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%I #87 Feb 09 2023 21:59:46
%S 1,5,15,34,64,108,169,249,351,478,632,816,1033,1285,1575,1906,2280,
%T 2700,3169,3689,4263,4894,5584,6336,7153,8037,8991,10018,11120,12300,
%U 13561,14905,16335,17854,19464,21168,22969,24869,26871,28978,31192,33516,35953
%N Number of partitions of 4n into 4 parts.
%C First differences of A238702. - _Wesley Ivan Hurt_, May 27 2014
%C Number of partitions of 4*(n-1) into at most 4 parts. - _Colin Barker_, Apr 01 2015
%H Iain Fox, <a href="/A238340/b238340.txt">Table of n, a(n) for n = 1..10000</a> (first 200 terms from Vincenzo Librandi)
%H Antonio Osorio, <a href="https://ideas.repec.org/p/pra/mprapa/56690.html">A Sequential Allocation Problem: The Asymptotic Distribution of Resources</a>, Munich Personal RePEc Archive, 2014.
%H <a href="/index/Par#part">Index entries for sequences related to partitions</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,2,-3,3,-1).
%F a(n) = A238328(n) / 4n.
%F G.f.: x*(x+1)*(2*x^2+x+1) / ((x-1)^4*(x^2+x+1)). - _Colin Barker_, Mar 10 2014
%F a(n) = 4/9*n^3 + 1/3*n^2 + O(1). - _Ralf Stephan_, May 29 2014
%F a(n) = A238702(n) - A238702(n-1), n>1. - _Wesley Ivan Hurt_, May 29 2014
%F Recurrence: Let b(1) = 4, with b(n) = (n/(n-1)) * b(n-1) + 4n * Sum_{i=0..2n} (floor((4n-2-i)/2)-i) * (floor((sign((floor((4n-2-i)/2)-i)) +2)/2)), for n>1. Then a(n) = b(n)/(4n). - _Wesley Ivan Hurt_, Jun 27 2014
%F Recurrence: (4*n^3 - 21*n^2 + 44*n - 33)*a(n) = 3*(4*n^2 - 10*n + 9)*a(n-1) + 3*(4*n^2 - 10*n + 9)*a(n-2) + (4*n^3 - 9*n^2 + 14*n - 6)*a(n-3). - _Vaclav Kotesovec_, Jul 04 2014
%F a(n) = round(((4n)^3 + 3*(4n)^2)/144). - _Giacomo Guglieri_, Jun 28 2020
%F E.g.f.: exp(-x/2)*(3*exp(3*x/2)*(1 + x*(7 + x*(15 + 4*x))) - 3*cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2))/27. - _Stefano Spezia_, Feb 09 2023
%e Count the partitions of 4*n into 4 parts:
%e 13 + 1 + 1 + 1
%e 12 + 2 + 1 + 1
%e 11 + 3 + 1 + 1
%e 10 + 4 + 1 + 1
%e 9 + 5 + 1 + 1
%e 8 + 6 + 1 + 1
%e 7 + 7 + 1 + 1
%e 11 + 2 + 2 + 1
%e 10 + 3 + 2 + 1
%e 9 + 4 + 2 + 1
%e 8 + 5 + 2 + 1
%e 7 + 6 + 2 + 1
%e 9 + 3 + 3 + 1
%e 8 + 4 + 3 + 1
%e 7 + 5 + 3 + 1
%e 6 + 6 + 3 + 1
%e 7 + 4 + 4 + 1
%e 6 + 5 + 4 + 1
%e 5 + 5 + 5 + 1
%e 9 + 1 + 1 + 1 10 + 2 + 2 + 2
%e 8 + 2 + 1 + 1 9 + 3 + 2 + 2
%e 7 + 3 + 1 + 1 8 + 4 + 2 + 2
%e 6 + 4 + 1 + 1 7 + 5 + 2 + 2
%e 5 + 5 + 1 + 1 6 + 6 + 2 + 2
%e 7 + 2 + 2 + 1 8 + 3 + 3 + 2
%e 6 + 3 + 2 + 1 7 + 4 + 3 + 2
%e 5 + 4 + 2 + 1 6 + 5 + 3 + 2
%e 5 + 3 + 3 + 1 6 + 4 + 4 + 2
%e 4 + 4 + 3 + 1 5 + 5 + 4 + 2
%e 5 + 1 + 1 + 1 6 + 2 + 2 + 2 7 + 3 + 3 + 3
%e 4 + 2 + 1 + 1 5 + 3 + 2 + 2 6 + 4 + 3 + 3
%e 3 + 3 + 1 + 1 4 + 4 + 2 + 2 5 + 5 + 3 + 3
%e 3 + 2 + 2 + 1 4 + 3 + 3 + 2 5 + 4 + 4 + 3
%e 1 + 1 + 1 + 1 2 + 2 + 2 + 2 3 + 3 + 3 + 3 4 + 4 + 4 + 4
%e 4(1) 4(2) 4(3) 4(4) .. 4n
%e ------------------------------------------------------------------------
%e 1 5 15 34 .. a(n)
%t CoefficientList[Series[(x + 1)*(2*x^2 + x + 1)/((x - 1)^4*(x^2 + x + 1)), {x, 0, 50}], x] (* _Wesley Ivan Hurt_, Jun 27 2014 *)
%t Table[2*n/9 + n^2/3 + 4*n^3/9 - Floor[n/3]/3 - Floor[(n+1)/3]/3, {n, 1, 50}] (* _Vaclav Kotesovec_, Jul 04 2014 *)
%t LinearRecurrence[{3, -3, 2, -3, 3, -1}, {1, 5, 15, 34, 64, 108}, 50] (* _Vincenzo Librandi_, Aug 29 2015 *)
%o (PARI) Vec(x*(x+1)*(2*x^2+x+1)/((x-1)^4*(x^2+x+1)) + O(x^100)) \\ _Colin Barker_, Mar 24 2014
%o (PARI) a(n)=n^2*(4*n+3)\/9 \\ _Charles R Greathouse IV_, Jun 29 2020
%o (Magma) I:=[1,5,15,34,64,108]; [n le 6 select I[n] else 3*Self(n-1)-3*Self(n-2)+2*Self(n-3)-3*Self(n-4)+3*Self(n-5)-Self(n-6): n in [1..45]]; // _Vincenzo Librandi_, Aug 29 2015
%Y Cf. A238328, A238702.
%K nonn,easy
%O 1,2
%A _Wesley Ivan Hurt_ and _Antonio Osorio_, Feb 24 2014
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