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a(n) = (n*(n+1))^3.
5

%I #38 Aug 28 2024 02:57:13

%S 0,8,216,1728,8000,27000,74088,175616,373248,729000,1331000,2299968,

%T 3796416,6028568,9261000,13824000,20123648,28652616,40001688,54872000,

%U 74088000,98611128,129554216,168196608,216000000,274625000,345948408

%N a(n) = (n*(n+1))^3.

%H Harry J. Smith, <a href="/A060459/b060459.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).

%F a(n) = A002378(n)^3. - _Zerinvary Lajos_, Apr 11 2006

%F G.f.: 8*x*(1 + 20*x + 48*x^2 + 20*x^3 + x^4)/(1-x)^7. - _Colin Barker_, Jan 29 2012

%F Sum_{n>=1} 1/a(n) = 10 - Pi^2. - _Vaclav Kotesovec_, Feb 14 2015

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 3*zeta(3)/2 + 12*log(2) - 10. - _Amiram Eldar_, Jul 15 2020

%F E.g.f.: exp(x)*x*(8 + 100*x + 184*x^2 + 98*x^3 + 18*x^4 +x^5). - _Stefano Spezia_, Jul 01 2024

%e (0)^3, (1+1)^3, (2+2+2)^3, (3+3+3+3)^3, (4+4+4+4+4)^3, (5+5+5+5+5+5)^3, ...

%p seq((n*(n+1))^3, n=0..66);

%t Table[(n*(n+1))^3, {n, 0, 50}] (* _Paolo Xausa_, Aug 28 2024 *)

%o (PARI) { for (n=0, 1000, write("b060459.txt", n, " ", (n*(n + 1))^3); ) } \\ _Harry J. Smith_, Jul 05 2009

%Y Cf. A002378.

%K easy,nonn

%O 0,2

%A _Jason Earls_, Apr 09 2001