OFFSET
1,2
COMMENTS
From Jonathan Vos Post, Mar 18 2006: (Start)
The key to this sequence is: 1^3 + 2^3 + 3^3 + ... + n^3 = (1+2+3+...+n)^2.
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^3 = A000537(n) = (A000217(n))^2 = (n*(n+1)/2)^2 = (C(n+1,2))^2, have a(A000537(n)) = a((A000217(n))^2) = n and thus a(1+A000537(n)) = a(1+(A000217(n))^2) = n+1.
The current sequence is, loosely, the inverse function of the square of the triangular number sequence. (End)
LINKS
Boris Putievskiy, Table of n, a(n) for n = 1..8281
Boris Putievskiy, Integer Sequences: Irregular Arrays and Intra-Block Permutations, arXiv:2310.18466 [math.CO], 2023.
FORMULA
a(n) = ceiling((1/2)*(sqrt(8*sqrt(n) + 1) - 1)). - Boris Putievskiy, Jun 19 2024
From Chai Wah Wu, Nov 04 2024: (Start)
a(n) = m+1 if n>(m(m+1))^2/4 and a(n) = m otherwise where m = floor((4n)^(1/4)).
More generally, for a sequence a_k(n) where n appears n^(k-1) times, a_k(n) = m+1 if n > Sum_{i=1..m} i^(k-1) and a_k(n) = m otherwise where m = floor((kn)^(1/k)).
Note that Sum_{i=1..m} i^(k-1) can be written as a k-th order polynomial of m using Faulhaber's formula. (End)
MATHEMATICA
Flatten @ Table[ Table[k, {k^3}], {k, 5}] (* Giovanni Resta, Jun 17 2016 *)
a[n_]:=Ceiling[1/2 (Sqrt[8 Sqrt[n]+1]-1)]
Nmax=225; Table[a[n], {n, 1, Nmax}] (* Boris Putievskiy, Jun 19 2024 *)
PROG
(Python)
from sympy import integer_nthroot
def A108582(n): return (m:=integer_nthroot(k:=n<<2, 4)[0])+(k>(m*(m+1))**2) # Chai Wah Wu, Nov 04 2024
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jonathan Vos Post, Jul 25 2005
EXTENSIONS
Two missing terms from Giovanni Resta, Jun 17 2016
STATUS
approved