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A384733 a(n) = 5*binomial(n,6) + 2*binomial(n,4).

%I A384733 #13 Jun 11 2025 11:33:36
%S A384733 0,0,0,0,2,10,35,105,280,672,1470,2970,5610,10010,17017,27755,43680,
%T A384733 66640,98940,143412,203490,283290,387695,522445,694232,910800,1181050,
%U A384733 1515150,1924650,2422602,3023685,3744335,4602880,5619680,6817272,8220520,9856770,11756010
%N A384733 a(n) = 5*binomial(n,6) + 2*binomial(n,4).
%C A384733 6*a(n) is the number of words of length n defined on 5 letters that have exactly one a, exactly one b, and exactly two c's and no d's or exactly four d's and no c's (see examples).
%C A384733 3*a(n) is the number of words of length n defined on 4 letters that have exactly two a's and exactly two b's and no c's or exactly two a's and exactly four c's and no b's. For example, 3*a(6) = 3*35 = 105 since the words are the 90 permutations of aabbdd and the 15 permutations of aacccc.
%H A384733 Paolo Xausa, <a href="/A384733/b384733.txt">Table of n, a(n) for n = 0..10000</a>
%H A384733 <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).
%F A384733 a(n) = -(4/3)*n + (203/72)*n^2 - (33/16)*n^3 + (97/144)*n^4 - (5/48)*n^5 + (1/144)*n^6.
%F A384733 G.f.: (7*x^6 - 4*x^5 + 2*x^4)/(1-x)^7.
%F A384733 E.g.f.: exp(x)*(x^4/12 + x^6/144).
%F A384733 a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n > 6. - _Chai Wah Wu_, Jun 11 2025
%e A384733 6*a(4) = 6*2 = 12 since the words are the 12 permutations of abcc.
%e A384733 6*a(7) = 6*105 = 630 since the words are the 420 permutations of abcceee and the 210 permutations of abdddde.
%t A384733 A384733[n_] := 5*Binomial[n, 6] + 2*Binomial[n, 4];
%t A384733 Array[A384733, 50, 0] (* _Paolo Xausa_, Jun 11 2025 *)
%o A384733 (Python)
%o A384733 def A384733(n): return n*(n*(n*(n*(n*(n-15)+97)-297)+406)-192)//144 # _Chai Wah Wu_, Jun 11 2025
%Y A384733 Cf. A384686.
%K A384733 nonn
%O A384733 0,5
%A A384733 _Enrique Navarrete_, Jun 08 2025