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a(n) = n^3 + 114 * n.
0

%I #6 Oct 18 2022 15:30:30

%S 115,236,369,520,695,900,1141,1424,1755,2140,2585,3096,3679,4340,5085,

%T 5920,6851,7884,9025,10280,11655,13156,14789,16560,18475,20540,22761,

%U 25144,27695,30420,33325,36416,39699,43180,46865,50760,54871,59204

%N a(n) = n^3 + 114 * n.

%H Paul Cooijmans, <a href="http://web.archive.org/web/20050302174449/http://members.chello.nl/p.cooijmans/gliaweb/tests/odds.html">Odds</a>.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(1)=115, a(2)=236, a(3)=369, a(4)=520, a(n)=4*a(n-1)-6*a(n-2)+ 4*a(n-3)- a(n-4). - _Harvey P. Dale_, Mar 14 2015

%t Table[n^3+114n,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{115,236,369,520},40] (* _Harvey P. Dale_, Mar 14 2015 *)

%o (PARI) for(n=1,50,print1(n^3+114*n,","))

%K nonn,easy

%O 1,1

%A Herman Jamke (hermanjamke(AT)fastmail.fm), Sep 20 2006

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