cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A064866 Write numbers 1, then 1 up to 2^2, then 1 up to 3^2, then 1 up to 4^2 and so on.

Original entry on oeis.org

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, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28
Offset: 1

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Author

Floor van Lamoen, Oct 08 2001

Keywords

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)

Crossrefs

Mini-index to these sequences: A064766, A064865, A064866, A065221-A655234 are all of the same type. See A064766 for a detailed explanation.

Programs

  • Mathematica
    Table[Range[n^2],{n,10}]//Flatten (* Harvey P. Dale, Mar 05 2018 *)
  • 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

Formula

a(n) = A237451(n) + (A237452(n)*A074279(n)) + 1. - M. F. Hasler, Feb 17 2014
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

Extensions

Edited by Ralf Stephan, Dec 17 2013