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A192129 a(n) is difference of indices of terms >= 9 in A187824. 3
16, 90, 55, 233, 55, 1, 54, 16, 56, 34, 55, 1, 70, 90, 128, 70, 56, 34, 55, 1, 70, 1, 55, 34, 56, 70, 128, 90, 70, 1, 55, 34, 56, 16, 54, 1, 55, 233, 55, 90, 16, 54, 1, 1, 54, 16, 90, 55, 233, 55, 1, 54, 16, 56, 34, 55, 1, 70, 90, 128, 70, 56, 34, 55, 1, 70 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
The repeat cycle contains 45 numbers which have 22 Fibonacci numbers symmetrically distributed as below.
1 F(1), F(2)
54
16
90
55 F(10)
233 F(13)
55 F(10)
1 F(1), F(2)
54
16
56
34 F(9)
55 F(10)
1 F(1), F(2)
70
90
128
70
56
34 F(9)
55 F(10)
1 F(1), F(2)
70
1 F(1), F(2)
55 F(10)
34 F(9)
56
70
128
90
70
1 F(1), F(2)
55 F(10)
34 F(9)
56
16
54
1 F(1), F(2)
55 F(10)
233 F(13)
55 F(10)
90
16
54
1 F(1), F(2)
LINKS
EXAMPLE
For A187824(m(i)) >= 9:
m(1) = 55, m(2) = 71, m(3) = 161, m(4) = 216,
...
a(1) = 71 - 55 = 16, a(2) = 161 - 71 = 90,
a(3) = 216 - 161 = 55,
...
MATHEMATICA
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 *)
PROG
(Small Basic)
j = 1
For n = 1 To 2000
For m = 1 To 100
I = Math.Round(n/m)
D = Math.Abs(n-I*m)
If D > 1 Then
a = m - 1
Goto OUT
Else
EndIf
EndFor
OUT:
If a >= 9 Then 'Filter for A187824 >= 9
mm[j] = n
j = j + 1
Else
EndIf
EndFor
For ii = 1 to j
aa[ii] = mm[ii+1] - mm[ii]
TextWindow.Write(ii+" ")
TextWindow.Write(aa[ii]+" ")
TextWindow.WriteLine(" ")
EndFor
CROSSREFS
Cf. A187824.
Sequence in context: A269078 A240262 A264531 * A253131 A119771 A055920
KEYWORD
nonn
AUTHOR
Kival Ngaokrajang, Dec 31 2012
STATUS
approved

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Last modified February 29 23:21 EST 2024. Contains 370428 sequences. (Running on oeis4.)