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A177454 ( binomial(2*p,p) - 2)/p where p = prime(n). 3
2, 6, 50, 490, 64130, 800046, 137270954, 1860277042, 357975249026, 1036802293087622, 15013817846943906, 47192717955016924590, 10360599532897359064118, 154361699651715243559786 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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

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

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).]

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..263

Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, J. Integer Sequ., Vol. 9 (2006), Article 06.2.4.

Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, J. Integer Sequ., Vol. 8 (2005), Article 05.4.5.

FORMULA

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

EXAMPLE

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

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

MAPLE

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):

MATHEMATICA

Table[(Binomial[2Prime[n], Prime[n]] - 2)/Prime[n], {n, 15}] (* Alonso del Arte, Feb 27 2013 *)

PROG

(MAGMA) [(Binomial(2*p, p)-2)/p where p is NthPrime(n):n in [1..14]]; // Marius A. Burtea, Aug 11 2019

CROSSREFS

Cf. A000984, A060842, A060545, A024498.

Sequence in context: A080310 A103990 A079835 * A052332 A238596 A134047

Adjacent sequences:  A177451 A177452 A177453 * A177455 A177456 A177457

KEYWORD

nonn

AUTHOR

Michel Lagneau, May 09 2010

STATUS

approved

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Last modified November 18 09:55 EST 2019. Contains 329261 sequences. (Running on oeis4.)