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A297826
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Solution (a(n)) of the near-complementary equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 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.
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17
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1, 2, 7, 9, 11, 15, 18, 21, 22, 24, 28, 29, 33, 34, 40, 42, 43, 45, 51, 51, 53, 59, 59, 61, 63, 65, 69, 74, 76, 77, 79, 81, 83, 87, 90, 91, 93, 95, 97, 101, 104, 107, 110, 111, 113, 117, 118, 120, 122, 126, 131, 133, 136, 139, 140, 142, 146, 147, 153, 155
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OFFSET
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0,2
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COMMENTS
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The sequence (a(n)) generated by the equation a(n) = a(1)*b(n-1) - a(0)*b(n-2) + 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) 0 <= a(k) - a(k-1) <= 6 for k>=1;
(2) if d is in {0,1,2,3,4,5,6}, then a(k) = a(k-1) + d for infinitely many k.
***
See A297830 for a guide to related sequences.
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LINKS
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Clark Kimberling, Table of n, a(n) for n = 0..10000
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EXAMPLE
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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, ...)
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MATHEMATICA
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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 *)
(* Peter J. C. Moses, Jan 03 2017 *)
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CROSSREFS
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Cf. A297997, A297830.
Sequence in context: A022424 A360944 A136498 * A288598 A277737 A082371
Adjacent sequences: A297823 A297824 A297825 * A297827 A297828 A297829
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KEYWORD
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nonn,easy
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AUTHOR
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Clark Kimberling, Feb 04 2018
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STATUS
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approved
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