A060842 -id:A060842 - cp's OEIS frontend

A087754 a(n) = (C(2p,p)-2) / p^3, where p = prime(n).

2, 10, 530, 4734, 474986, 5153122, 676701794, 1232820800342, 15623119507746, 34472401720246110, 6163354867874693078, 83483882991733501114, 15658391111267929558466, 42132263940113324754864134

Offset: 3

Henry Bottomley, Oct 02 2003

nonn

			a(6)=4734 since 13 is the sixth prime and (C(26,13)-2)/13^3 = (10400600-2)/2197 = 4734.

R. R. Aidagulov, M. A. Alekseyev. On p-adic approximation of sums of binomial coefficients. Journal of Mathematical Sciences 233:5 (2018), 626-634. doi:10.1007/s10958-018-3948-0 arXiv:1602.02632
R. P. Stanley, MIT Course OCW 18.S66, The Art of Counting, Spring 2003. (Problem 18)

Cf. A268512, A268589, A268590.

Table[(Binomial[2p,p]-2)/p^3,{p,Prime[Range[3,20]]}] (* Harvey P. Dale, Oct 23 2017 *)

a(n) = A060842(n) / A000040(n).

a(n) = 2 * A034602(n).

A177454 ( binomial(2*p,p) - 2)/p where p = prime(n).

Original entry on oeis.org

2, 6, 50, 490, 64130, 800046, 137270954, 1860277042, 357975249026, 1036802293087622, 15013817846943906, 47192717955016924590, 10360599532897359064118, 154361699651715243559786

Offset: 1

Michel Lagneau, May 09 2010

nonn

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).]

			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.

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.

Cf. A000984, A060842, A060545, A024498.

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

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):

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

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

cp's OEIS Frontend

A087754 a(n) = (C(2p,p)-2) / p^3, where p = prime(n).

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