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A283207
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a(n) = a(floor(n/a(n-1))) + a(floor(n/a(n-2))) with a(1) = a(2) = 2.
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4
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2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 6, 4, 6, 6, 4, 8, 6, 6, 8, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8
(list;
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listen;
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internal format)
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OFFSET
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1,1
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COMMENTS
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For the first 10^6 terms, the maximum value of a(n) is 64 and the values of b(n) = least k such that a(k) = 2*n are 1, 3, 13, 12, 105, 97, 126, 96, 1681, 1552, 1746, 1537, 1734, 1926, 4050, 1536, 53793, 49665, 53890, 49185, 53862, 57024, 55616, 49153, 55488, 55302, 81249, 61446, 83619, 115214, 162000, 49152; note that b(2^n) = 3*2^((n+2)*(n-1)/2) for n = 1 to 5.
This sequence is a_{1,2}(n) where a_{r,s}(n) = a_{r,s}(floor(n/a_{r,s}(n-r))) + a_{r,s}(floor(n/a_{r,s}(n-s))) with a_{r,s}(n) = 2 for n <= s (r < s). - Altug Alkan, Jun 28 2020
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LINKS
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EXAMPLE
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a(5) = 4 because a(5) = a(floor(5/a(4))) + a(floor(5/a(3))) = a(floor(5/4)) + a(floor(5/4)) = a(1) + a(1) = 4.
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MAPLE
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A:= Vector(100):
A[1]:= 2: A[2]:= 2:
for n from 3 to 100 do A[n]:= A[floor(n/A[n-1])] + A[floor(n/A[n-2])] od:
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MATHEMATICA
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a[1] = a[2] = 2; a[n_] := a[n] = a[Floor[n/a[n - 1]]] + a[Floor[n/a[n - 2]]]; Array[a, 120] (* Michael De Vlieger, Mar 06 2017 *)
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PROG
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(PARI) a=vector(100); a[1]=a[2]=2; for(n=3, #a, a[n]=a[n\a[n-1]]+a[n\a[n-2]]); a
(Python)
from functools import lru_cache
@lru_cache(maxsize=None)
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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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