A061718 - OEIS

%I #29 Dec 25 2023 15:09:24

%S 1,9,216,10000,759375,85766121,13492928512,2821109907456,

%T 756680642578125,253295162119140625,103510234140112521216,

%U 50714860157241037295616,29345269354638035222576971

%N a(n) = (n*(n+1)/2)^n.

%C a(n) is the number of n X n matrices with nonnegative integer entries such that every row sum equals 2. - Sharon Sela (sharonsela(AT)hotmail.com), May 08 2002

%C Resultant of the polynomials P(n,x) and Q(n,x) where P(n,x)=sum(k=1,n,k*(-x)^k) and Q(n,x)=x^n-1. - _Benoit Cloitre_, Jan 26 2003

%C a(n) is also the number of positive-volume, axis-aligned, n-dimensional rectangular solids that have vertices in the set {0,1,...,n}^n. Proof: If (M_1,...,M_n) is the corner with the maximum coordinate values for such a solid, then there are (M_1)*...*(M_n) possibilities for the corner with the minimum coordinate values. The sum over all possibilities for M_1, ..., M_n can be factored into the product of n sums; each of the n sums simplifies to n(n+1)/2. - _Lee A. Newberg_, Aug 31 2009

%H Harry J. Smith, <a href="/A061718/b061718.txt">Table of n, a(n) for n = 1..100</a>

%F Sum(i=1..n,j=1..n,k=1..n,...,(i*j*k*...)). E.g., a(2) = 9 because 1*1 + 1*2 + 2*1 + 2*2 = 9. - _Ben Paul Thurston_, Aug 15 2006

%F a(n) = [x^n] 1/(1 - (n*(n + 1)/2)*x). - _Ilya Gutkovskiy_, Oct 10 2017

%p a:=n->mul(sum(j, j=0..n),k=1..n): seq(a(n), n=1..13); # _Zerinvary Lajos_, Jun 02 2007

%p a:=n->mul(binomial(n+2,2), k=0..n): seq(a(n), n=0..12); # _Zerinvary Lajos_, Oct 02 2007

%t Table[((n(n+1))/2)^n,{n,20}] (* _Harvey P. Dale_, Dec 09 2022 *)

%t With[{nn=15},#[[1]]^#[[2]]&/@Thread[{Accumulate[Range[nn]],Range[nn]}]] (* _Harvey P. Dale_, Dec 25 2023 *)

%o (C++) /* e.g. n = 6: */

%o int main()

%o {

%o     int sum = 0;

%o     for(int i = 1; i < 7; i++)

%o         for(int j = 1; j < 7; j++)

%o             for(int k=1;k<7; k++)

%o                 for(int l = 1; l < 7; l++)

%o                     for(int m = 1; m < 7; m++)

%o                         for(int n = 1; n < 7; n++)

%o                             sum += i*j*k*l*m*n;

%o     cout << sum << endl;

%o     return 0;

%o } // _Ben Paul Thurston_, Aug 15 2006

%o (PARI) { for (n=1, 100, write("b061718.txt", n, " ", (n*(n + 1)/2)^n) ) } \\ _Harry J. Smith_, Jul 26 2009

%Y Cf. A066300.

%K easy,nonn

%O 1,2

%A _Jason Earls_, Jun 20 2001