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a(n) = binomial(4*n, 4).
3

%I #28 Mar 08 2022 05:06:29

%S 1,70,495,1820,4845,10626,20475,35960,58905,91390,135751,194580,

%T 270725,367290,487635,635376,814385,1028790,1282975,1581580,1929501,

%U 2331890,2794155,3321960,3921225,4598126,5359095,6210820,7160245,8214570,9381251,10668000,12082785

%N a(n) = binomial(4*n, 4).

%H Harry J. Smith, <a href="/A060541/b060541.txt">Table of n, a(n) for n=1..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F a(n) = n*(2n-1)*(4n-1)*(4n-3)/3.

%F a(n) = n * A015219(n-1) = A000332(4n) = A060539(n, 4).

%F G.f.: x*(1+65*x+155*x^2+35*x^3) / (1-x)^5. - _R. J. Mathar_, Oct 03 2011

%F From _Amiram Eldar_, Mar 08 2022: (Start)

%F Sum_{n>=1} 1/a(n) = 6*log(2) - Pi.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 2*sqrt(2)*log(sqrt(2)-1) - log(2) + (2*sqrt(2) - 3/2)*Pi. (End)

%t Table[Binomial[4n, 4], {n, 100}] (* _Wesley Ivan Hurt_, Sep 27 2013 *)

%t LinearRecurrence[{5,-10,10,-5,1},{1,70,495,1820,4845},40] (* _Harvey P. Dale_, Jan 13 2015 *)

%o (PARI) a(n) = n*(2*n - 1)*(4*n - 1)*(4*n - 3)/3; \\ _Harry J. Smith_, Jul 06 2009

%o (Magma) [Binomial(4 n, 4): n in [1..40]]; // _Vincenzo Librandi_, Jan 20 2015

%Y Cf. A000027, A000332, A000384, A006566, A015219, A060539.

%K nonn,easy

%O 1,2

%A _Henry Bottomley_, Apr 02 2001

%E Offset changed from 0 to 1 by _Harry J. Smith_, Jul 06 2009