A336683 Sum of 2^k for all residues k found in the Fibonacci sequence mod n.
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
Examples
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
...
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..3322
- Michael De Vlieger, Plot of bits of a(n) beginning with 2^0 at left for 1 <= n <= 5000.
Programs
-
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 *)
Comments