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 1-5 of 5 results.

A219238 Coefficient table for the first differences of table A047971: Coefficients of the difference of Gauss polynomials [n+3,3]_q - [n+2,3]_q.

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 0, 1, 1, 2, 1, 1, 0, 0, 0, 1, 1, 2, 2, 2, 1, 1, 0, 0, 0, 0, 1, 1, 2, 2, 3, 2, 2, 1, 1, 0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 3, 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 3, 3, 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 4, 3, 3, 2, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 4, 4, 3, 3, 2, 2, 1, 1
Offset: 0

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Author

Wolfdieter Lang, Dec 06 2012

Keywords

Comments

The row lengths sequence is A016777 (3*n+1). The sum for row n is A000217(n+1) = binomial(n+2,2).
The coefficients of the Gauss polynomial [n+3,3]_q are given in A047971.
a(n,k) = [q^k]([n+3,3]_q - [n+2,3]_q). One can use the identity [n+3,3]_q - [n+2,3]_q = q^n*[n+2,2]_q (see the Andrews reference given in A047971, p. 35, (3.3.3)). Therefore, the present array is obtained from A008967 after a shift of row n by n units to the right, inserting zeros for the first n entries.
The o.g.f. of the row polynomials in q of degree 3*n is 1/((1-q)*(1-q^2)*(1-q^3)) (multiply the o.g.f. of A047971 by (1-z)). a(n,k) determines therefore the number of partitions of k with precisely n parts, each <= 3. Alternatively, a(n,k) determines the number of partitions of k with at most 3 parts, with each part <= n but not each part <= (n-1), i.e., part n, maybe more than once, is present besides possibly smaller ones.

Examples

			The table a(n,k) begins:
n\k 0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18...
0:  1
1:  0  1  1  1
2:  0  0  1  1  2  1  1
3:  0  0  0  1  1  2  2  2  1  1
4:  0  0  0  0  1  1  2  2  3  2  2  1  1
5:  0  0  0  0  0  1  1  2  2  3  3  3  2  2  1  1
6:  0  0  0  0  0  0  1  1  2  2  3  3  4  3  3  2  2  1  1
...
Row n=1 is 0,1,1,1 because [3,2]_q = 1 + q + q^2 and the coefficient of q^{-1} is 0, the one of q^0 is 1, the one of q^1 is 1 and the one of q^2 is 1. A shift of row n=1 of A008967 by one unit to the right.
a(n,k) = 0 if n > k because a partition of k never has more than k parts.
a(n,k) = 0 if k > 3*n because there is no partition of 3*n+m, with m >= 1, and exactly n parts, each <= 3.
a(2,4) = 2 because the partitions of 4 with 2 parts are 1,3 and 2,2, and the parts in both are <= 3.
a(2,4) = 2 because the partitions of 4 with number of parts <= 3, each <= 2, are 2,2 and 1,1,2, and part 2 is present in both of them. Note the conjugacy of partitions 1,3 and 1,1,2.
		

Crossrefs

Cf. A047971, A008967 (with shifted rows).

Formula

a(n,k) = [q^k]([n+3,3]_q - [n+2,3]_q), = [q^(k-n)] [n+2,2]_q , n >= 0, 0 <= k <= 3*n. For the Gauss polynomial (q-binomial) [n+m,m]_q = [m+n,n]_q see a comment on A219237 where also the Andrews reference and a link to Mathworld is found.

A008967 Coefficients of Gaussian polynomials q_binomial(n-2, 2). Also triangle of distribution of rank sums: Wilcoxon's statistic. Irregular triangle read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 3, 2, 2, 1, 1, 1, 1, 2, 2, 3, 3, 3, 2, 2, 1, 1, 1, 1, 2, 2, 3, 3, 4, 3, 3, 2, 2, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 3, 3, 2, 2, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 4, 4, 3, 3, 2, 2, 1, 1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 5, 4, 4, 3, 3, 2, 2, 1, 1
Offset: 4

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Author

Keywords

Comments

Rows are numbers of dominoes with k spots where each half-domino has zero to n spots (in standard domino set: n=6, there are 28 dominoes and row is 1,1,2,2,3,3,4,3,3,2,2,1,1). - Henry Bottomley, Aug 23 2000
These numbers appear in the solution of Cayley's counting problem on covariants as N(p,2,w) = [x^p,q^w] Phi(q,x) with the o.g.f. Phi(q,x) = 1/((1-x)(1-qx)(1-q^2x)) given by Peter Bala in the formula section. See the Hawkins reference, p. 264, were also references are given. - Wolfdieter Lang, Nov 30 2012
The entry a(p,w), p >= 0, w = 0,1,...,2*p, of this irregular triangle is the number of nonnegative solutions of m_0 + m_1 + m_2 = p and 1*m_1 + 2*m_2 = w. See the Hawkins reference p. 264, (4.8). N(p,2,w) there is a(p,w). See also the Cayley reference p. 110, 35. with m = 2, Theta = p and q = w. - Wolfdieter Lang, Dec 01 2012
From Gus Wiseman, Sep 20 2023: (Start)
Also the number of unordered pairs of distinct positive integers up to n with sum k. For example, row n = 9 counts the following pairs:
12 13 14 15 16 17 18 19 29 39 49 59 69 79 89
23 24 25 26 27 28 38 48 58 68 78
34 35 36 37 47 57 67
45 46 56
Allowing repeated parts (x,x) gives A004737.
For strict partitions instead of just pairs we have A053632.
(End)

Examples

			1;
1,1,1;
1,1,2,1,1;
1,1,2,2,2,1,1;
1,1,2,2,3,2,2,1,1;
1,1,2,2,3,3,3,2,2,1,1;
...
Partitions: row p=2 and column w=2 has entry 2 because the 2 solutions of the two equations mentioned in a comment above are: m_0 = 0, m_1 = 2, m_2 = 0 and m_0 = 1, m_1 = 0, m_2 = 1. - _Wolfdieter Lang_, Dec 01 2012
		

References

  • G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 242.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 236.
  • T. Hawkins, Emergence of the Theory of Lie Groups, Springer 2000, ch. 7.4, p. 260-5.

Crossrefs

A version with zeros is A219238.
This is the case of A365541 counting only length-2 subsets.

Programs

  • Maple
    qBinom := proc(n,m,q)
            mul((1-q^(n-i))/(1-q^(i+1)),i=0..m-1) ;
            factor(%) ;
            expand(%) ;
    end proc:
    A008967 := proc(n,k)
            coeftayl( qBinom(n,2,q),q=0,k ) ;
    end proc:
    seq(seq( A008967(n,k),k=0..2*n-4),n=2..10) ; # assumes offset 2. R. J. Mathar, Oct 13 2011
  • Mathematica
    rmax = 11; f[r_] := Product[(x^i - x^(r+1))/(1-x^i), {i, 1, r-2}]/  x^((r-1)*(r-2)/2); row[r_] := CoefficientList[ Series[ f[r], {x, 0, 2rmax}], x]; Flatten[ Table[ row[r], {r, 2, rmax}]] (* Jean-François Alcover, Oct 13 2011, after given formula *)
    T[n_, k_] := SeriesCoefficient[QBinomial[n - 2, 2, q], {q, 0, k}];
    Table[T[n, k], {n, 4, 13}, {k, 0, 2 n - 8}] // Flatten (* Jean-François Alcover, Aug 20 2019 *)
    Table[Length[Select[Subsets[Range[n],{2}],Total[#]==k&]],{n,2,15},{k,3,2n-1}] (* Gus Wiseman, Sep 20 2023 *)
  • SageMath
    print(flatten([q_binomial(n-2, 2).list() for n in (4..13)])) # Peter Luschny, Oct 23 2019

Formula

Let f(r) = Product( (x^i-x^(r+1))/(1-x^i), i = 1..r-2) / x^((r-1)*(r-2)/2); then expanding f(r) in powers of x and taking coefficients gives the successive rows of this triangle (with a different offset).
Expanding (q^n - 1)(q^(n+1) - 1)/((q - 1)(q^2 - 1)) in powers of q and taking coefficients gives the n-th row of the triangle. Ordinary generating function: 1/((1-x)(1-qx)(1-q^2x)) = 1 + x(1 + q + q^2) + x^2(1 + q + 2q^2 + q^3 + q^4) + .... - Peter Bala, Sep 23 2007
For n >= 2, let a(n,i) denote the i-th entry of the (n-1)-st row of this triangle; for every 0 <= i <= n-2, a(n,i) = a(n,2(n-2)-i) = ceiling((i+1)/2). - Christian Barrientos, Aug 08 2019

Extensions

More terms from Christian Barrientos, Aug 08 2019

A063746 Triangle read by rows giving number of partitions of k (k=0 .. n^2) with Ferrers plot fitting in an n X n box.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 3, 3, 3, 2, 1, 1, 1, 1, 2, 3, 5, 5, 7, 7, 8, 7, 7, 5, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 7, 9, 11, 14, 16, 18, 19, 20, 20, 19, 18, 16, 14, 11, 9, 7, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 7, 11, 13, 18, 22, 28, 32, 39, 42, 48, 51, 55, 55, 58, 55, 55, 51, 48, 42, 39, 32, 28
Offset: 0

Views

Author

Wouter Meeussen, Aug 14 2001

Keywords

Comments

Seems to approximate a Gaussian distribution, the sum of all 1+n^2 terms in a row equals the central binomial coefficients.
a(n,k) is the number of sequences of n 0's and n 1's having major index equal to k (the major index is the sum of the positions of the 1's that are immediately followed by 0's). Equivalently, a(n,k) is the number of Grand Dyck paths of length 2n for which the sum of the positions of the valleys is k. Example: a(3,7)=2 because the only sequences of three 0's and three 1's with major index 7 are 010110 and 110010. The corresponding Grand Dyck paths are obtained by replacing a 0 by a U=(1,1) step and a 1 by a D=(1,-1) step. - Emeric Deutsch, Oct 02 2007
Also, number of n-multisets in [0..n] whose elements sum up to n. - M. F. Hasler, Apr 12 2012
Let P be the poset [n] X [n] ordered by the product order. Let J(P) be the set of all order ideals of P, ordered by inclusion. Then J(P) is a finite sublattice of Young's lattice and T(n,k) is the number of elements in J(P) that have rank k. - Geoffrey Critzer, Mar 26 2020

Examples

			From _M. F. Hasler_, Apr 12 2012: (Start)
The table reads:
n=0: 1  _  (k=0)
n=1: 1 1  _  (k=0..1)
n=2: 1 1 2 1 1  _  (k=0..4)
n=3: 1 1 2 3 3 3 3  2  1  1  _  (k=0..9)
n=4: 1 1 2 3 5 5 7  7  8  7  7  5  5  3  2  1  1  _  (k=0..16)
n=5: 1 1 2 3 5 7 9 11 14 16 18 19 20 20 19 18 16 ...  _  (k=0..25)
etc. (End)
Cycle index of S(3) is (1/6)*(x(1)^3+3*x(1)*x(2)+2*x(3)), so g.f. for 3rd row is (1/6)*((1+x+x^2+x^3)^3+3*(1+x+x^2+x^3)*(1+x^2+x^4+x^6)+2*(1+x^3+x^6+x^9)) = x^9+x^8+2*x^7+3*x^6+3*x^5+3*x^4+3*x^3+2*x^2+x+1.
a(3,7)=2 because the only partitions of 7 with Ferrers plot fitting into a 3 X 3 box are [3,3,1] and [3,2,2].
		

References

  • G. E. Andrews and K. Eriksson, Integer partitions, Cambridge Univ. Press, 2004, pp. 67-69.
  • D. M. Bressoud, Proofs and Confirmations, Camb. Univ. Press, 1999; exercise 3.2.3.
  • A. V. Yurkin, New binomial and new view on light theory, (book), 2013, 78 pages, no publisher listed.

Crossrefs

Row lengths are given by A002522. - M. F. Hasler, Apr 14 2012
Antidiagonal sums are given by A260894.
Row sums give A000984.

Programs

  • Maple
    for n from 0 to 15 do QBR[n]:=sum(q^i,i=0..n-1) od: for n from 0 to 15 do QFAC[n]:=product(QBR[j],j=1..n) od: qbin:=(n,k)->QFAC[n]/QFAC[k]/QFAC[n-k]: for n from 0 to 7 do P[n]:=sort(expand(simplify(qbin(2*n,n)))) od: for n from 0 to 7 do seq(coeff(P[n],q,j),j=0..n^2) od; # yields sequence in triangular form - Emeric Deutsch, Apr 23 2007
    # second Maple program:
    b:= proc(n, i, k) option remember;
          `if`(n=0, 1, `if`(i<1 or k<1, 0, b(n, i-1, k)+
          `if`(i>n, 0, b(n-i, i, k-1))))
        end:
    T:= n-> seq(b(k, min(n, k), n), k=0..n^2):
    seq(T(n), n=0..8); # Alois P. Heinz, Apr 05 2012
  • Mathematica
    Table[nn=n^2;CoefficientList[Series[Product[(1-x^(n+i))/(1-x^i),{i,1,n}],{x,0,nn}],x],{n,0,6}]//Grid (* Geoffrey Critzer, Sep 27 2013 *)
    Table[CoefficientList[QBinomial[2n,n,q] // FunctionExpand, q], {n,0,6}] // Flatten (* Peter Luschny, Jul 22 2016 *)
    b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i < 1 || k < 1, 0, b[n, i - 1, k] + If[i > n, 0, b[n - i, i, k - 1]]]];
    T[n_] := Table[b[k, Min[n, k], n], {k, 0, n^2}];
    Table[T[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Nov 27 2020, after Alois P. Heinz *)
  • PARI
    T(n,k)=polcoeff(prod(i=0,n,sum(j=0,n,x^(j*i*(n^2+n+1)+j),O(x^(k*(n^2+n+1)+n+1)))),k*(n^2+n+1)+n)  /* Based on a more general formula due to R. Gerbicz. M. F. Hasler, Apr 12 2012 */

Formula

Table[T[k, n, n], {n, 0, 9}, {k, 0, n^2}] with T[ ] defined as in A047993.
G.f.: Consider a function; f(n) = 1 + sum(i_1=1, n, sum(i_2=0, i_1, ..., sum(i_n=0, i_(n-1), x^(sum(j=1, n, i_j))*(1+...+x^i_n))...)) Then the GF is f(1)+x^3.f(2)+x^8.f(3)+..., where after x^3 the increase is n^2+1 from f(n). - Jon Perry, Jul 13 2004
G.f. for n-th row is obtained if we set x(i) = 1+x^i+x^(2*i)+...+x^(n*i), i=1, 2, ..., n, in the cycle index Z(S(n);x(1), x(2), ..., x(n)) of the symmetric group S(n) of degree n. - Vladeta Jovovic, Dec 17 2004
G.f. of row n: the q-binomial coefficient [2n,n]. - Emeric Deutsch, Apr 23 2007
T(n,k)=1 for k=0,1,n^2-1,n^2. For all m>n, T(m,n)=T(n,n)=A000041(n), i.e., below the diagonal the columns remain constant, because there cannot be more than n nonzero elements with sum <= n. - M. F. Hasler, Apr 12 2012
T(n,2n) = A128552(n-2). - Geoffrey Critzer, Sep 27 2013
From Alois P. Heinz, Jan 09 2025: (Start)
Sum_{k=0..n} T(n,k) = A000070(n).
Sum_{k=0..n} k * T(n,k) = A182738(n).
Sum_{k=0..n^2} k * T(n,k) = A002544(n-1) for n>=1.
Sum_{k=0..n^2} (-1)^k * T(n,k) = A126869(n). (End)

A219237 Coefficient of Gauss polynomials [n+4,4]_q (q-binomials).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 2, 1, 1, 1, 1, 2, 3, 4, 4, 5, 4, 4, 3, 2, 1, 1, 1, 1, 2, 3, 5, 5, 7, 7, 8, 7, 7, 5, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 6, 8, 9, 11, 11, 12, 11, 11, 9, 8, 6, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 6, 9, 10, 13, 14, 16, 16, 18, 16, 16, 14, 13, 10, 9, 6, 5, 3, 2, 1, 1
Offset: 0

Views

Author

Wolfdieter Lang, Dec 04 2012

Keywords

Comments

The length of row n of this table is 4*n + 1 = A016813(n).
The sum of row n is binomial(n+4,4) = A000332(n+4), n>= 0.
The Gauss polynomial [n+4,4]q := [n+4]_q/([n]_q*[4]_q), with [n]_q = Product{j=1..n} (1-q^j) = (q;q)n (in q-shifted factorials notation), n>=0. [n+4,4]_q = (Product{j=(n+1)..(n+4)} (1-q^j))/(Product_{j=1..4} (1-q^j)). This is a polynomial in q (of degree 4*n) because it is the o.g.f. of the numbers p(n,4,k), the number of partitions of k into at most 4 parts, each <= n (see Andrews, p. 33 and 35). p(n,4,k) is also the number of partitions of k into at most n parts, each <= 4, due to the symmetry property [n+4,4]q = [n+4,n]_q (Andrews, (3,3,2), p.35). With the latter interpretation p(n,4,k) is the number of solutions of the two Diophantine equations Sum{j=1..4} j*m(j) = k and Sum_{j=0..m} m(j) = n, i.e. Sum_{j=1..m} m(j) = n - m(0), with 0 <= m(j) <= n. Therefore p(n,4,k) = [q^k] [x^n] G(4;x,q) with o.g.f. G(4;x,q) = 1/Product_{j=0..4} (1-x*q^j). Here we will call p(n,4,k) = T(n,k), n >= 0, 0 <= k <= 4*n.
See the comments in A008967 concerning a counting problem of Cayley (there m = 4, Theta = n and q = k), described also in the Hawkins reference (N(p->n,4,w->k) = T(n,k)) given there.

Examples

			The triangle T(n,k) begins:
  n\k 0  1  2  3  4  5  6   7   8   9  10  11  12  13  14  15  16 ...
  0:  1
  1:  1  1  1  1  1
  2:  1  1  2  2  3  2  2   1   1
  3:  1  1  2  3  4  4  5   4   4   3   2   1   1
  4:  1  1  2  3  5  5  7   7   8   7   7   5   5   3   2   1   1
  5:  1  1  2  3  5  6  8   9  11  11  12  11  11   9   8   6   5   3  2  1  1
  6:  1  1  2  3  5  6  9  10  13  14  16  16  18  16  16  14  13  10  9  6  5  3  2  1 1
Partition interpretation: T(3,5) = 4 because there are 4 partitions of 5 into at most 4 parts, each <= 3, namely 23, 113, 122 and 1112. here are also 4 partitions of 5 into at most 3 parts, each <= 4, namely 14, 23, 113 and 122. Note the conjugacy of the partitions 1112 and 14.
The 4 solutions of the two Diophantine equations given in a comment, with k=5 and n=3, are for (m(0), m(1), m(2), m(3), m(4)): (1,1,0,0,1), (1,0,1,1,0), (0,2,0,1,0) and (0,1,2,0,0).
		

References

  • G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 240, 242-3.

Crossrefs

Cf. A000012 (as triangle for m=1), A008967 (m=2), A047971 (m=3).

Programs

  • Mathematica
    a[0, 0] = 1; a[n_, k_] := SeriesCoefficient[ QBinomial[n+4, 4, q], {q, 0, k}]; Table[a[n, k], {n, 0, 6}, {k, 0, 4*n}] // Flatten (* Jean-François Alcover, Dec 04 2013 *)

Formula

T(n,k) = [q^k] [x^n](1/Product_{j=0..4} (1-x*q^j)), n >= 0, 0 <= k <= 4*n.
T(n,k) = [q^k]([n+4,4]_q), n >= 0, 0 <= k <= 4*n.
See the comments above.

A089789 Number of irreducible factors of Gauss polynomials.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 2, 2, 0, 0, 1, 2, 2, 1, 0, 0, 3, 3, 4, 3, 3, 0, 0, 1, 3, 3, 3, 3, 1, 0, 0, 3, 3, 5, 4, 5, 3, 3, 0, 0, 2, 4, 4, 5, 5, 4, 4, 2, 0, 0, 3, 4, 6, 5, 7, 5, 6, 4, 3, 0, 0, 1, 3, 4, 5, 5, 5, 5, 4, 3, 1, 0, 0, 5, 5, 7, 7, 9, 7, 9, 7, 7, 5, 5, 0, 0, 1, 5, 5, 6, 7, 7, 7, 7, 6, 5, 5, 1, 0
Offset: 0

Views

Author

Paul Boddington, Jan 09 2004

Keywords

Comments

T(n,k) is the number of irreducible factors of the (separable) polynomial [n]!/([k]![n-k]!). Here [n]! denotes the product of the first n quantum integers, the n-th quantum integer being defined as (1-q^n)/(1-q).
T(n,k) gives the number of positive integers m <= n such that (n mod m) < (k mod m). - Tom Edgar, Aug 21 2014

Examples

			The triangle T(n,k) begins:
n\k  0  1  2  3  4  5  6  7  8  9  10  11  12  13 ...
0:   0
1:   0  0
2:   0  1  0
3:   0  1  1  0
4:   0  2  2  2  0
5:   0  1  2  2  1  0
6:   0  3  3  4  3  3  0
7:   0  1  3  3  3  3  1  0
8:   0  3  3  5  4  5  3  3  0
9:   0  2  4  4  5  5  4  4  2  0
10:  0  3  4  6  5  7  5  6  4  3   0
11:  0  1  3  4  5  5  5  5  4  3   1   0
12:  0  5  5  7  7  9  7  9  7  7   5   5   0
13:  0  1  5  5  6  7  7  7  7  6   5   5   1   0
... Formatted by _Wolfdieter Lang_, Dec 07 2012
T(8,3) equals the number of irreducible factors of (1-q^8)(1-q^7)(1-q^6)/((1-q^3)(1-q^2)(1-q)), which is a product of 5 cyclotomic polynomials in q, namely the 2nd, 4th, 6th, 7th and 8th. Thus T(8,3)=5.
		

Crossrefs

Formula

T(n, k) = T(n-1, k-1) + d(n) - d(k), where d(n) is the number of divisors of n.
Showing 1-5 of 5 results.