OFFSET
0,3
COMMENTS
From M. F. Hasler, Oct 05 2009: (Start)
For each k > 0, the term k^2 is listed 2k - 1 times.
a(n+1) is the least square greater than n. (End)
REFERENCES
Krassimir Atanassov, On the 40th and 41st Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 4, No. 3 (1998), 101-104.
J. Castillo, Other Smarandache Type Functions: Inferior/Superior Smarandache f-part of x, Smarandache Notions Journal, Vol. 10, No. 1-2-3, 1999, 202-204.
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Krassimir Atanassov, On Some of Smarandache's Problems, American Research Press, 1999, 27-32.
Florentin Smarandache, Only Problems, Not Solutions!.
FORMULA
a(n) = (ceiling(sqrt(n)))^2. - Alonso del Arte, Jun 21 2015
Sum_{n>=1} 1/a(n)^2 = 2*zeta(3) - Pi^4/90. - Amiram Eldar, Aug 15 2022
MAPLE
MATHEMATICA
(Ceiling[Sqrt[Range[0, 99]]])^2 (* Alonso del Arte, Jun 21 2015 *)
PROG
(PARI) A048761(n)=if(n, (sqrtint(n-1)+1)^2, 0) \\ M. F. Hasler, Oct 05 2009
(Haskell)
a048761 n = (a000196 n + 1 - a010052 n) ^ 2
a048761_list = 0 : concat (f 1 1) where
f u v = (take v $ repeat u) : f (u + v + 2) (v + 2)
-- Reinhard Zumkeller, Mar 16 2014
(Magma) [Ceiling(Sqrt(n))^2: n in [0..80]]; // Vincenzo Librandi, Jun 21 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Charles T. Le (charlestle(AT)yahoo.com)
EXTENSIONS
Missing a(49) = 49 inserted by Reinhard Zumkeller, Mar 16 2014
STATUS
approved