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5 times heptagonal numbers: a(n) = 5*n*(5*n-3)/2.
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%I #21 Oct 05 2024 14:38:42

%S 0,5,35,90,170,275,405,560,740,945,1175,1430,1710,2015,2345,2700,3080,

%T 3485,3915,4370,4850,5355,5885,6440,7020,7625,8255,8910,9590,10295,

%U 11025,11780,12560,13365,14195,15050,15930,16835,17765

%N 5 times heptagonal numbers: a(n) = 5*n*(5*n-3)/2.

%H G. C. Greubel, <a href="/A153785/b153785.txt">Table of n, a(n) for n = 0..1000</a>

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

%F a(n) = (25*n^2 - 15*n)/2 = A000566(n)*5.

%F a(n) = 25*n + a(n-1) - 20 (with a(0)=0). - _Vincenzo Librandi_, Aug 03 2010

%F From _G. C. Greubel_, Aug 28 2016: (Start)

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

%F G.f.: 5*x*(1 + 4*x)/(1 - x)^3.

%F E.g.f.: (5/2)*x*(2 + 5*x)*exp(x). (End)

%t s=0;lst={s};Do[s+=n;AppendTo[lst,s],{n,5,8!,25}];lst (* _Vladimir Joseph Stephan Orlovsky_, Apr 03 2009 *)

%t Table[5*n*(5*n - 3)/2, {n,0,25}] (* or *) LinearRecurrence[{3,-3,1}, {0,5,35}, 25] (* _G. C. Greubel_, Aug 28 2016 *)

%o (PARI) a(n) = 5*n*(5*n-3)/2; \\ _Michel Marcus_, Aug 28 2016

%Y Cf. A000566, A153784, A154786.

%K nonn,easy

%O 0,2

%A _Omar E. Pol_, Jan 07 2009