login
A177456
a(n) = binomial(n^2,n+1)/n.
5
2, 42, 1092, 35420, 1391280, 64425438, 3442573064, 208710267480, 14162980464360, 1063958304188780, 87677864005521636, 7865449972066576656, 763126447532235966816, 79629871834780293333510
OFFSET
2,1
COMMENTS
n divides binomial(n^2,n+1).
Proof 1 :(n+1)*binomial(n^2,n+1) = n*(n-1)*binomial(n^2,n) => n divide binomial(n^2,n+1) because gcd(n,n+1) = 1.
Proof 2 : a(n) = binomial(n^2,n+1)/n = (n-1)*binomial(n^2-2,n-1)=> a(n) is an integer. - Michel Lagneau, May 13 2010
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
FORMULA
a(n) = binomial(n^2,n+1)/n.
From G. C. Greubel, Apr 29 2024: (Start)
a(n) = (n-1)*A177234(n).
a(n) = (n-1)*A177788(n)/n.
a(n) = n*(n-1)*A177784(n).
a(n) = A014062(n)/n. (End)
EXAMPLE
For n=4, 1092 is in the sequence because binomial(16,5)/4 = 4368/4 = 1092.
MAPLE
with(numtheory):n0:=30:T:=array(1..n0-1):for n from 2 to n0 do:T[n-1]:= (binomial(n*n, n+1))/n:od:print(T):
MATHEMATICA
Table[Binomial[n^2, n+1]/n, {n, 2, 30}] (* G. C. Greubel, Apr 29 2024 *)
PROG
(Magma) [Binomial(n^2, n+1)/n: n in [2..30]]; // G. C. Greubel, Apr 29 2024
(SageMath) [binomial(n^2, n+1)/n for n in range(2, 31)] # G. C. Greubel, Apr 29 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, May 09 2010
STATUS
approved