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a(n) = binomial(n+1, 2)^5.
8

%I #31 May 15 2022 07:31:12

%S 1,243,7776,100000,759375,4084101,17210368,60466176,184528125,

%T 503284375,1252332576,2887174368,6240321451,12762815625,24883200000,

%U 46525874176,83841135993,146211169851,247609900000,408410100000,657748550151,1036579476493,1601568101376

%N a(n) = binomial(n+1, 2)^5.

%C Number of 5-dimensional cage assemblies.

%C See Chap. 61, "Hyperspace Prisons", of C. Pickover's book "Wonders of Numbers" for full explanation of "cage numbers."

%D Clifford A. Pickover, Wonders of Numbers, Oxford University Press, 2001, p. 325.

%H Harry J. Smith, <a href="/A059860/b059860.txt">Table of n, a(n) for n = 1..1000</a>

%H Clifford A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," <a href="http://www.zentralblatt-math.org/zmath/en/search/?q=an:0983.00008&amp;format=complete">Zentralblatt review</a>.

%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

%F L(n) = ((n^m)(n + 1)^m)/(2^m) where m is the dimension.

%F G.f.: x * (x^8 +232*x^7 +5158*x^6+ 27664*x^5 +47290*x^4 +27664*x^3 +5158*x^2 +232*x +1) / (1-x)^11. - _Colin Barker_, Jun 28 2012

%F From _Amiram Eldar_, May 15 2022: (Start)

%F Sum_{n>=1} 1/a(n) = 4032 - 1120*Pi^2/3 - 32*Pi^4/9.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 4480*log(2) + 720*zeta(3) + 60*zeta(5) - 4032. (End)

%p for n from 1 to 100 do printf(`%d,`,((n^5)*(n + 1)^5)/(2^5) ) od:

%p with (combinat):seq(mul(stirling2(n+1,n),k=1..5),n=1..21); # _Zerinvary Lajos_, Dec 14 2007

%t m = 5; Table[ ( (n^m)(n + 1)^m )/(2^m), {n, 1, 26} ]

%t Table[Binomial[n+1,2]^5,{n,20}] (* _Harvey P. Dale_, May 04 2018 *)

%o (PARI) { for (n=1, 1000, write("b059860.txt", n, " ", (n*(n + 1)/2)^5); ) } \\ _Harry J. Smith_, Jun 29 2009

%Y Cf. A059827.

%K easy,nonn

%O 1,2

%A _Jason Earls_, Feb 28 2001

%E More terms from _James A. Sellers_, Feb 28 2001

%E Better definition from _Zerinvary Lajos_, May 23 2006

%E Corrected the definition from binomial(n+2,2)^5 to binomial(n+1,2)^5. - _Harry J. Smith_, Jun 29 2009