A255977 The number of numbers j+k*r <= n, where r = golden ratio and j and k are nonnegative integers.

1, 2, 4, 6, 9, 13, 17, 22, 27, 33, 40, 47, 55, 64, 73, 83, 93, 104, 116, 128, 141, 154, 168, 183, 198, 214, 231, 248, 266, 284, 303, 323, 343, 364, 386, 408, 431, 454, 478, 503, 528, 554, 580, 607, 635, 663, 692, 722, 752, 783, 814, 846, 879, 912, 946, 980

Offset: 1

Author

Clark Kimberling, Mar 14 2015

Keywords

• nonn
• easy

Comments

The difference sequence is A019446.

From Thomas Anton, Oct 22 2018: (Start)

It appears that this sequence (apart from the first term) can be obtained through the following sieving process. Start with the positive integers. Then, at each stage, circle the first remaining number that has not already been circled, and delete all terms in the subsequence of terms that were not circled in previous stages with circled indices that have not yet been deleted. E.g., the first few iterations are

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...

We circle 1, and take the subsequence of previously uncircled numbers, which is the entire sequence, and delete all terms with circled indices that have not been deleted, in this case, just the 1st, 1.

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...

We circle 2, and take the previously uncircled subsequence, which is again the entire sequence, and delete all terms with circled indices that have not been deleted, just the 2nd, 3.

(2), 4, 5, 6, 7, 8, 9, 10, 11, 12, ...

We circle 4, and take the previously uncircled subsequence, which is all terms of the sequence except 2, and delete all terms in that subsequence with circled indices (as terms of the subsequence) that have not been deleted, the 2nd and 4th, respectively the 3rd and 5th terms of the entire sequence, 5 and 7.

(2), (4), 6, 8, 9, 10, 11, 12, 13, 14, ...

etc.

(End)

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A022776, A022778, A022780, A054347.

Programs

Maple

t:=(1+sqrt(5))/2: a:=n->n+1+add(floor((n-k)/t),k=0..n): seq(a(n),n=0..55); # Muniru A Asiru, Oct 24 2018

Mathematica

Table[n + 1 + Sum[Floor[(n - k)/GoldenRatio], {k, 0, n}], {n, 0, 200}]

Formula

a(n) = n + 1 + Sum{floor[(n - k)/tau], k = 0..n}, where tau = (1 + sqrt(5))/2.

a(n) = A054347(n-1) - (n^2 - 3*n)/2. - Alan Michael Gómez Calderón, Nov 21 2024