cp's OEIS Frontend

A001222(a(n)) = A036043(n).

A001221(a(n)) = A103921(n).

A001222(a(n)) = A036043(n).

a(n) = A001221(A185974(n)). - Gus Wiseman, May 20 2020

For a partition P = [p_1,...,p_n] with the p_i in increasing order, define b(i,j) to be the number of subpartitions of [p_1,...,p_i] with the i-th part = j (b(i,0) is subpartitions with less than i parts). Then b(1,j)=1 for j<=p_1, b(i+1,j) = Sum_{k=0..j} b(i,k) for 0<=j<=p_{i+1}; and the total number of subpartitions is sum_{k=1..p_n} b(n,k).

For a partition P = {p(n)}, the number of subpartitions {s(n)} of P can be determined by the g.f.: 1/(1-x) = Sum_{n>=0} s(n)*x^n*(1-x)^p(n). - Paul D. Hanna, Jul 03 2006

A001222(a(n)) appears to be A049085(n).

A001222(a(n)) = A333486(n).

A056239(a(n)) <= 2*A001222(a(n)).

a(n) = A322136(n)/4.

a(n,k) = n!/(Product_{j=1..n} e(n,k,j)!) with the exponent e(n,k,j) of j in the k-th partition of n in the A-St ordering of the partitions of n. Exponents 0 can be omitted due to 0!=1.

From Tom Copeland, Sep 18 2011: (Start)

Raising and lowering operators are given for the partition polynomials formed from A130561 in the Copeland link in "Lagrange a la Lah Part I" on pp. 22-23.

An e.g.f. for the partition polynomials is on page 3:

exp[t*:c.*x/(1-c.*x):] = exp[t*(c_1*x + c_2*x^2 + c_3*x^3 + ...)] where :(...): denotes umbral evaluation of the enclosed expression and c. is an umbral coefficient. (End)

From Tom Copeland, Sep 07 2016: (Start)

The row partition polynomials of this array P(n,x_1,x_2,...,x_n), given in the Lang link, are n! * S(n,x_1,x_2,...,x_n), where S(n,x_1,...,x_n) are the elementary Schur polynomials, for which d/d(x_m) S(n,x_1,...,x_n) = S(n-m,x_1,...,x_(n-m)) with S(k,...) = 0 for k < 0, so d/d(x_m) P(n,x_1,...,x_n) = (n!/(n-m)!) P(n-m,x_1,...,x_(n-m)), confirming that the row polynomials form an Appell sequence in the indeterminate x_1 with P(0,...) = 1. See p. 127 of the Ernst paper for more on these Schur polynomials.

With the e.g.f. exp[t * P(.,x_1,x_2,..)] = exp(t*x_1) * exp(x_2 t^2 + x_3 t^3 + ...), the e.g.f. for the partition polynomials that form the umbral compositional inverse sequence U(n,x_1,...,x_n) in the indeterminate x_1 is exp[t * U(.,x_1,x_2,...)] = exp(t*x_1) exp[-(x_2 t^2 + x_3 t^3 + ...)]; therefore, U(n,x_1,x_2,...,x_n) = P(n,x_1,-x_2,.,-x_n), so umbrally P[n,P(.,x_1,-x_2,-x_3,...),x_2,x_3,...,x_n] = (x_1)^n = P[n,P(.,x_1,x_2,...),-x_2,-x_3,...,-x_n]. For example, P(1,x_1) = x_1, P2(x_1,x_2) = 2 x_2 + x_1^2, and P(3,x_1,x_2,x_3) = 6 x_3 + 6 x_2 x_1 + x_1^3, then P[3,P(.,x_1,-x_2,...),x_2,x_3] = 6 x_3 + 6 x_2 P(1,x_1) + P(3,x_1,-x_2,-x_3) = 6 x_3 + 6 x_2 x_1 + 6 (-x_3) + 6 (-x_2) x_1 + x_1^3 = x_1^3.

From the Appell formalism, umbrally [P(.,0,x_2,x_3,...) + y]^n = P(n,y,x_2,x_3,...,x_n).

The indeterminates of the partition polynomials can also be extracted using the Faber polynomials of A263916 with -n * x_n = F(n,S(1,x_1),...,S(n,x_1,...,x_n)) = F(n,P(1,x_1),...,P(n,x_1,...,x_n)/n!). Compare with A263634.

Also P(n,x_1,...,x_n) = ST1(n,x_1,2*x_2,...,n*x_n), where ST1(n,...) are the row partition polynomials of A036039.

(End)