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Sum of the smallest parts of the partitions of 4n into 4 parts.
16

%I #61 Jun 19 2024 22:07:47

%S 1,6,21,55,119,227,396,645,996,1474,2106,2922,3955,5240,6815,8721,

%T 11001,13701,16870,20559,24822,29716,35300,41636,48789,56826,65817,

%U 75835,86955,99255,112816,127721,144056,161910,181374,202542,225511,250380,277251,306229

%N Sum of the smallest parts of the partitions of 4n into 4 parts.

%C Partial sums of A238340. - _Wesley Ivan Hurt_, May 27 2014

%H Vincenzo Librandi, <a href="/A238702/b238702.txt">Table of n, a(n) for n = 1..100</a>

%H Antonio Osorio, <a href="https://mpra.ub.uni-muenchen.de/id/eprint/56690">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_07">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,5,-5,6,-4,1).

%F G.f.: -x*(x+1)*(2*x^2+x+1) / ((x-1)^5*(x^2+x+1)). - _Colin Barker_, Mar 10 2014

%F a(n) = (1/9)*n^4 + (1/3)*n^3 + (5/18)*n^2 + (1/6)*n + O(1). - _Ralf Stephan_, May 29 2014

%F a(n) = Sum_{i=1..n} A238340(i). - _Wesley Ivan Hurt_, May 29 2014

%F a(n) = (1/4) * Sum_{i=1..n} A238328(i)/i. - _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)). Then a(1) = 1, with a(n) = b(n)/(4n) + a(n-1), for n>1. - _Wesley Ivan Hurt_, Jun 27 2014

%F E.g.f.: (exp(x)*(4 + 3*x*(16 + x*(37 + 2*x*(9 + x)))) - 4*exp(-x/2)*cos(sqrt(3)*x/2))/54. - _Stefano Spezia_, Feb 09 2023

%F a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - 5*a(n-4) + 6*a(n-5) - 4*a(n-6) + a(n-7). - _Wesley Ivan Hurt_, Jun 19 2024

%e Add the numbers in the last column for a(n):

%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 6 21 55 .. a(n)

%t CoefficientList[Series[(x + 1)*(2*x^2 + x + 1)/((1 - x)^5*(x^2 + x + 1)), {x, 0, 50}], x] (* _Wesley Ivan Hurt_, Jun 27 2014 *)

%t LinearRecurrence[{4, -6, 5, -5, 6, -4, 1}, {1, 6, 21, 55, 119, 227, 396}, 50] (* _Vincenzo Librandi_, Aug 29 2015 *)

%o (PARI) Vec(-x*(x+1)*(2*x^2+x+1)/((x-1)^5*(x^2+x+1)) + O(x^100)) \\ _Colin Barker_, Mar 23 2014

%Y Cf. A238328, A238340.

%K nonn,easy

%O 1,2

%A _Wesley Ivan Hurt_ and _Antonio Osorio_, Mar 03 2014