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A304799
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Solution (a(n)) of the complementary equation a(n) = b(n) + b(2n); see Comments.
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19
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2, 7, 10, 14, 18, 23, 26, 31, 34, 38, 43, 46, 50, 55, 59, 62, 66, 71, 74, 79, 82, 86, 90, 95, 98, 103, 106, 110, 115, 118, 122, 126, 131, 134, 139, 142, 146, 151, 154, 158, 162, 167, 170, 174, 179, 182, 187, 191, 194, 199, 203, 206, 210, 215, 218, 223, 226
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OFFSET
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0,1
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COMMENTS
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Define complementary sequences a(n) and b(n) recursively: both are strictly increasing, b(0) = 1, and a(n) = b(n) + b(2n) for n >= 0. Empirically,
(1) {a(n) - 4*n: n >= 0} = {2,3} and {3*b(n) - 4*n: n >= 0} = {2,3,4,5}.
(2) If the equation for a(n) is generalized to a(n) = b(h*n) + b(k*n), where 1 <= h < k, then {a(n) - (h + k + 1)*n: n >= 0} = {2,3} and {(h + k)*b(n) - (h + k + 1)*n : n >= 0} = {k + h - 1, k + h, ..., 2*k + 2*h - 1}.
(3) {a(n) - a(n-1): n >= 1) = {h+k, h+k+1, h+k+2}.
(4) {k*b(n)-b(k*n): n >= 0} = {k-2, k-1, ..., 2*k-2}
***
Guide to related sequences:
h k (a(n)) (b(n))
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LINKS
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EXAMPLE
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b(0) = 1, so that a(0) = 2. Since a(1) = b(1) + b(2), we must have a(1) >= 7, so that b(1) = 3, b(2) = 4, b(3) = 5, b(4) = 6, and a(1) = 7.
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MATHEMATICA
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mex[list_, start_] := (NestWhile[# + 1 &, start, MemberQ[list, #] &]);
h = 1; k = 2; a = {}; b = {1};
AppendTo[a, mex[Flatten[{a, b}], 1]];
Do[Do[AppendTo[b, mex[Flatten[{a, b}], Last[b]]], {k}];
AppendTo[a, Last[b] + b[[1 + (Length[b] - 1)/k h]]], {500}];
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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