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%I #49 Jan 25 2013 16:25:21
%S 16,90,55,233,55,1,54,16,56,34,55,1,70,90,128,70,56,34,55,1,70,1,55,
%T 34,56,70,128,90,70,1,55,34,56,16,54,1,55,233,55,90,16,54,1,1,54,16,
%U 90,55,233,55,1,54,16,56,34,55,1,70,90,128,70,56,34,55,1,70
%N a(n) is difference of indices of terms >= 9 in A187824.
%C The repeat cycle contains 45 numbers which have 22 Fibonacci numbers symmetrically distributed as below.
%C 1 F(1), F(2)
%C 54
%C 16
%C 90
%C 55 F(10)
%C 233 F(13)
%C 55 F(10)
%C 1 F(1), F(2)
%C 54
%C 16
%C 56
%C 34 F(9)
%C 55 F(10)
%C 1 F(1), F(2)
%C 70
%C 90
%C 128
%C 70
%C 56
%C 34 F(9)
%C 55 F(10)
%C 1 F(1), F(2)
%C 70
%C 1 F(1), F(2)
%C 55 F(10)
%C 34 F(9)
%C 56
%C 70
%C 128
%C 90
%C 70
%C 1 F(1), F(2)
%C 55 F(10)
%C 34 F(9)
%C 56
%C 16
%C 54
%C 1 F(1), F(2)
%C 55 F(10)
%C 233 F(13)
%C 55 F(10)
%C 90
%C 16
%C 54
%C 1 F(1), F(2)
%e For A187824(m(i)) >= 9:
%e m(1) = 55, m(2) = 71, m(3) = 161, m(4) = 216,
%e ...
%e a(1) = 71 - 55 = 16, a(2) = 161 - 71 = 90,
%e a(3) = 216 - 161 = 55,
%e ...
%t A187824[n_ /; n > 1] := Catch[For[k = 4, True, k++, m = Mod[n, k, -Floor[k/2]]; If[m != m^3, Throw[k - 1]]]]; Differences[Select[Range[2, 6000], A187824[#] >= 9 &] ] (* _Jean-François Alcover_, Jan 09 2013 *)
%o (Small Basic)
%o j = 1
%o For n = 1 To 2000
%o For m = 1 To 100
%o I = Math.Round(n/m)
%o D = Math.Abs(n-I*m)
%o If D > 1 Then
%o a = m - 1
%o Goto OUT
%o Else
%o EndIf
%o EndFor
%o OUT:
%o If a >= 9 Then 'Filter for A187824 >= 9
%o mm[j] = n
%o j = j + 1
%o Else
%o EndIf
%o EndFor
%o For ii = 1 to j
%o aa[ii] = mm[ii+1] - mm[ii]
%o TextWindow.Write(ii+" ")
%o TextWindow.Write(aa[ii]+" ")
%o TextWindow.WriteLine(" ")
%o EndFor
%Y Cf. A187824.
%K nonn
%O 1,1
%A _Kival Ngaokrajang_, Dec 31 2012
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