A177466 - OEIS

%I #27 Sep 16 2024 12:48:57

%S 10,360525,23263187479980,4195317468983232014706855,

%T 3118254010126197540790713959812283024388,

%U 13329519847131745416659896296893907619682838146506167497550

%N a(n) = binomial(n^3, n^2) / (n^2 + n + 1).

%C All entries are integers. [Proof: binomial(n^3, n^2) / (n^2 + n + 1) = n^3 (n^3 - 1) (n^3 - 2)*...*(n^3- n^2 +1) / ( (n^2)! *(n^2 + n + 1)). With n^3 - 1 = (n-1)*(n^2 + n + 1), we obtain a(n) = n* binomial(n^3-2, n^2-2) / (n+1). Finally: (n+1) * binomial(n^3, n^2) * 1/ (n^2 + n + 1) = n*binomial(n^3-2, n^2-2). QED]

%C The step after "finally" seems to demonstrate merely that (n+1)*a(n) is an integer, but not that a(n) is itself an integer. Is the proof incomplete? - _R. J. Mathar_, Dec 06 2010

%C So far all that has been shown is that (n+1)*a(n) is an integer. To complete the proof, note that a(n) = n^3*(n-1)*(n^3-2)*...*(n^3-n^2) / (n^2*(n^2-1)!*(n^3-n^2)) = binomial(n^3-2,n^2-1)/n. Hence n*a(n) is also an integer, and so (n+1)*a(n) - n*a(n) = a(n) is an integer. Q.E.D. - _N. J. A. Sloane_, Dec 09 2010

%H G. C. Greubel, <a href="/A177466/b177466.txt">Table of n, a(n) for n = 2..23</a>

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL9/Barry/barry91.html">On Integer-Sequence-Based Constructions of Generalized Pascal Triangles</a>, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

%e For n = 2, a(2) = binomial(8,4)/7 = 70/7 = 10.

%p A177466 := proc(n) binomial(n^3,n^2)/(n^2+n+1); end proc:

%p seq(A177466(n),n=2..10) ; # _R. J. Mathar_, Dec 06 2010

%t Table[Binomial[n^3,n^2]/(n^2+n+1),{n,2,7}] (* _Harvey P. Dale_, Jan 24 2019 *)

%o (Magma) [Binomial(n^3,n^2)/(n^2+n+1): n in [2..12]]; // _G. C. Greubel_, Jul 18 2024

%o (SageMath) [binomial(n^3,n^2)/(n^2+n+1) for n in range(2,13)] # _G. C. Greubel_, Jul 18 2024

%Y Cf. A000108, A000984, A060545, A177234, A177456.

%K nonn

%O 2,1

%A _Michel Lagneau_, May 09 2010