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%I #71 Mar 14 2024 03:16:09
%S 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,
%T 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,
%U 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
%N Coefficients of Gaussian polynomials q_binomial(n-2, 2). Also triangle of distribution of rank sums: Wilcoxon's statistic. Irregular triangle read by rows.
%C 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
%C The Gaussian polynomial (or Gaussian binomial) [n,2]_q is an example of a q-binomial coefficient (see the link) and may be defined for n >= 2 by [n,2]_q = ([n]_q * [n-1]_q)/([1]_q * [2]_q), where [n]_q := q^n - 1. The first few values are: [2,2]_q = 1; [3,2]_q = 1 + q + q^2; [4,2]_q = 1 + q + 2q^2 + q^3 + q^4. - _Peter Bala_, Sep 23 2007
%C 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
%C 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
%C From _Gus Wiseman_, Sep 20 2023: (Start)
%C 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:
%C 12 13 14 15 16 17 18 19 29 39 49 59 69 79 89
%C 23 24 25 26 27 28 38 48 58 68 78
%C 34 35 36 37 47 57 67
%C 45 46 56
%C Allowing repeated parts (x,x) gives A004737.
%C For strict partitions instead of just pairs we have A053632.
%C (End)
%D G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 242.
%D F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 236.
%D T. Hawkins, Emergence of the Theory of Lie Groups, Springer 2000, ch. 7.4, p. 260-5.
%H John Tyler Rascoe, <a href="/A008967/b008967.txt">Rows n = 4..103, flattened</a>
%H A. Cayley, <a href="https://doi.org/10.1098/rstl.1856.0008">A Second Memoir Upon Quantics</a>, Phil. Trans. R. Soc. London, 146 (1856) 101-126.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/q-BinomialCoefficient.html">q-Binomial Coefficient</a>.
%H <a href="/index/Do#domino">Index entries for sequences related to dominoes</a>
%F 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).
%F 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
%F 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
%e 1;
%e 1,1,1;
%e 1,1,2,1,1;
%e 1,1,2,2,2,1,1;
%e 1,1,2,2,3,2,2,1,1;
%e 1,1,2,2,3,3,3,2,2,1,1;
%e ...
%e 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
%p qBinom := proc(n,m,q)
%p mul((1-q^(n-i))/(1-q^(i+1)),i=0..m-1) ;
%p factor(%) ;
%p expand(%) ;
%p end proc:
%p A008967 := proc(n,k)
%p coeftayl( qBinom(n,2,q),q=0,k ) ;
%p end proc:
%p seq(seq( A008967(n,k),k=0..2*n-4),n=2..10) ; # assumes offset 2. _R. J. Mathar_, Oct 13 2011
%t 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 T[n_, k_] := SeriesCoefficient[QBinomial[n - 2, 2, q], {q, 0, k}];
%t Table[T[n, k], {n, 4, 13}, {k, 0, 2 n - 8}] // Flatten (* _Jean-François Alcover_, Aug 20 2019 *)
%t Table[Length[Select[Subsets[Range[n],{2}],Total[#]==k&]],{n,2,15},{k,3,2n-1}] (* _Gus Wiseman_, Sep 20 2023 *)
%o (SageMath)
%o print(flatten([q_binomial(n-2, 2).list() for n in (4..13)])) # _Peter Luschny_, Oct 23 2019
%Y Cf. A008968, A047971, A106822.
%Y A version with zeros is A219238.
%Y This is the case of A365541 counting only length-2 subsets.
%Y Cf. A000009, A004526, A004737, A053632, A068911, A167762, A238628, A365544.
%K tabf,nonn,nice,easy
%O 4,7
%A _N. J. A. Sloane_
%E More terms from _Christian Barrientos_, Aug 08 2019
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