A207386 -id:A207386 - cp's OEIS frontend

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

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

Paul D. Hanna, Jul 17 2010

nonn

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.

			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 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..360
Kathrin Bringmann, Yingkun Li, Robert C. Rhoades, Asymptotics for the number of row-Fishburn matrices, European Journal of Combinatorics, vol.41, pp.183-196, (October-2014); preprint.
Stuart A. Hannah, Sieved Enumeration of Interval Orders and Other Fishburn Structures, arXiv:1502.05340 [math.CO], (18-February-2015).
Hsien-Kuei Hwang, Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.
Vít Jelínek, Counting general and self-dual interval orders, Journal of Combinatorial Theory, Series A, Volume 119, Issue 3, April 2012, pp. 599-614.
Sherry H. F. Yan and Yuexiao Xu, Self-dual interval orders and row-Fishburn matrices, arXiv:1111.4723 [math.CO], 2011.

Cf. variants: A022493, A122400, A158691, A158690, A207386, A207397, A207433.

Cf. A207434 (log).

a[ n_] := SeriesCoefficient[ Sum[ Product[ (1 + x)^j - 1, {j, k}], {k, 0, n}], {x, 0, n}]; (* Michael Somos, Jun 27 2017 *)

{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), ", "))

/* 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), ", "))

{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), ", "))

/* 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

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)

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

Paul D. Hanna, Feb 17 2012

nonn

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.

			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 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..240
Hsien-Kuei Hwang, Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.

Cf. A158690, A158691, A179525, A207386, A207433.

{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)}

{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),", "))

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

Paul D. Hanna, Feb 17 2012

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

			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)

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.

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

{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)}

{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),", "))

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

A207384 A206815(n+1)-A206815(n).

Original entry on oeis.org

1, 3, 2, 2, 2, 3, 1, 2, 2, 3, 1, 3, 2, 2, 1, 3, 2, 3, 1, 2, 2, 3, 2, 1, 2, 2, 2, 2, 2, 3, 2, 2, 1, 2, 2, 3, 2, 1, 2, 3, 2, 2, 2, 2, 2, 3, 2, 1, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 1, 2, 3, 2, 2, 1, 3, 2, 3, 2, 1, 2, 2, 2, 3, 2, 1, 2, 3, 2, 2, 1, 2, 2, 3, 2, 2, 1, 2, 2, 2, 2, 3, 1, 2, 2

Offset: 1

Clark Kimberling, Feb 17 2012

nonn

The sequences A206815, A206818, A207384, A207835 illustrate the closeness of {j+pi(j)} to {k+(k+1)/log(k+1)}, as suggested by the prime number theorem and the conjecture that all the terms of A207384 and A207835 are in the set {1,2,3}.

			The joint ranking is represented by
1 < 3 < 3.8 < 4.7 < 5 < 5.8 < 6 <7.1 < 8 < 8.3 < 9 < ...
Positions of numbers j+pi(j): 1,2,5,7,9,...
Positions of numbers k+(k+1)/log(k+1): 3,4,6,8,10,..

Cf. A000720, A206815, A206818.

f[1, n_] := n + PrimePi[n];
f[2, n_] := n + N[(n + 1)/Log[n + 1]]; z = 500;
t[k_] := Table[f[k, n], {n, 1, z}];
t = Sort[Union[t[1], t[2]]];
p[k_, n_] := Position[t, f[k, n]];
Flatten[Table[p[1, n], {n, 1, z}]]    (* A206815 *)
Flatten[Table[p[2, n], {n, 1, z}]]    (* A206818 *)
d1[n_] := p[1, n + 1] - p[1, n]
Flatten[Table[d1[n], {n, 1, z - 1}]]  (* A207385 *)
d2[n_] := p[2, n + 1] - p[2, n]
Flatten[Table[d2[n], {n, 1, z - 1}]]  (* A207386 *)

A207385 A206818(n+1)-A206818(n).

Original entry on oeis.org

1, 2, 2, 2, 2, 1, 3, 2, 2, 1, 3, 1, 2, 2, 3, 1, 2, 2, 1, 3, 2, 2, 1, 2, 3, 2, 2, 2, 2, 2, 1, 2, 2, 3, 2, 2, 1, 2, 3, 2, 1, 2, 2, 2, 2, 2, 2, 1, 2, 3, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 3, 2, 1, 2, 2, 3, 1, 2, 2, 1, 2, 3, 2, 2, 2, 1, 2, 3, 2, 1, 2, 2, 3, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1

Offset: 1

Clark Kimberling, Feb 17 2012

nonn

			The joint ranking is represented by
1 < 3 < 3.8 < 4.7 < 5 < 5.8 < 6 <7.1 < 8 < 8.3 < 9 < ...
Positions of numbers j+pi(j): 1,2,5,7,9,...
Positions of numbers k+(k+1)/log(k+1): 3,4,6,8,10,..

Cf. A000720, A206815, A206818.

f[1, n_] := n + PrimePi[n];
f[2, n_] := n + N[(n + 1)/Log[n + 1]]; z = 500;
t[k_] := Table[f[k, n], {n, 1, z}];
t = Sort[Union[t[1], t[2]]];
p[k_, n_] := Position[t, f[k, n]];
Flatten[Table[p[1, n], {n, 1, z}]]    (* A206815 *)
Flatten[Table[p[2, n], {n, 1, z}]]    (* A206818 *)
d1[n_] := p[1, n + 1] - p[1, n]
Flatten[Table[d1[n], {n, 1, z - 1}]]  (* A207385 *)
d2[n_] := p[2, n + 1] - p[2, n]
Flatten[Table[d2[n], {n, 1, z - 1}]]  (* A207386 *)

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

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

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

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A207384 A206815(n+1)-A206815(n).

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A207385 A206818(n+1)-A206818(n).

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