%I #64 Dec 07 2019 12:18:29

%S 1,2,8,12,36,80,175,441,972,2304,5376,12348,29008,67081,156065,363350,

%T 843144,1962396,4560200,10600000,24648975,57288465,133194600,

%U 309636096,719790336,1673379352,3890033728,9043304417,21023197601,48872682810,113615800200,264124052396

%N Volume of each square prism building the next 3-dimensional box in A100538 where side lengths form the Padovan spiral number sequence (A134816), starting with 1 X 1 X 1, 1 X 1 X 2, 2 X 2 X 2, 2 X 2 X 3, 4 X 4 X 5, ...

%H Vincenzo Librandi, <a href="/A285551/b285551.txt">Table of n, a(n) for n = 1..1000</a>

%H Robert Dickau, <a href="http://demonstrations.wolfram.com/PadovansSpiralNumbers/">Padovan's Spiral Numbers</a>, Wolfram Demonstrations Project Published: August 19, 2010.

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,3,-2, 4,-4,-1,-1,0,-1).

%F a(n) = A000931(n+5)^2*A000931(n+6).

%F a(n) = A100538(n+1) - A100538(n).

%t A[n_]:=Sum[Binomial[k, n - 2k], {k, 0, Floor[n/2]}]; a000931[n_]:=If[n==0, 1, If[n<3, 0, A[n - 3]]]; a[n_]:=a000931[n + 5]^2*a000931[n + 6]; Table[a[n], {n, 0, 50}] (* _Indranil Ghosh_, Apr 26 2017 *)

%t LinearRecurrence[{1, 2, 3, -2, 4, -4, -1, -1, 0, -1}, {1, 2, 8, 12, 36, 80, 175, 441, 972, 2304}, 40] (* _Vincenzo Librandi_, Jul 19 2017 *)

%o (PARI) A(n) = sum(k=0, n\2, binomial(k, n - 2*k));

%o a000931(n) = if(n==0, 1, if(n<3, 0, A(n - 3)));

%o a(n) = a000931(n + 5)^2*a000931(n + 6); \\ _Indranil Ghosh_, Apr 26 2017

%o (Python)

%o from sympy import binomial

%o def A(n): return sum([binomial(k, n - 2*k) for k in range(int(n/2) + 1)])

%o def a000931(n): return 1 if n==0 else 0 if n<3 else A(n - 3)

%o def a(n): return a000931(n + 5)**2*a000931(n + 6) # _Indranil Ghosh_, Apr 26 2017

%Y Cf. A000931, A100538, A134816.

%K nonn

%O 1,2

%A _Peter M. Chema_, Apr 25 2017