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A372983
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a(n) = floor((a(n-1)+a(n-2)+n)/(a(n-3)+1)), with a(1)=1, a(2)=2, a(3)=4.
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4
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1, 2, 4, 5, 4, 3, 2, 2, 3, 5, 6, 5, 4, 3, 3, 4, 6, 7, 6, 4, 3, 4, 6, 8, 7, 5, 4, 4, 6, 8, 9, 7, 5, 4, 5, 7, 9, 9, 7, 5, 5, 6, 9, 9, 9, 6, 6, 6, 8, 9, 9, 7, 6, 6, 8, 10, 10, 8, 7, 6, 8, 9, 11, 9, 8, 6, 8, 9, 12, 10, 9, 7, 8, 8, 11, 10, 10, 8, 8, 8, 10, 11, 11
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OFFSET
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1,2
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COMMENTS
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This sequence is unbounded.
Probably a(n) is around sqrt(n). Can this be made precise and proven?
Is there some n > 3 such that a(n) > a(k) + 1 for all k < n? In other words, do the record values increase by 1? (End)
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LINKS
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MATHEMATICA
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a[1] = 1; a[2] = 2; a[3] = 4;
a[n_] := a[n] = Floor[(a[n - 1] + a[n - 2] + n)/(a[n - 3] + 1)];
Table[a[n], {n, 1, 100}]
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PROG
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(PARI) first(n)=if(n<4, return(powers(2, n-1))); my(v=List([1, 2, 4])); for(k=4, n, listput(v, (v[k-1]+v[k-2]+k)\(v[k-3]+1))); Vec(v) \\ Charles R Greathouse IV, Jul 16 2024
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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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STATUS
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approved
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