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A177466 a(n) = binomial(n^3, n^2) / (n^2 + n + 1). 1
10, 360525, 23263187479980, 4195317468983232014706855, 3118254010126197540790713959812283024388, 13329519847131745416659896296893907619682838146506167497550 (list; graph; refs; listen; history; text; internal format)
OFFSET
2,1
COMMENTS
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]
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
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
LINKS
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
EXAMPLE
For n = 2, a(2) = binomial(8,4)/7 = 70/7 = 10.
MAPLE
A177466 := proc(n) binomial(n^3, n^2)/(n^2+n+1) ; end proc:
seq(A177466(n), n=2..10) ; # R. J. Mathar, Dec 06 2010
MATHEMATICA
Table[Binomial[n^3, n^2]/(n^2+n+1), {n, 2, 7}] (* Harvey P. Dale, Jan 24 2019 *)
CROSSREFS
Sequence in context: A201548 A363197 A281361 * A069878 A235029 A013854
KEYWORD
nonn
AUTHOR
Michel Lagneau, May 09 2010
STATUS
approved

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Last modified June 20 03:21 EDT 2024. Contains 373512 sequences. (Running on oeis4.)