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%I #46 Feb 26 2022 04:25:07
%S 0,5,25,60,110,175,255,350,460,585,725,880,1050,1235,1435,1650,1880,
%T 2125,2385,2660,2950,3255,3575,3910,4260,4625,5005,5400,5810,6235,
%U 6675,7130,7600,8085,8585,9100,9630,10175,10735,11310,11900,12505,13125,13760,14410
%N 5 times pentagonal numbers: 5*n*(3*n-1)/2.
%C a(n) can be represented as a figurate number using n concentric pentagons (see example). - _Omar E. Pol_, Aug 21 2011
%H Ivan Panchenko, <a href="/A152734/b152734.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) = 5*A000326(n).
%F a(n) = a(n-1)+15*n-10 (with a(0)=0). - _Vincenzo Librandi_, Nov 26 2010
%F G.f.: 5*x*(1+2*x)/(1-x)^3. a(n) = 4*A000217(n)+A051865(n). - _Bruno Berselli_, Feb 11 2011
%F E.g.f.: (5/2)*(3*x^2 + 2*x)*exp(x). - _G. C. Greubel_, Jul 17 2017
%F From _Amiram Eldar_, Feb 26 2022: (Start)
%F Sum_{n>=1} 1/a(n) = (9*log(3) - sqrt(3)*Pi)/15.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(sqrt(3)*Pi- 6*log(2))/15. (End)
%e From _Omar E. Pol_, Aug 22 2011 (Start):
%e Illustration of initial terms as concentric pentagons (in a precise representation the pentagons should be strictly concentric):
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%e . 5 25 60
%e (End)
%p A152734:=n->5*n*(3*n-1)/2: seq(A152734(n), n=0..50); # _Wesley Ivan Hurt_, Sep 19 2014
%t Table[5 n (3 n - 1)/2, {n, 0, 50}] (* _Wesley Ivan Hurt_, Sep 19 2014 *)
%t 5*PolygonalNumber[5,Range[0,50]] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Oct 13 2020 *)
%o (Magma) [5*n*(3*n-1)/2 : n in [0..50]]; // _Wesley Ivan Hurt_, Sep 19 2014
%o (PARI) a(n)=5*n*(3*n-1)/2 \\ _Charles R Greathouse IV_, Sep 24 2015
%Y Cf. A000326, A033581, A085250, A152751, A194275.
%Y Cf. sequences of the form n*(d*n+10-d)/2 indexed in A140090.
%K easy,nonn
%O 0,2
%A _Omar E. Pol_, Dec 11 2008
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