login
a(n) = n*(9*n+2).
4

%I #29 Nov 03 2024 02:15:19

%S 0,11,40,87,152,235,336,455,592,747,920,1111,1320,1547,1792,2055,2336,

%T 2635,2952,3287,3640,4011,4400,4807,5232,5675,6136,6615,7112,7627,

%U 8160,8711,9280,9867,10472,11095,11736,12395,13072,13767,14480,15211,15960

%N a(n) = n*(9*n+2).

%C For n >= 1, the continued fraction expansion of sqrt(4*a(n)) is [6n; {1, 1, 1, 3n-1, 1, 1, 1, 12n}]. - _Magus K. Chu_, Sep 17 2022

%H Reply to V. Librandi, <a href="http://list.seqfan.eu/oldermail/seqfan/2009-March/001017.html">A147296 (SeqFan list)</a> - _M. F. Hasler_, Mar 01 2009

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 1).

%F a(n) = n*(9*n + 2), as conjectured by V. Librandi. - _M. F. Hasler_, Mar 01 2009

%F G.f.: x*(11+7*x)/(1-x)^3. - Jaume Oliver Lafont, Aug 30 2009

%F a(n) = floor((3*n + 1/3)^2). - _Reinhard Zumkeller_, Apr 14 2010

%t Table[n(9n+2),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,11,40},50] (* _Harvey P. Dale_, Dec 19 2014 *)

%o (PARI) A147296(n) = n*(9*n + 2) \\ _M. F. Hasler_, Mar 01 2009

%Y Equals first 9-fold decimation of A144454.

%Y Cf. A010701, A017173, A185019.

%K nonn,easy

%O 0,2

%A _Paul Curtz_, Nov 05 2008

%E More terms from _M. F. Hasler_, Mar 01 2009