cp's OEIS Frontend

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

%I A336685 #5 Feb 21 2021 10:56:26
%S A336685 1,3,7,15,30,63,127,190,511,990,1183,3582,8190,16383,18590,47806,
%T A336685 131070,247967,298911,854686,1453502,2423967,8362495,10366142,
%U A336685 31738014,67100670,134217727,262073758,302254239,609175710,1779923167,3133061822,4962151582,16855148990
%N A336685 Sum of 2^k for residue k in among Lucas numbers mod n.
%C A336685 Row n of A336684 compactified as a binary number.
%C A336685 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.
%F A336685 a(3^j) = 2^(3^j+1) - 1 for all j.
%F A336685 A066981(n) = binary weight of a(n).
%F A336685 A223487(n) = n - A066981(n) = number of zeros in the binary expansion of a(n).
%F A336685 a(m) = 2^(m+1) - 1 for m = A224482(n).
%e A336685 a(1) = 1 by convention.
%e A336685 a(2) = 3 = 2^0 + 2^1, since the Lucas sequence contains both even and odd numbers.
%e A336685 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.
%e A336685 Binary equivalents of first terms:
%e A336685    n    a(n)   a(n) in binary
%e A336685    --------------------------
%e A336685     1      1                 1
%e A336685     2      3                11
%e A336685     3      7               111
%e A336685     4     15              1111
%e A336685     5     30             11110
%e A336685     6     63            111111
%e A336685     7    127           1111111
%e A336685     8    190          10111110
%e A336685     9    511         111111111
%e A336685    10    990        1111011110
%e A336685    11   1183       10010011111
%e A336685    12   3582      110111111110
%e A336685    13   8190     1111111111110
%e A336685    14  16383    11111111111111
%e A336685    15  18590   100100010011110
%e A336685    16  47806  1011101010111110
%e A336685    ...
%t A336685 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]
%Y A336685 Cf. A000032, A066981, A106291, A223487, A336684. Analogous to A336683.
%K A336685 nonn,easy
%O A336685 1,2
%A A336685 _Michael De Vlieger_, Oct 07 2020