A192129 a(n) is difference of indices of terms >= 9 in A187824.

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

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

Table of n, a(n) for n=1..66.

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

Adjacent sequences: A192126 A192127 A192128 * A192130 A192131 A192132

Keyword

nonn

Author

Kival Ngaokrajang, Dec 31 2012

Status

approved