0,6

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 0s and n 1s having major index equal to k (the major index is the sum of the positions of the 1s that are immediately followed by 0s). 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 3 0s and 3 1s 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 (deutsch(AT)duke.poly.edu), Oct 02 2007

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.

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 (perry(AT)globalnet.co.uk), 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 (vladeta(AT)eunet.rs), Dec 17 2004

G.f. of row n = the q-binomial coefficient [2n,n]. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 23 2007

{1}, {1,1}, {1,1,2,1,1}, {1,1,2,3,3,3,3,2,1,1}, ...

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 by 3 box are [3,3,1] and [3,2,2].

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 (deutsch(AT)duke.poly.edu), Apr 23 2007

(PARI) f1(x)=1+x*sum(j=0, 1, x^j); f2(x)=1+sum(i=1, 2, x^i*sum(j=0, i, x^j)); f3(x)=1+sum(i=1, 3, sum(k=0, i, x^(i+k)*sum(j=0, k, x^j))); f4(x)=1+sum(i=1, 4, sum(i1=0, i, sum(k=0, i1, x^(i+i1+k)*sum(j=0, k, x^j)))) f(x)=f1(x)+x^3*f2(x)+x^8*f3(x)+x^18*f4(x); for (i=0, 30, print1(", "polcoeff(f(x), i))) (Perry)

A008968, A047971, A047993.

Sequence in context: A083671 A029384 A094102 this_sequence A131338 A106498 A093466

Adjacent sequences: A063743 A063744 A063745 this_sequence A063747 A063748 A063749

nonn,tabf

Wouter Meeussen (wouter.meeussen(AT)pandora.be), Aug 14 2001