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A336683
Sum of 2^k for all residues k found in the Fibonacci sequence mod n.
2
1, 3, 7, 15, 31, 63, 127, 175, 511, 1023, 1327, 4031, 7471, 16383, 32767, 43951, 127807, 238895, 502063, 1048575, 1319215, 2719023, 7798639, 10692015, 33554431, 61209903, 134217727, 259173375, 337649967, 1073741823, 1571892655, 2880154543, 5417565487, 15638470959
OFFSET
1,2
COMMENTS
Row n of A079002 compactified as a binary number.
LINKS
Michael De Vlieger, Plot of bits of a(n) beginning with 2^0 at left for 1 <= n <= 5000.
FORMULA
a(n) = Sum(2^k) for all k in row n of A189768.
a(n) = 2^(n+1) - 1 for n in A079002.
EXAMPLE
a(1) = 1 by convention.
a(2) = 3 = 2^0 + 2^1, since the Fibonacci sequence mod 2 is {0,1,1} repeated, and 0 and 1 appear in the sequence.
a(8) = 175 = 2^0 + 2^1 + 2^2 + 2^3 + 2^5 + 2^7, since the Fibonacci sequence mod 8 is {0,1,1,2,3,5,0,5,5,2,7,1} repeated, and we are missing 4 and 6, leaving the exponents of 2 as shown.
Binary equivalents of first terms:
n a(n) a(n) in binary
--------------------------
1 1 1
2 3 11
3 7 111
4 15 1111
5 31 11111
6 63 111111
7 127 1111111
8 175 10101111
9 511 111111111
10 1023 1111111111
11 1327 10100101111
12 4031 111110111111
13 7471 1110100101111
14 16383 11111111111111
15 32767 111111111111111
16 43951 1010101110101111
...
MATHEMATICA
{1}~Join~Array[Block[{w = {0, 1}}, Do[If[SequenceCount[w, {0, 1}] == 1, AppendTo[w, Mod[Total@ w[[-2 ;; -1]], #]], Break[]], {i, 2, Infinity}]; Total[2^Union@ w]] &, 33, 2]
(* Second program: generate the first n terms using the plot in Links *)
With[{n = 34, img = ImageData@ ColorNegate@ Import["https://oeis.org/A336683/a336683.png"]}, Map[FromDigits[#, 2] &@ Drop[#, LengthWhile[#, # == 0 &]] &@ Reverse[IntegerPart[#]] &, img[[1 ;; n]]]] (* Michael De Vlieger, Oct 05 2020 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Michael De Vlieger, Oct 04 2020
STATUS
approved