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a(n) = ceiling((n+1)^4/2).
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%I #33 Sep 08 2022 08:45:50

%S 1,8,41,128,313,648,1201,2048,3281,5000,7321,10368,14281,19208,25313,

%T 32768,41761,52488,65161,80000,97241,117128,139921,165888,195313,

%U 228488,265721,307328,353641,405000,461761,524288,592961,668168,750313,839808

%N a(n) = ceiling((n+1)^4/2).

%C Number of compositions of even natural numbers into 4 parts <=n.

%C Number of ways of placings of an even number of indistinguishable objects into 4 distinguishable boxes with the condition that in each box there can be at most n objects.

%H Vincenzo Librandi, <a href="/A171714/b171714.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = 1/2*((n + 1)^4 + ((1 + (-1)^n)*1/2)^4).

%F a(n) = +4*a(n-1) -5*a(n-2) +5*a(n-4) -4*a(n-5) +1*a(n-6).

%F G.f.: (1 + 4*x + 14*x^2 + 4*x^3 + x^4)/((1 + x)*(1 - x)^5).

%F a(n) = (n+1)^4 - floor((n+1)^4/2). - _Bruno Berselli_, Jan 18 2017

%e a(1)=8: there are 8 compositions of even natural numbers into 4 parts <=1

%e (0,0,0,0);

%e (0,0,1,1), (0,1,0,1), (0,1,1,0), (1,0,0,1), (1,0,1,0), (1,1,0,0);

%e (1,1,1,1).

%e a(2)=41: there are 41 compositions of even natural numbers into 4 parts <=2

%e for 0: (0,0,0,0);

%e for 2: (0,0,0,2), (0,0,2,0), (0,2,0,0), (2,0,0,0), (0,0,1,1), (0,1,0,1), (0,1,1,0), (1,0,0,1), (1,0,1,0), (1,1,0,0);

%e for 4: (0,0,2,2), (0,2,0,2), (0,2,2,0), (2,0,0,2), (2,0,2,0), (2,2,0,0), (0,1,1,2), (0,1,2,1), (0,2,1,1), (1,0,1,2), (1,0,2,1), (1,1,0,2), (1,1,2,0), (1,2,0,1), (1,2,1,0), (2,0,1,1), (2,1,0,1), (2,1,1,0), (1,1,1,1);

%e for 6: (0,2,2,2), (2,0,2,2), (2,2,0,2), (2,2,2,0), (1,1,2,2), (1,2,1,2), (1,2,2,1), (2,1,1,2), (2,1,2,1), (2,2,1,1);

%e for 8: (2,2,2,2).

%t Table[1/2((n + 1)^4 + ((1 + (-1)^n)*1/2)^4), {n, 0, 25}]

%t Ceiling[Range[40]^4/2] (* _Bruno Berselli_, Jan 18 2017 *)

%o (Magma) [1/2*((n+1)^4+((1+(-1)^n)*1/2)^4): n in [0..40]]; // _Vincenzo Librandi_, Jun 16 2011

%o (PARI) a(n) = ceil(n^4/2); \\ _Michel Marcus_, Dec 14 2013

%Y Cf. A175110, A191903.

%K nonn,easy

%O 0,2

%A _Adi Dani_, May 29 2011

%E Better name from _Enrique Pérez Herrero_, Dec 14 2013