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%I #18 Oct 22 2013 12:42:48
%S 1,3,2,6,7,9,12,16,17,23,26,30,35,41,44,52,57,63,70,78,83,93,100,108,
%T 117,127,134,146,155,165,176,188,197,211,222,234,247,261,272,288,301,
%U 315,330,346,359,377,392,408,425,443
%N Coins left after packing 5 curves coins patterns into fountain of coins base n.
%C Refer to arrangement same as A005169: "A fountain is formed by starting with a row of coins, then stacking additional coins on top so that each new coin touches two in the previous row". The 5 curves coins patterns consist of a part of circumference and forms continuous area. There is total 13 distinct patterns. I would like to call "5C4S" type as it cover 4 coins and symmetry. When packing 5C4S into fountain of coins base n, the total number of 5C4S is A001399, the coins left is a(n) and void is A230276. See illustration in links.
%H Kival Ngaokrajang, <a href="/A230267/a230267.pdf">Illustration of initial terms (U)</a>
%F G.f.: x*(x^3 - 2*x^2 + 2*x + 1)/((1-x)*(1-x^2)*(1-x^3)) (conjectured). - _Ralf Stephan_, Oct 17 2013
%o (Small Basic)
%o a[1]=1
%o d[2]=2
%o For n = 1 To 100
%o If n+1 >= 3 Then
%o If Math.Remainder(n+1,3)=math.Remainder(n+1,6) Then
%o d2=1
%o Else
%o d2=Math.Remainder(n+1,3)+math.Remainder(n+1,6)*Math.Power(-1,math.Remainder(n+1,2))
%o EndIf
%o d[n+1]=d[n]+d2
%o EndIf
%o a[n+1]=a[n]+d[n+1]
%o TextWindow.Write(a[n]+", ")
%o EndFor
%Y Cf. A008795 (3-curves coins patterns), A074148, A229093, A229154 (4-curves coins patterns), A001399 (5-curves coins patterns), A229593 (6-curves coins patterns).
%K nonn
%O 1,2
%A _Kival Ngaokrajang_, Oct 15 2013
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