A259361 n occurs 2n+2 times.

0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 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, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset

0,7

Comments

Define the oblong root obrt(x) to be the (larger) solution of y * (y+1) = x; i.e., obrt(x) = sqrt(x+1/4) - 1/2. So obrt(x) is an integer iff x is an oblong number (A002378). Then a(n) = floor(obrt(n)).

a(n) gives (from the preceding comment) also the maximal number of parts of partitions of n with no part 1 and difference of parts at least two. See A003106, with the combinatorial interpretation of the sum of the Rogers-Ramanujan identity. - Wolfdieter Lang, Oct 29 2016

Links

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Formula

a(n) = A000194(n+1)-1.

a(n) = floor((-1 + sqrt(1+4*n))/2). See the first comment above. - Wolfdieter Lang, Oct 29 2016

Mathematica

Flatten[Table[PadLeft[{}, 2n + 2, n], {n, 0, 8}]] (* Alonso del Arte, Jun 30 2015 *)
Table[Floor[(-1 + Sqrt[1 + 4 n])/2], {n, 0, 120}] (* Michael De Vlieger, Oct 31 2016 *)

Prog

(Haskell)
a259361 = floor . subtract (1 / 2) . sqrt . (+ 1 / 4) . fromIntegral
a259361_list = concat xss where
   xss = iterate (\(xs@(x:_)) -> map (+ 1) (x : x : xs)) [0, 0]
-- Reinhard Zumkeller, Jul 09 2015
(Magma) [Floor((-1+Sqrt(1+4*n))/2): n in [0..85]]; // Vincenzo Librandi, Oct 30 2016
(Python)
from math import isqrt
def A259361(n): return (m:=isqrt(n-1)-1)+(n-1>m*(m+3)) if n else 0 # Chai Wah Wu, Nov 07 2024

Crossrefs

Cf. A000194, A002378.

Cf. A003056.

Sequence in context: A056556 A111651 A151982 * A111854 A132085 A111654

Adjacent sequences: A259358 A259359 A259360 * A259362 A259363 A259364

Keyword

nonn,easy

Author

Franklin T. Adams-Watters, Jun 24 2015

Status

approved