

A297997


Solution (b(n)) of the nearcomplementary equation a(n) = a(1)*b(n1)  a(0)*b(n2) + n, where a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, and (b(n)) is the increasing sequence of positive integers not in (a(n)). See Comments.


6



3, 4, 5, 6, 8, 10, 12, 13, 14, 16, 17, 19, 20, 23, 25, 26, 27, 30, 31, 32, 35, 36, 37, 38, 39, 41, 44, 46, 47, 48, 49, 50, 52, 54, 55, 56, 57, 58, 60, 62, 64, 66, 67, 68, 70, 71, 72, 73, 75, 78, 80, 82, 84, 85, 86, 88, 89, 92, 94, 96, 98, 99, 100, 102, 103
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OFFSET

0,1


COMMENTS

The sequence (a(n)) generated by the equation a(n) = a(1)*b(n1)  a(0)*b(n2) + n, with initial values as shown, includes duplicates; e.g. a(18) = a(19) = 51. If the duplicates are removed from (a(n)), the resulting sequence and (b(n)) are complementary. Conjectures:
(1) 1 <= b(k)  b(k1) <= 3 for k>=1;
(2) if d is in {1,2,3}, then b(k) = b(k1) + d for infinitely many k.
***
See A297830 for a guide to related sequences.


LINKS



EXAMPLE

a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4, so that a(2) = 7.
Complement: (b(n)) = (3, 4, 5, 6, 8,10,12,13,14,16, ...)


MATHEMATICA

mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
tbl = {}; a[0] = 1; a[1] = 2; b[0] = 3; b[1] = 4;
a[n_] := a[n] = a[1]*b[n  1]  a[0]*b[n  2] + n;
b[n_] := b[n] = mex[tbl = Join[{a[n], a[n  1], b[n  1]}, tbl], b[n  1]];
Table[a[n], {n, 0, 300}] (* A297826 *)
Table[b[n], {n, 0, 300}] (* A297997 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



