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a(n) = n*(5*n^2 - 2)/3.
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%I #46 Sep 08 2022 08:44:33

%S 0,1,12,43,104,205,356,567,848,1209,1660,2211,2872,3653,4564,5615,

%T 6816,8177,9708,11419,13320,15421,17732,20263,23024,26025,29276,32787,

%U 36568,40629,44980,49631,54592

%N a(n) = n*(5*n^2 - 2)/3.

%C 3-dimensional analog of centered polygonal numbers.

%C Also as a(n)=(1/6)*(10*n^3-4*n), n>0: structured pentagonal anti-diamond numbers (vertex structure 11) (Cf. A051673 = alternate vertex A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004

%C a(n+1)-10*a(n) = (n+1)*(5*(n+1)^2-2)/3 - (10n(n+1)(n+2)/6) = n. The unit digits are 0,1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,... . - _Eric Desbiaux_, Aug 18 2008

%D E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

%H Vincenzo Librandi, <a href="/A004466/b004466.txt">Table of n, a(n) for n = 0..5000</a>

%H T. P. Martin, <a href="http://dx.doi.org/10.1016/0370-1573(95)00083-6">Shells of atoms</a>, Phys. Reports, 273 (1996), 199-241, eq. (11).

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

%F G.f.: x*(1+8*x+x^2)/(1-x)^4. - _Colin Barker_, Jan 08 2012

%F E.g.f.: (x/3)*(3 + 15*x + 5*x^2)*exp(x). - _G. C. Greubel_, Sep 01 2017

%p A004466:=n->n*(5*n^2 - 2)/3; seq(A004466(n), n=0..50); # _Wesley Ivan Hurt_, Mar 10 2014

%t Table[n(5n^2-2)/3,{n,0,80}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 18 2011 *)

%o (Magma) [n*(5*n^2-2)/3: n in [0..50]]; // _Vincenzo Librandi_, May 15 2011

%o (PARI) a(n)=n*(5*n^2-2)/3 \\ _Charles R Greathouse IV_, Sep 24 2015

%Y Cf. A062786 (first differences), A264853 (partial sums).

%Y 1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

%K nonn,easy

%O 0,3

%A Albert D. Rich (Albert_Rich(AT)msn.com)