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a(n) = binomial(n^2, n)/n.
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%I #30 Oct 29 2023 18:14:16

%S 1,3,28,455,10626,324632,12271512,553270671,28987537150,1731030945644,

%T 116068178638776,8634941152058949,705873715441872264,

%U 62895036884524942320,6067037854078498539696,629921975126394617164575,70043473196734767582082230

%N a(n) = binomial(n^2, n)/n.

%H Harry J. Smith, <a href="/A060545/b060545.txt">Table of n, a(n) for n = 1..100</a>

%H Romeo Meštrović, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv preprint arXiv:1111.3057 [math.NT], 2011.

%F a(n) = A060543(n, n) = A014062(n)/n.

%F a(n+1) = C(A005563(n), n) for n >= 0. - _Fred Daniel Kline_, Sep 27 2016

%F From _Peter Bala_, Oct 22 2023: (Start)

%F a(p^r) == 1 (mod p^(3+r)) for all positive integers r and all primes p >= 5 (apply Meštrović, Remark 17, p. 12).

%F Conjecture: a(2*p^r) == 4*p^r - 1 (mod p^(3+r)) for all positive integers r and all primes p >= 5. (End)

%p A060545:=n->binomial(n^2,n)/n: seq(A060545(n), n=1..20); # _Wesley Ivan Hurt_, Sep 28 2016

%t Table[Binomial[n^2, n]/n, {n, 15}] (* _Michael De Vlieger_, Sep 28 2016 *)

%o (PARI) a(n) = binomial(n^2, n)/n; \\ _Harry J. Smith_, Jul 06 2009

%Y Cf. A005563, A014062, A060543.

%K nonn,easy

%O 1,2

%A _Henry Bottomley_, Apr 02 2001

%E More terms from _Fred Daniel Kline_, Sep 28 2016