OFFSET
0,5
COMMENTS
If {(X,Y,Z)} are triples of nonnegative integers with X >= Y >= Z ordered by X, Y and Z, then X=A056556(n), Y=A056557(n) and Z=A056558(n)
From Gus Wiseman, Jul 03 2019: (Start)
Also the maximum number of distinct multiplicities among integer partitions of n. For example, random partitions of 56 realizing each number of distinct multiplicities are:
1: (24,17,6,5,3,1)
2: (10,9,9,5,5,4,4,3,3,2,1,1)
3: (6,5,5,5,4,4,4,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1)
4: (28,5,5,3,3,3,2,2,1,1,1,1,1)
5: (13,4,4,4,4,4,3,3,3,2,2,2,2,2,2,1,1)
6: (6,5,5,4,4,4,3,3,3,3,2,2,2,2,2,1,1,1,1,1,1)
The maximum number of distinct multiplicities is 6, so a(56) = 6.
(End)
FORMULA
a(n) = floor(x) where x is the (largest real) solution to x^3 + 3x^2 + 2x - 6n = 0; a(A000292(n)) = n+1.
a(n+1) = a(n)+1 if a(n) = A056558(n), otherwise a(n). - Graeme McRae, Jan 09 2007
a(n) = floor(t/3 + 1/t - 1), where t = (81*n + 3*sqrt(729*n^2 - 3))^(1/3). - Ridouane Oudra, Mar 21 2021
a(n) = floor(t + 1/(3*t) - 1), where t = (6*n)^(1/3), for n>=1. - Ridouane Oudra, Nov 04 2022
EXAMPLE
3 is (3+1) * (3+2)/2 = 10 times in the sequence all these occurrences are in consecutive places. The first 3 is at position binomial(3 + 2, 3) = 10, the last one at binomial((3 + 1) + 2, 3) - 1. - David A. Corneth, Oct 14 2022
MATHEMATICA
Table[Table[m, {(m+1)(m+2)/2}], {m, 0, 7}] // Flatten (* Jean-François Alcover, Feb 28 2019 *)
PROG
(PARI) a(n)=my(t=polrootsreal(x^3+3*x^2+2*x-6*n)); t[#t]\1 \\ Charles R Greathouse IV, Feb 22 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Henry Bottomley, Jun 26 2000
EXTENSIONS
Incorrect formula deleted by Ridouane Oudra, Nov 04 2022
STATUS
approved