1, 1, 3, 18, 151, 1640, 21825, 343763, 6253234, 128993019, 2975165831, 75866604098, 2119310099700, 64361149952242, 2111222815441491, 74391641880144734, 2802300974537717340, 112379709083552152423, 4780136025081921948194, 214954914688567198802759
Offset: 0

Views

Author

Paul D. Hanna, Feb 18 2012

Keywords

Comments

Compare g.f. to Sum_{n>=0} Product_{k=1..n} ((1+x)^k - 1), which is the g.f. of A179525.
Compare g.f. to Sum_{n>=0} Product_{k=1..n} (1 - (1 - x)^(2*k-1)), which is the g.f. of A158691. - Peter Bala, Dec 04 2020

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 18*x^3 + 151*x^4 + 1640*x^5 + 21825*x^6 + ...
such that, by definition,
A(x) = 1 + ((1+x)-1) + ((1+x)-1)*((1+x)^3-1) + ((1+x)-1)*((1+x)^3-1)*((1+x)^5-1) + ((1+x)-1)*((1+x)^3-1)*((1+x)^5-1)*((1+x)^7-1) + ...
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Sum[Product[(1+x)^(2*k-1)-1, {k, 1, n}], {n, 0, 20}], {x, 0, 20}], x] (* Vaclav Kotesovec, May 06 2014 *)
  • PARI
    {a(n)=polcoeff(sum(m=0,n,prod(k=1,m,(1+x)^(2*k-1)-1) +x*O(x^n)),n)}
    for(n=0,25,print1(a(n),", "))

Formula

a(n) ~ sqrt(12) * 24^n * n^n / (exp(n+Pi^2/48) * Pi^(2*n+1)). - Vaclav Kotesovec, May 06 2014
G.f.: 1/2*( 1 + Sum_{n>=0} (1 + x)^(2*n+1) * Product_{k = 1..n} ((1 + x)^(2*k-1) - 1) ). Cf. A053250 and A215066. - Peter Bala, May 15 2017
Conjectural g.f.: Sum_{n>=0} (-1)^n*Product_{k = 1..n} 1 + ( -1/(1 + x) )^k. - Peter Bala, Dec 04 2020
From Peter Bala, Jan 29 2021: (Start)
Conjectural g.f.s: Sum_{n >= 0} (-1)^n*(1 + x)^(n+1)*Product_{k = 1..n} (1 + (-1)^k*(1 + x)^k)^2. Also
(1/2)*( 1 + Sum_{n >= 0} 1/(1 + x)^(n+1)*Product_{k = 1..n} (1 + (-1)^k/(1 + x)^k) ). (End)

A207397 G.f.: Sum_{n>=0} Product_{k=1..n} (q^k - 1) where q = (1+x)/(1+x^2).

Original entry on oeis.org

1, 1, 1, 2, 11, 74, 557, 4799, 47004, 516717, 6302993, 84502346, 1235198136, 19552296646, 333212892221, 6083009119262, 118433569748072, 2449663066933397, 53643715882853914, 1239875630317731463, 30163779836127304106, 770476745704778418686
Offset: 0

Views

Author

Paul D. Hanna, Feb 17 2012

Keywords

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.

Examples

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 11*x^4 + 74*x^5 + 557*x^6 + 4799*x^7 +...
Let q = (1+x)/(1+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) +...
which also is proposed to equal:
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 +...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+x,q=(1+x)/(1+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)/(1+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)/(1+x^2). [Based on Peter Bala's conjecture in A158690]
a(n) ~ c * 12^n * n! / Pi^(2*n), where c = 6*sqrt(2) / (Pi^2 * exp(Pi^2/8)) = 0.250367043877216848533826021231826... . - Vaclav Kotesovec, May 06 2014, updated Aug 22 2017

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

Original entry on oeis.org

1, 1, 3, 11, 56, 350, 2609, 22582, 222625, 2462969, 30219676, 407276420, 5981197376, 95073427910, 1626294895274, 29788176027819, 581704672430937, 12064521684969823, 264843222932272690, 6135057298705027024, 149559103545555671423, 3827360866024134614644
Offset: 0

Views

Author

Paul D. Hanna, Feb 17 2012

Keywords

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)
		

Crossrefs

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

A289313 The number of upper-triangular matrices with integer entries whose absolute sum is equal to n and such that each row contains a nonzero entry.

Original entry on oeis.org

1, 2, 10, 74, 722, 8786, 128218, 2182554, 42456226, 929093538, 22590839466, 604225121258, 17630145814898, 557285515817970, 18970857530674554, 691929648113663802, 26919562120779248962, 1112769248605003393858, 48704349211392743606602
Offset: 0

Views

Author

Peter Bala, Jul 02 2017

Keywords

Comments

A row-Fishburn matrix of size n is defined to be an upper-triangular matrix with nonnegative integer entries which sum to n and such that each row contains a nonzero entry. See A158691.
Here we consider generalized row-Fishburn matrices where we allow the row_Fishburn matrices to have positive and negative nonzero entries. We define the size of a generalized row-Fishburn matrix to be the absolute sum of the matrix entries. This sequence gives the number of generalized row-Fishburn matrices of size n.
Alternatively, this sequence gives the number of 2-colored row-Fishburn matrices of size n, that is, ordinary row-Fishburn matrices of size n where each nonzero entry in the matrix can have one of two different colors.
More generally, if F(x) = Sum_{n >= 0} ( Product_{i = 1..n} (1 + x)^i - 1 ) is the o.g.f. for primitive row-Fishburn matrices A179525 (i.e., row-Fishburn matrices with entries restricted to the set {0,1}) and C(x) := c_1*x + c_2*x^2 + ..., where c_i is a sequence of nonnegative integers, then the composition F(C(x)) is the o.g.f. for colored row-Fishburn matrices where entry i in the matrix can have one of c_i different colors: c_i = 0 for some i means i does not appear as an entry in the Fishburn matrix. This result is an application of Lemma 2.2.22 of Goulden and Jackson.

Examples

			a(2) = 10: The ten generalized row-Fishburn matrices of size 2 are
  (+-2),
  /+-1  0\ and  /0 +-1\
  |      |      |     |
  \0  +-1/      \0 +-1/.
		

References

  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, p. 42.

Crossrefs

Programs

  • Maple
    G:= add(mul( ((1 + x)/(1 - x))^i - 1, i=1..n),n=0..20):
    S:= series(G,x,21):
    seq(coeff(S,x,j),j=0..20);
    # Peter Bala, Jul 24 2017

Formula

O.g.f.: Sum_{n >= 0} ( Product_{i = 1..n} ((1 + x)/(1 - x))^i - 1 ).
The o.g.f. has several alternative forms:
Sum_{n >= 0} ( Product_{i = 1..n} ( 1 - ((1 - x)/(1 + x))^(2*i-1) ) );
Sum_{n >= 0} ((1 - x)/(1 + x))^(n+1) * ( Product_{i = 1..n} 1 - ((1 - x)/(1 + x))^(2*i) );
1/2*( 1 + Sum_{n >= 0} ((1 + x)/(1 - x))^((n+1)*(n+2)/2) * Product_{i = 1..n} ( 1 - ((1 - x)/(1 + x))^i ) ).
Conjectural g.f.: Sum_{n >= 0} ((1 + x)/(1 - x))^((n+1)*(2*n+1)) * Product_{i = 1..2*n} ( ((1 - x)/(1 + x))^i - 1 ).
a(n) ~ 2^(3*n+2) * 3^(n+1) * n^(n + 1/2) / (exp(n) * Pi^(2*n + 3/2)). - Vaclav Kotesovec, Aug 31 2023

A289316 The number of upper-triangular matrices whose nonzero entries are positive odd numbers summing to n and each row contains a nonzero entry.

Original entry on oeis.org

1, 1, 2, 8, 37, 219, 1557, 12994, 124427, 1344506, 16178891, 214522339, 3107144562, 48805300668, 826268787588, 14998055299920, 290550119360174, 5983278021430064, 130512410617529321, 3006012061455129053, 72900477505718600661
Offset: 0

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Author

Peter Bala, Jul 24 2017

Keywords

Comments

A row-Fishburn matrix of size n is defined to be an upper-triangular matrix with nonnegative integer entries which sum to n and each row contains a nonzero entry. See A158691. Here we are considering row-Fishburn matrices where the nonzero entries are all odd.
The g.f. F(x) for primitive row_Fishburn matrices (i.e., row_Fishburn matrices with entries restricted to the set {0,1}), is F(x) = Sum_{n>=0} Product_{k=1..n} ( (1 + x)^k - 1 ). See A179525. Let C(x) = x/(1 - x^2) = x + x^3 + x^5 + x^7 + .... Then appplying Lemma 2.2.22 of Goulden and Jackson gives the g.f. for the present sequence as the composition F(C(x)).

Examples

			a(3) = 8: The eight row-Fishburn matrices of size 3 with odd nonzero entries are
(3) /1 1\
    \0 1/
/1 0 0\  /0 1 0\  /0 0 1\
|0 1 0|  |0 1 0|  |0 1 0|
\0 0 1/  \0 0 1/  \0 0 1/
/1 0 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/
		

References

  • I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, p. 42.

Crossrefs

Programs

  • Maple
    C:= x -> x/(1 - x^2):
    G:= add(mul( (1 + C(x))^k - 1, k=1..n),n=0..20):
    S:= series(G,x,21):
    seq(coeff(S,x,j),j=0..20);

Formula

G.f.: A(x) = Sum_{n >= 0} Product_{k = 1..n} ( (1 + x/(1 - x^2))^k - 1 ).
a(n) ~ 12^(n+1) * n^(n + 1/2) / (exp(n + Pi^2/24) * Pi^(2*n + 3/2)). - Vaclav Kotesovec, Aug 31 2023
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