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A177454
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( binomial(2*p,p) - 2)/p where p = prime(n).
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3
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2, 6, 50, 490, 64130, 800046, 137270954, 1860277042, 357975249026, 1036802293087622, 15013817846943906, 47192717955016924590, 10360599532897359064118, 154361699651715243559786
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OFFSET
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1,1
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COMMENTS
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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).]
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LINKS
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FORMULA
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EXAMPLE
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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.
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MAPLE
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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):
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MATHEMATICA
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Table[(Binomial[2Prime[n], Prime[n]] - 2)/Prime[n], {n, 15}] (* Alonso del Arte, Feb 27 2013 *)
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PROG
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(Magma) [(Binomial(2*p, p)-2)/p where p is NthPrime(n):n in [1..14]]; // Marius A. Burtea, Aug 11 2019
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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