A204262 Permanent of the n-th principal submatrix of A003983.

1, 1, 3, 19, 209, 3545, 85803, 2807723, 119377321, 6397099105, 421772316915, 33552418294339, 3168847554832961, 350514662908385321, 44885099167514403963, 6587836555407268741019, 1098597117953239632728089, 206564512095561068049417265, 43495029251774783469442768323

Offset: 0

Author

Clark Kimberling, Jan 14 2012

Keywords

• nonn

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..264
• Discussion at dxdy.ru, Permanent of a matrix, (in Russian) (2023).
• Terence Tao, Remarkable recursions for the A204262, answer to question on MathOverflow (2023).

Crossrefs

Cf. A003983, A204264.

Programs

Maple

with(LinearAlgebra):
a:= n-> `if`(n=0, 1, Permanent(Matrix(n, ()-> min(args)))):
seq(a(n), n=0..16);  # Alois P. Heinz, Nov 14 2016

Mathematica

f[i_, j_] := Min[i, j];
m[n_] := Table[f[i, j], {i, 1, n}, {j, 1, n}]
TableForm[m[8]] (* 8x8 principal submatrix *)
Flatten[Table[f[i, n + 1 - i],
  {n, 1, 12}, {i, 1, n}]]      (* A003983 *)
Permanent[m_] :=
  With[{a = Array[x, Length[m]]},
   Coefficient[Times @@ (m.a), Times @@ a]];
Table[Permanent[m[n]], {n, 1, 15}]    (* A204262 *)

PARI

a(n)={my(S,z,v=vector(n));for(i=0,n!-1,v=numtoperm(n,i);z=1;for(j=1,n,z*= n+1-max(j,v[j]));S+=z);return(S)} \\ R. J. Cano, Nov 14 2016

PARI

upto(n)=my(v1, x='x); v1=vector(n+1, i, i--; i!*x^i); for(i=1, n, for(j=i, n, my(A=intformal((j-i)^2*v1[j])); v1[j+1] = A + subst(v1[j+1] - A, x, i))); v1 \\ Mikhail Kurkov, Aug 03 2023

Formula

From Mikhail Kurkov, Aug 03 2023: (Start)

a(n) = f(n, n, 0) for n >= 0 where f(n, q, x) = g(n, q, x) + f(n, q-1, q) - g(n, q, q) for n >= 0, q > 0 with f(n, 0, x) = n!*x^n for n >= 0 and where g(n, q, x) = Integral (n-q)^2*f(n-1, q, x) dx for n > 0, q > 0 (formula due to user with the nickname Null on a scientific forum dxdy.ru).

a(n) = R(n-1, 0) for n > 0 with a(0) = 1 where R(n, q) = Sum_{j=0..q+1} binomial(q+1, j)*(j+1)*R(n-1, j) for n > 0, q >= 0 with R(0, q) = 1 for q >= 0.

Both results were proved by Terence Tao, see Links section. (End)

Conjecture: Limit_{n->oo} (a(n)/n!^2)^(1/n) = 2/Pi. - Vaclav Kotesovec, Aug 05 2023

Extensions

a(0)=1 prepended and more terms added by Alois P. Heinz, Nov 14 2016