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.

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A227543 Triangle defined by g.f. A(x,q) such that: A(x,q) = 1 + x*A(q*x,q)*A(x,q), as read by terms k=0..n*(n-1)/2 in rows n>=0.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 3, 2, 1, 1, 1, 4, 6, 7, 7, 5, 5, 3, 2, 1, 1, 1, 5, 10, 14, 17, 16, 16, 14, 11, 9, 7, 5, 3, 2, 1, 1, 1, 6, 15, 25, 35, 40, 43, 44, 40, 37, 32, 28, 22, 18, 13, 11, 7, 5, 3, 2, 1, 1, 1, 7, 21, 41, 65, 86, 102, 115, 118, 118, 113, 106, 96, 85, 73, 63, 53, 42, 34, 26, 20, 15, 11, 7, 5, 3, 2, 1, 1
Offset: 0

Views

Author

Paul D. Hanna, Jul 15 2013

Keywords

Comments

See related triangle A138158.
Row sums are the Catalan numbers (A000108), set q=1 in the g.f. to see this.
Antidiagonal sums equal A005169, the number of fountains of n coins.
The maximum in each row of the triangle is A274291. - Torsten Muetze, Nov 28 2018
The area between a Dyck path and the x-axis may be decomposed into unit area triangles of two types - up-triangles with vertices at the integer lattice points (x, y), (x+1, y+1) and (x+2, y) and down-triangles with vertices at the integer lattice points (x, y), (x-1, y+1) and (x+1, y+1). The table entry T(n,k) equals the number of Dyck paths of semilength n containing k down triangles. See the illustration in the Links section. Cf. A239927. - Peter Bala, Jul 11 2019
The row polynomials of this table are a q-analog of the Catalan numbers due to Carlitz and Riordan. For MacMahon's q-analog of the Catalan numbers see A129175. - Peter Bala, Feb 28 2023

Examples

			G.f.: A(x,q) = 1 + x*(1) + x^2*(1 + q) + x^3*(1 + 2*q + q^2 + q^3)
 + x^4*(1 + 3*q + 3*q^2 + 3*q^3 + 2*q^4 + q^5 + q^6)
 + x^5*(1 + 4*q + 6*q^2 + 7*q^3 + 7*q^4 + 5*q^5 + 5*q^6 + 3*q^7 + 2*q^8 + q^9 + q^10)
 + x^6*(1 + 5*q + 10*q^2 + 14*q^3 + 17*q^4 + 16*q^5 + 16*q^6 + 14*q^7 + 11*q^8 + 9*q^9 + 7*q^10 + 5*q^11 + 3*q^12 + 2*q^13 + q^14 + q^15) +...
where g.f.A(x,q) = Sum_{k=0..n*(n-1)/2, n>=0} T(n,k)*x^n*q^k
satisfies A(x,q) = 1 + x*A(q*x,q)*A(x,q).
This triangle of coefficients T(n,k) in A(x,q) begins:
 1;
 1;
 1, 1;
 1, 2, 1, 1;
 1, 3, 3, 3, 2, 1, 1;
 1, 4, 6, 7, 7, 5, 5, 3, 2, 1, 1;
 1, 5, 10, 14, 17, 16, 16, 14, 11, 9, 7, 5, 3, 2, 1, 1;
 1, 6, 15, 25, 35, 40, 43, 44, 40, 37, 32, 28, 22, 18, 13, 11, 7, 5, 3, 2, 1, 1;
 1, 7, 21, 41, 65, 86, 102, 115, 118, 118, 113, 106, 96, 85, 73, 63, 53, 42, 34, 26, 20, 15, 11, 7, 5, 3, 2, 1, 1;
 1, 8, 28, 63, 112, 167, 219, 268, 303, 326, 338, 338, 331, 314, 293, 268, 245, 215, 190, 162, 139, 116, 97, 77, 63, 48, 38, 28, 22, 15, 11, 7, 5, 3, 2, 1, 1; ...
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := Module[{P, Q},
    P = Sum[q^(m^2) (-x)^m/Product[1-q^j, {j, 1, m}] + x O[x]^n, {m, 0, n}];
    Q = Sum[q^(m(m-1)) (-x)^m/Product[1-q^j, {j, 1, m}] + x O[x]^n, {m, 0, n}];
    SeriesCoefficient[P/Q, {x, 0, n}, {q, 0, k}]
    ];
    Table[T[n, k], {n, 0, 10}, {k, 0, n(n-1)/2}] // Flatten (* Jean-François Alcover, Jul 27 2018, from PARI *)
  • PARI
    /* From g.f. A(x,q) = 1 + x*A(q*x,q)*A(x,q): */
    {T(n, k)=local(A=1); for(i=1, n, A=1+x*subst(A, x, q*x)*A +x*O(x^n)); polcoeff(polcoeff(A, n, x), k, q)}
    for(n=0, 10, for(k=0, n*(n-1)/2, print1(T(n, k), ", ")); print(""))
    
  • PARI
    /* By Ramanujan's continued fraction identity: */
    {T(n,k)=local(P=1,Q=1);
    P=sum(m=0,n,q^(m^2)*(-x)^m/prod(k=1,m,1-q^k)+x*O(x^n));
    Q=sum(m=0,n,q^(m*(m-1))*(-x)^m/prod(k=1,m,1-q^k)+x*O(x^n));
    polcoeff(polcoeff(P/Q,n,x),k,q)}
    for(n=0, 10, for(k=0, n*(n-1)/2, print1(T(n, k), ", ")); print(""))
    
  • PARI
    P(x, n) =
    {
        if ( n<=1, return(1) );
        return( sum( i=0, n-1, P(x, i) * P(x, n-1 -i) * x^((i+1)*(n-1 -i)) ) );
    }
    for (n=0, 10, print( Vec( P(x, n) ) ) ); \\ Joerg Arndt, Jan 23 2024
    
  • PARI
    \\ faster with memoization:
    N=11;
    VP=vector(N+1);  VP[1] =VP[2] = 1;  \\ one-based; memoization
    P(n) = VP[n+1];
    for (n=2, N, VP[n+1] = sum( i=0, n-1, P(i) * P(n-1 -i) * x^((i+1)*(n-1-i)) ) );
    for (n=0, N, print( Vec( P(n) ) ) ); \\ Joerg Arndt, Jan 23 2024

Formula

G.f.: A(x,q) = 1/(1 - x/(1 - q*x/(1 - q^2*x/(1 - q^3*x/(1 - q^4*x/(1 -...)))))), a continued fraction.
G.f. satisfies: A(x,q) = P(x,q)/Q(x,q), where
P(x,q) = Sum_{n>=0} q^(n^2) * (-x)^n / Product_{k=1..n} (1-q^k),
Q(x,q) = Sum_{n>=0} q^(n*(n-1)) * (-x)^n / Product_{k=1..n} (1-q^k),
due to Ramanujan's continued fraction identity.
...
Sum_{k=0..n*(n-1)/2} T(n,k)*k = 2^(2*n-1) - C(2*n+1,n) + C(2*n-1,n-1) = A006419(n-1) for n>=1.
Logarithmic derivative of the g.f. A(x,q), wrt x, yields triangle A227532.
From Peter Bala, Jul 11 2019: (Start)
(n+1)th row polynomial R(n+1,q) = Sum_{k = 0..n} q^k*R(k,x)*R(n-k,q), with R(0,q) = 1.
1/A(q*x,q) is the generating function for the triangle A047998. (End)
Conjecture: b(n) = P(n, n) where b(n) is an integer sequence with g.f. B(x) = 1/(1 - f(0)*x/(1 - f(1)*x/(1 - f(2)*x/(1 - f(3)*x/(1 - f(4)*x/(1 -...)))))), P(n, k) = P(n-1, k) + f(n-k)*P(n, k-1) for 0 < k <= n with P(n, k) = 0 for k > n, P(n, 0) = 1 for n >= 0 and where f(n) is an arbitrary function. In fact for this sequence we have f(n) = q^n. - Mikhail Kurkov, Sep 26 2024

A145855 Number of n-element subsets of {1,2,...,2n-1} whose elements sum to a multiple of n.

Original entry on oeis.org

1, 1, 4, 9, 26, 76, 246, 809, 2704, 9226, 32066, 112716, 400024, 1432614, 5170604, 18784169, 68635478, 252085792, 930138522, 3446167834, 12815663844, 47820414962, 178987624514, 671825133644, 2528212128776, 9536894664376
Offset: 1

Views

Author

T. D. Noe, Oct 21 2008, Oct 22 2008, Oct 24 2008

Keywords

Comments

It is easy to see that {1,2,...,2n-1} can be replaced by any 2n-1 consecutive numbers and the results will be the same. Erdos, Ginzburg and Ziv proved that every set of 2n-1 numbers -- not necessarily consecutive -- contains a subset of n elements whose sum is a multiple of n.

Examples

			a(3)=4 because, of the 10 3-element subsets of 1..7, only {1,2,3}, {1,3,5}, {2,3,4} and {3,4,5} have sums that are multiples of 3.
L.g.f.: L(x) = x + x^2/2 + 4*x^3/3 + 9*x^4/4 + 26*x^5/5 + 76*x^6/6 + 246*x^7/7 +...
where exponentiation yields the g.f. of A000571:
exp(L(x)) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 9*x^5 + 22*x^6 + 59*x^7 + 167*x^8 +...
		

Crossrefs

Column k=2 of A309148.

Programs

  • Mathematica
    Table[Length[Select[Plus@@@Subsets[Range[2n-1],{n}], Mod[ #,n]==0&]], {n,10}]
    Table[d=Divisors[n]; Sum[(-1)^(n+d[[i]]) EulerPhi[n/d[[i]]] Binomial[2d[[i]], d[[i]]]/2/n, {i,Length[d]}], {n,30}] (* T. D. Noe, Oct 24 2008 *)
  • PARI
    {a(n)=sumdiv(n, d, (-1)^(n+d)*eulerphi(n/d)*binomial(2*d, d)/(2*n))}
    
  • PARI
    {A227532(n, k)=local(G=1); for(i=1, n, G=1+x*subst(G, x, q*x)*G +x*O(x^n)); n*polcoeff(polcoeff(log(G), n, x), k, q)}
    {a(n)=sum(k=0,n\2, A227532(n, n*k))} \\ Paul D. Hanna, Jul 17 2013

Formula

a(n) = (1/(2*n))*Sum_{d|n} (-1)^(n+d)*phi(n/d)*binomial(2*d,d). Conjectured by Vladeta Jovovic, Oct 22 2008; proved by Max Alekseyev, Oct 23 2008 (see link).
a(2n+1) = A003239(2n+1) and a(2n) = A003239(2n) - A003239(d), where d is the largest odd divisor of n. - T. D. Noe, Oct 24 2008
a(n) = Sum_{d|n} (-1)^(n+d)*d*A131868(d). - Vladeta Jovovic, Oct 28 2008
a(n) = Sum_{k=0..[n/2]} A227532(n,n*k), where A227532 is the logarithmic derivative, wrt x, of the g.f. G(x,q) = 1 + x*G(q*x,q)*G(x,q) of triangle A227543. - Paul D. Hanna, Jul 17 2013
Logarithmic derivative of A000571, the number of different scores that are possible in an n-team round-robin tournament. - Paul D. Hanna, Jul 17 2013
G.f.: -Sum_{m >= 1} (phi(m)/m) * log((1 + sqrt(1 + 4*(-y)^m))/2). - Petros Hadjicostas, Jul 15 2019
a(n) ~ 2^(2*n-1) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Mar 28 2023

Extensions

Extension T. D. Noe, Oct 24 2008
Showing 1-2 of 2 results.