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%I #29 Sep 04 2022 04:11:37
%S 49,147,294,490,735,1029,1372,1764,2205,2695,3234,3822,4459,5145,5880,
%T 6664,7497,8379,9310,10290,11319,12397,13524,14700,15925,17199,18522,
%U 19894,21315,22785,24304,25872,27489,29155,30870,32634,34447,36309,38220,40180
%N a(n) = 49*(n-1)*(n-2)/2.
%H Vincenzo Librandi, <a href="/A027469/b027469.txt">Table of n, a(n) for n = 3..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F Numerators of sequence a[ n, n-2 ] in (a[ i, j ])^3 where a[ i, j ] = binomial(i-1, j-1)/2^(i-1) if j <= i, 0 if j > i.
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(3)=49, a(4)=147, a(5)=294. - _Harvey P. Dale_, Aug 24 2011
%F G.f.: 49*x^3/(1-x)^3. - _Harvey P. Dale_, Aug 24 2011
%F From _Amiram Eldar_, Sep 04 2022: (Start)
%F a(n) = A162942(n-2).
%F Sum_{n>=3} 1/a(n) = 2/49.
%F Sum_{n>=3} (-1)^(n+1)/a(n) = 2*(2*log(2)-1)/49. (End)
%t Table[49(n-1)(n-2)/2,{n,3,70}] (* or *) LinearRecurrence[{3,-3,1},{49,147,294},70] (* _Harvey P. Dale_, Aug 24 2011 *)
%o (Magma) [49*(n-1)*(n-2)/2: n in [3..50]]; // _Vincenzo Librandi_, Aug 25 2011
%o (PARI) a(n)=49*(n-1)*(n-2)/2 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Third diagonal of A027466.
%Y Cf. A162942.
%K nonn,easy
%O 3,1
%A _Olivier Gérard_
%E More terms from _Harvey P. Dale_, Aug 24 2011
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