A065048 Largest unsigned Stirling number of the first kind: max_k(s(n+1,k)); i.e., largest coefficient of polynomial x*(x+1)*(x+2)*(x+3)*...*(x+n).

1, 1, 3, 11, 50, 274, 1764, 13132, 118124, 1172700, 12753576, 150917976, 1931559552, 26596717056, 392156797824, 6165817614720, 102992244837120, 1821602444624640, 34012249593822720, 668609730341153280, 13803759753640704000, 298631902863216384000

Offset: 0

Author

Henry Bottomley, Nov 06 2001

Keywords

• nonn

Comments

n! <= a(n) <= (n+1)!; n <= a(n+1)/a(n) <= (n+1). - Max Alekseyev, Jul 17 2019

Examples

			a(4)=50 since polynomial is x^4 + 10*x^3 + 35*x^2 + 50*x + 24.
		

Links

• Robert Israel, Table of n, a(n) for n = 0..448

Crossrefs

Cf. A000254, A000399, A002870, A008275, A309237.

Programs

Maple

P:= x: A[0]:= 1:
for n from 1 to 50 do
  P:= expand(P*(x+n));
  A[n]:= max(coeffs(P,x));
od:
seq(A[i],i=0..50); # Robert Israel, Jul 04 2016

Mathematica

a[n_] := Max[Array[Abs[StirlingS1[n+1, #]]&, n+1]];
Array[a, 100, 0] (* Griffin N. Macris, Jul 03 2016 *)

PARI

a(n) = if (n==0, 1, vecmax(vector(n, k, abs(stirling(n+1, k, 1))))); \\ Michel Marcus, Jul 04 2016; corrected Jun 12 2022

Python

from collections import Counter
def A065048(n):
    c = {1:1}
    for k in range(1,n+1):
        d = Counter()
        for j in c:
            d[j] += k*c[j]
            d[j+1] += c[j]
        c = d
    return max(c.values()) # Chai Wah Wu, Jan 31 2024

Formula

For n in the interval [A309237(k)-1, A309237(k+1)-2], a(n) = |Stirling1(n+1,k)|. - Max Alekseyev, Jul 17 2019