OFFSET
1,3
COMMENTS
This is a fractal sequence: if the first instance of each number is deleted, the original sequence is recovered. - Franklin T. Adams-Watters, Dec 14 2013
Subsequences start at indices A000330 + 1. - Ralf Stephan, Dec 17 2013
When sequence fills a triangular array by rows, the main diagonal is A064865:
This triangle begins:
1
1 2
3 4 1
2 3 4 5
6 7 8 9 1
From Antti Karttunen, Feb 17 2014: (Start)
A more natural way of organizing this sequence is as an irregular table consisting of successively larger square matrices:
1;
1, 2;
3, 4;
1, 2, 3;
4, 5, 6;
7, 8, 9;
1, 2, 3, 4;
5, 6, 7, 8;
9,10,11,12;
13,14,15,16;
etc.
(End)
FORMULA
For 1 <= n <= 650, a(n) = n - t(t-1)(2t-1)/6, where t = floor((3*n)^(1/3)+1/2). - Mikael Aaltonen, Jan 17 2015
a(n) = n-k(k-1)(2k-1)/6 where k = m+1 if n>m(m+1)(2m+1)/6 and k = m otherwise and m = floor((3n)^(1/3)). - Chai Wah Wu, Nov 05 2024
MATHEMATICA
Table[Range[n^2], {n, 10}]//Flatten (* Harvey P. Dale, Mar 05 2018 *)
PROG
(PARI) A064866_vec(N=9)=concat(vector(N, i, vector(i^2, j, j))) \\ Note: This creates a vector; use A064866_vec()[n] to get the n-th term. - M. F. Hasler, Feb 17 2014
(Python)
from sympy import integer_nthroot
def A064866(n): return n-(k:=(m:=integer_nthroot(3*n, 3)[0])+(6*n>m*(m+1)*((m<<1)+1)))*(k-1)*((k<<1)-1)//6 # Chai Wah Wu, Nov 04 2024
CROSSREFS
KEYWORD
AUTHOR
Floor van Lamoen, Oct 08 2001
EXTENSIONS
Edited by Ralf Stephan, Dec 17 2013
STATUS
approved