A336685 Sum of 2^k for residue k in among Lucas numbers mod n.

1, 3, 7, 15, 30, 63, 127, 190, 511, 990, 1183, 3582, 8190, 16383, 18590, 47806, 131070, 247967, 298911, 854686, 1453502, 2423967, 8362495, 10366142, 31738014, 67100670, 134217727, 262073758, 302254239, 609175710, 1779923167, 3133061822, 4962151582, 16855148990

Offset: 1

Author

Michael De Vlieger, Oct 07 2020

Keywords

• nonn
• easy

Comments

Row n of A336684 compactified as a binary number.

a(n) contains even numbers whereas A336683 (pertaining to the Fibonacci sequence) is strictly odd, since 0 is a Fibonacci number but not a Lucas number.

Examples

			a(1) = 1 by convention.
a(2) = 3 = 2^0 + 2^1, since the Lucas sequence contains both even and odd numbers.
a(5) = 30 = 2^1 + 2^2 + 2^3 + 2^4, since the Lucas numbers mod 5 is {2,1,3,4,2,1} repeated, and we are missing 0, 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     30             11110
    6     63            111111
    7    127           1111111
    8    190          10111110
    9    511         111111111
   10    990        1111011110
   11   1183       10010011111
   12   3582      110111111110
   13   8190     1111111111110
   14  16383    11111111111111
   15  18590   100100010011110
   16  47806  1011101010111110
   ...
		

Crossrefs

Cf. A000032, A066981, A106291, A223487, A336684. Analogous to A336683.

Programs

Mathematica

Total /@ {Most@ #, #} &[2^Range[0, 1]]~Join~Array[Block[{w = {2, 1}}, Do[If[SequenceCount[w, {2, 1}] == 1, AppendTo[w, Mod[Total@ w[[-2 ;; -1]], #]], Break[]], {i, 2, Infinity}]; Total[2^Union@ w]] &, 32, 3]

Formula

a(3^j) = 2^(3^j+1) - 1 for all j.

A066981(n) = binary weight of a(n).

A223487(n) = n - A066981(n) = number of zeros in the binary expansion of a(n).

a(m) = 2^(m+1) - 1 for m = A224482(n).