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A078098
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Let u(1)=u(2)=1, u(3)=2n+1, u(k)=abs(u(k-1)-u(k-2)-u(k-3)); then for any n (u(k),u(k+1)) = (v(n),w(n)) for k large enough; sequence gives values of Max(v(n),w(n)).
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2
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3, 7, 11, 13, 21, 29, 39, 39, 49, 69, 67, 69, 69, 79, 83, 87, 81, 101, 111, 115, 133, 141, 139, 151, 187, 157, 191, 187, 199, 213, 223, 211, 221, 241, 255, 275, 309, 293, 287, 279, 295, 293, 303, 283, 325, 345, 357, 367, 403, 393, 419, 419, 477, 457, 519, 487
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OFFSET
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1,1
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COMMENTS
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a(n) is necessarily odd. Starting with u(1)=u(2)=1 u(3)=2n then u(k) seems unbounded and there seems to be 2 integer values x(n) y(n) such that for any m>x(n), Max( u(k) : 1<=k<=m) = sqrtint(m+y(n))
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LINKS
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FORMULA
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Conjecture : a(n)/n is bounded
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EXAMPLE
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Map of 2*2+1=5 under u(k) is : 1->1->5 ->3->3->5->1->7->1->7>->1->7->1....Hence a(2)=Max(1,7)=7
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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