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( binomial(2*p,p) - 2)/p where p = prime(n).
3

%I #19 Sep 08 2022 08:45:53

%S 2,6,50,490,64130,800046,137270954,1860277042,357975249026,

%T 1036802293087622,15013817846943906,47192717955016924590,

%U 10360599532897359064118,154361699651715243559786

%N ( binomial(2*p,p) - 2)/p where p = prime(n).

%C All entries are integer because binomial(2p, p) == 2 (mod p). [Proof: p!*binomial(2p, p) = 2p(2p - 1)(2p - 2) ... (p + 1) .

%C Therefore (p - 1)!*binomial(2p, p) = 2(2p - 1) ... (p + 1) == 2(p - 1)! (mod p).

%C Since p is prime: (p - 1)! <> 0 (mod p). Because Z/pZ is a finite field, we conclude that binomial(2p, p) == 2 (mod p).]

%H Amiram Eldar, <a href="/A177454/b177454.txt">Table of n, a(n) for n = 1..263</a>

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

%H Paul Barry, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Barry/barry84.html">A Catalan Transform and Related Transformations on Integer Sequences</a>, J. Integer Sequ., Vol. 8 (2005), Article 05.4.5.

%F a(n) = (A000984(p) - 2) / p with p = A000040(n).

%e a(1) = 2 because prime(1) = 2 and (binomial(4, 2) - 2)/2 = (6 - 2)/2 = 2.

%e a(4) = 490 because prime(4) = 7 and (binomial(14, 7) - 2)/7 = (3432 - 2)/7 = 490.

%p with(numtheory): n0:=20: T:=array(1..n0): k:=1: for n from 1 to 72 do:if type(n,prime)=true then T[k]:= (binomial(2*n,n)-2)/n: k:=k+1: fi: od: print(T):

%t Table[(Binomial[2Prime[n], Prime[n]] - 2)/Prime[n], {n, 15}] (* _Alonso del Arte_, Feb 27 2013 *)

%o (Magma) [(Binomial(2*p,p)-2)/p where p is NthPrime(n):n in [1..14]]; // _Marius A. Burtea_, Aug 11 2019

%Y Cf. A000984, A060842, A060545, A024498.

%K nonn

%O 1,1

%A _Michel Lagneau_, May 09 2010