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Expansion of o.g.f. 1/(1 - 5*x - 10*x^2 - 10*x^3 - 5*x^4 - x^5).
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%I #24 Oct 28 2024 13:40:47

%S 1,5,35,235,1580,10626,71460,480570,3231845,21734235,146163251,

%T 982951365,6610371480,44454906580,298960311840,2010515259876,

%U 13520763292345,90927457083265,611489327404315,4112280377388895,27655184063541876,185981775414350150,1250731895575163300

%N Expansion of o.g.f. 1/(1 - 5*x - 10*x^2 - 10*x^3 - 5*x^4 - x^5).

%C a(n) is the number of generalized compositions of n using parts of size at most 5 where there are binomial(5,i) types of i (see example).

%C The coefficients of 1/(1 - C(k,1)*x - C(k,2)*x^2 - C(k,3)*x^3 - ... - C(k,k)*x^k) give the number of generalized compositions of n using parts of size at most k where there are binomial(k,i) types of i.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,10,10,5,1).

%F a(n) = 5*a(n-1) + 10*a(n-2) + 10*a(n-3) + 5*a(n-4) + a(n-5), n=>5.

%F a(n) = Sum_{k>=0} (1/2)^(k+1) * binomial(5*k,n). - _Seiichi Manyama_, Aug 03 2024

%e The following table gives the type of composition, the number of such compositions, and the total number of compositions of n = 5 using parts of size at most 5 where there are binomial(5,i) types of i (ie. 5 types of 1, 10 types of 2, 10 types of 3, 5 types of 4, and 1 type of 5):

%e Type Number Total

%e 5 1 1

%e 4+1 2 50

%e 3+2 2 200

%e 3+1+1 3 750

%e 2+2+1 3 1500

%e 2+1+1+1 4 5000

%e 1+1+1+1+1+1 1 3125,

%e adding to a(5)=10626.

%t CoefficientList[Series[1/(1-5*x-10*x^2-10*x^3-5*x^4-x^5),{x,0,22}],x] (* _Stefano Spezia_, Jul 09 2024 *)

%Y Cf. A000012, A000129, A108368, A374454.

%Y Cf. A145841.

%K nonn,easy

%O 0,2

%A _Enrique Navarrete_, Jul 08 2024

%E a(20) corrected by _Georg Fischer_, Oct 28 2024