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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 (list; graph; refs; listen; history; text; internal format)
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 A054347(n)/(n*(n+1)) tends to phi/2, as n tends to infinity (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) -> (phi/2 + 1/2)/phi/2, which equals phi.

(End)

LINKS

Table of n, a(n) for n=0..49.

FORMULA

a(n) = Sum(1, n) floor(n*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 *)

CROSSREFS

Cf. A001950, A054347, A000201, A000217 (triangular numbers).

Sequence in context: A051640 A119354 A249547 * A194111 A343859 A102999

Adjacent sequences:  A088204 A088205 A088206 * A088208 A088209 A088210

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

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Last modified May 12 09:16 EDT 2021. Contains 343821 sequences. (Running on oeis4.)