A074279 n appears n^2 times.

1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

Author

Jon Perry, Sep 21 2002

Keywords

• nonn
• tabf

Comments

Since the last occurrence of n comes one before the first occurrence of n+1 and the former is at Sum_{i=0..n} i^2 = A000330(n), we have a(A000330(n)) = a(n*(n+1)*(2n+1)/6) = n and a(1+A000330(n)) = a(1+(n*(n+1)*(2n+1)/6)) = n+1. So the current sequence is, loosely speaking, the inverse function of the square pyramidal sequence A000330. A000330 has many alternative formulas, thus yielding many alternative formulas for the current sequence. - Jonathan Vos Post, Mar 18 2006

Partial sums of A253903. - Jeremy Gardiner, Jan 14 2018

Examples

			This can be viewed also as an irregular table consisting of successively larger square matrices:
  1;
  2, 2;
  2, 2;
  3, 3, 3;
  3, 3, 3;
  3, 3, 3;
  4, 4, 4, 4;
  4, 4, 4, 4;
  4, 4, 4, 4;
  4, 4, 4, 4;
  etc.
When this is used with any similarly organized sequence, a(n) is the index of the matrix in whose range n is. A121997(n) (= A237451(n)+1) and A238013(n) (= A237452(n)+1) would then yield the index of the column and row within that matrix.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..9455 (Table of squares from 1 X 1 to 30 X 30, flattened)
• Y.-F. S. Petermann, J.-L. Remy and I. Vardi, Discrete derivatives of sequences, Adv. in Appl. Math. 27 (2001), 562-84.

Crossrefs

Cf. A000217, A000330, A002024, A003881, A006331, A050446, A050447, A000537, A006003, A005900, A064866, A237451, A237452.

Programs

Mathematica

Table[n, {n, 0, 6}, {n^2}] // Flatten (* Arkadiusz Wesolowski, Jan 13 2013 *)

PARI

A074279_vec(N=9)=concat(vector(N,i,vector(i^2,j,i))) \\ Note: This creates a vector; use A074279_vec()[n] to get the n-th term. - M. F. Hasler, Feb 17 2014

Python

from sympy import integer_nthroot
def A074279(n): return (m:=integer_nthroot(3*n,3)[0])+(6*n>m*(m+1)*((m<<1)+1)) # Chai Wah Wu, Nov 04 2024

Formula

For 1 <= n <= 650, a(n) = floor((3n)^(1/3)+1/2). - Mikael Aaltonen, Jan 05 2015

a(n) = 1 + floor( t(n) + 1 / ( 12 * t(n) ) - 1/2 ), where t(n) = (sqrt(3888*(n-1)^2-1) / (8*3^(3/2)) + 3 * (n-1)/2 ) ^(1/3). - Mikael Aaltonen, Mar 01 2015

a(n) = floor(t + 1/(12*t) + 1/2), where t = (3*n - 1)^(1/3). - Ridouane Oudra, Oct 30 2023

a(n) = m+1 if n > m(m+1)(2m+1)/6 and a(n) = m otherwise where m = floor((3n)^(1/3)). - Chai Wah Wu, Nov 04 2024

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/4 (A003881). - Amiram Eldar, Jun 30 2025

Extensions

Offset corrected from 0 to 1 by Antti Karttunen, Feb 08 2014