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%I A134163 #19 Sep 08 2022 08:45:32
%S A134163 1,187,1531,5977,16441,36811,71947,127681,210817,329131,491371,707257,
%T A134163 987481,1343707,1788571,2335681,2999617,3795931,4741147,5852761,
%U A134163 7149241,8650027,10375531,12347137,14587201,17119051,19966987,23156281,26713177
%N A134163 1 + 12*n + 81*n^3 + n*(105*n + 81*n^3)/2.
%C A134163 A000541(n) is divisible by A000537(n) if and only n is congruent to 1 mod 3 (see A016777).
%H A134163 Vincenzo Librandi, <a href="/A134163/b134163.txt">Table of n, a(n) for n = 0..1000</a>
%H A134163 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A134163 a(n) = (3(3n + 1)^4 + 6(3n + 1)^3 - (3n + 1)^2 - 4 (3n + 1) + 2)/6.
%F A134163 a(n) = ( sum_{k=1..3n+1} k^7 ) / ( sum_{k=1..3n+1} k^3 ).
%F A134163 G.f.: (1+182*x+606*x^2+182*x^3+x^4)/(1-x)^5. - _R. J. Mathar_, Nov 14 2007
%F A134163 a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - _Wesley Ivan Hurt_, Oct 23 2014
%p A134163 A134163:=n->1 + 12*n + 81*n^3 + n*(105*n + 81*n^3)/2: seq(A134163(n), n=0..30); # _Wesley Ivan Hurt_, Oct 23 2014
%t A134163 Table[(3(3n + 1)^4 + 6(3n + 1)^3 - (3n + 1)^2 - 4 (3n + 1) + 2)/6, {n, 0, 100}] (* or *) Table[Sum[k^7, {k, 1, 3n + 1}]/Sum[k^3, {k, 1, 3n + 1}], {n, 0, 100}]
%o A134163 (Magma) [1 + 12*n + 81*n^3 + n*(105*n+ 81*n^3)/2: n in [0..30]]; // _Vincenzo Librandi_, May 09 2011
%Y A134163 Cf. A000537, A000541, A119617, A134153, A134154, A133180, A134158, A134159, A134160.
%K A134163 nonn,easy
%O A134163 0,2
%A A134163 _Artur Jasinski_, Oct 10 2007

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