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A088207
a(n) = Sum_{k=0..n} floor(k*phi^2) where phi=(1+sqrt(5))/2.
1
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
OFFSET
0,2
COMMENTS
Partial sums of A001950.
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...
From Michel Dekking, Aug 19 2019: (Start)
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)
FORMULA
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
EXAMPLE
A001950(1) = 2, then 5, 7, 10, 13, ...; partial sums are 2, 7, 14, 24, 37, ...
MATHEMATICA
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 *)
PROG
(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))
A088207_list = list(islice(A088207_gen(), 10)) # Chai Wah Wu, Aug 29 2022
CROSSREFS
Cf. A001622 (phi), A001950, A054347, A000201, A000217 (triangular numbers).
Sequence in context: A051640 A119354 A249547 * A194111 A343859 A102999
KEYWORD
nonn
AUTHOR
Gary W. Adamson, Sep 23 2003
EXTENSIONS
More terms from Robert G. Wilson v and Benoit Cloitre, Sep 27 2003
STATUS
approved