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Centered truncated cube numbers: a(n) = (46*n^3 - 69*n^2 + 29*n - 3)/3.
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%I #9 Oct 26 2025 17:10:08

%S 1,49,235,651,1389,2541,4199,6455,9401,13129,17731,23299,29925,37701,

%T 46719,57071,68849,82145,97051,113659,132061,152349,174615,198951,

%U 225449,254201,285299,318835,354901,393589,434991,479199,526305,576401,629579,685931,745549,808525

%N Centered truncated cube numbers: a(n) = (46*n^3 - 69*n^2 + 29*n - 3)/3.

%D Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing (2012), pages 136-137.

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

%F a(n) = (2*n - 1)*(23*n^2 - 23*n + 3)/3.

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 4.

%F G.f.: x*(1 + 45*x + 45*x^2 + x^3)/(1 - x)^4.

%F E.g.f.: 1 + exp(x)*(46*x^3/3 + 23*x^2 + 2*x - 1).

%t a[n_]:=(46n^3-69n^2+29n-3)/3; Array[a,38]

%Y Partial sums of A005911.

%K nonn,easy

%O 1,2

%A _Stefano Spezia_, Oct 26 2025