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A088207
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a(n) = Sum_{k=0..n} floor(k*phi^2) where phi=(1+sqrt(5))/2.
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1
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0, 2, 7, 14, 24, 37, 52, 70, 90, 113, 139, 167, 198, 232, 268, 307, 348, 392, 439, 488, 540, 594, 651, 711, 773, 838, 906, 976, 1049, 1124, 1202, 1283, 1366, 1452, 1541, 1632, 1726, 1822, 1921, 2023, 2127, 2234, 2343, 2455, 2570, 2687, 2807, 2930, 3055, 3183
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OFFSET
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0,2
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COMMENTS
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A001950 is the upper Beatty sequence for the constant phi^2, where phi = (1 + sqrt(5))/2 and the sequence is generated by floor(n*phi). A054347 = partial sums of the lower Beatty sequence (A000201).
Conjecture: a(n)/A054347(n) tends to phi. Example: a(28)/A054347(28) = 1049/643 = 1.6314...
Proof of Adamson's conjecture. We know that lim_{n->oo} A054347(n)/(n*(n+1)) = phi/2 (see A054347).
Using that floor(k*phi^2) = floor(k*phi)+k, for k=1,...,n, we obtain a(n)/A054347(n) = (A054347(n)+(n*(n+1)/2))/((n*(n+1))))/(A054347(n)/(n*(n+1)/2) [Ambiguous, unbalanced parens - Editors of OEIS] -> (phi/2 + 1/2)/(phi/2), which equals phi.
(End)
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LINKS
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FORMULA
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a(n) = Sum_{k=1..n} floor(k*phi^2).
a(n) = floor((n*(n+1)/2)*phi^2 - n/2) + (0 or 1). - Benoit Cloitre, Sep 27 2003
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EXAMPLE
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A001950(1) = 2, then 5, 7, 10, 13, ...; partial sums are 2, 7, 14, 24, 37, ...
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MATHEMATICA
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a[0] = 0; a[n_] := a[n] = (a[n - 1] + Floor[n*(1 + Sqrt[5])^2/4]); Table[ a[n], {n, 1, 50}] (* Robert G. Wilson v, Sep 27 2003 *)
Accumulate[Floor[GoldenRatio^2 Range[0, 50]]] (* Harvey P. Dale, Aug 11 2021 *)
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PROG
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(Python)
from math import isqrt
from itertools import islice, count, accumulate
def A088207_gen(): # generator of terms
return accumulate((n+isqrt(5*n**2)>>1)+n for n in count(0))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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