%I #36 Sep 08 2022 08:45:51
%S 0,4,22,64,140,260,434,672,984,1380,1870,2464,3172,4004,4970,6080,
%T 7344,8772,10374,12160,14140,16324,18722,21344,24200,27300,30654,
%U 34272,38164,42340,46810,51584,56672,62084,67830,73920,80364,87172,94354,101920,109880
%N a(n) = n*(n+1)*(5*n+1)/3.
%C Also zero followed by bisection (even part) of A088003.
%C Numbers ending in 0, 2 or 4 (cf. 2*A053796(n)). Therefore we can easily see that a(m)^(2*k+1)==-1 (mod 5) only for m in A047219, while a(m)^(2*k)==-1 (mod 5) only for m in A016873 and k odd.
%H Vincenzo Librandi, <a href="/A174814/b174814.txt">Table of n, a(n) for n = 0..2000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: 2*x*(2+3*x)/(1-x)^4.
%F a(n) = 2*A033994(n) for n>0.
%F a(n) = n*A147875(n+1)-sum(k=1..n, A147875(k)) for n>0.
%F a(-n) = -A144945(n).
%t Table[1/3 n (1+n) (1+5 n), {n,0,50}] (* _Harvey P. Dale_, Feb 25 2011 *)
%o (Magma) [n*(n+1)*(5*n+1)/3: n in [0..50]]; // _Vincenzo Librandi_, May 30 2011
%o (PARI) a(n) = n*(n+1)*(5*n+1)/3 \\ _Charles R Greathouse IV_, May 30 2011
%K nonn,easy
%O 0,2
%A _Bruno Berselli_, Dec 01 2010 - Dec 02 2010
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