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A232091
Smallest square or promic (oblong) number greater than or equal to n.
3
0, 1, 2, 4, 4, 6, 6, 9, 9, 9, 12, 12, 12, 16, 16, 16, 16, 20, 20, 20, 20, 25, 25, 25, 25, 25, 30, 30, 30, 30, 30, 36, 36, 36, 36, 36, 36, 42, 42, 42, 42, 42, 42, 49, 49, 49, 49, 49, 49, 49, 56, 56, 56, 56, 56, 56, 56, 64, 64, 64, 64, 64, 64, 64, 64, 72, 72, 72, 72, 72, 72, 72, 72, 81
OFFSET
0,3
COMMENTS
Result attributed to the students Daring, et al., in the links section.
a(n) appears in floor(sqrt(a(n))) = A000194(n) successive terms.
Counting successive equal terms give sequence: 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ... (see A008619). - Michel Marcus, Jan 10 2014
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
E. Daring, I. Guadarrama, S. Sprague, and C. Winterer, WhaleConjecture.
Casey Douglas, The Next Square or Pronic, June 2012. [Wayback Machine copy]
FORMULA
a(n) = ceiling(n/ceiling(sqrt(n)))*ceiling(sqrt(n)).
a(n) = min(k : k >= n, k in A002620).
a(k^2) = k^2; a(k*(k+1)) = k*(k+1).
It appears that a(n) = A216607(n) + n. (Verified for all n<10^9 by Lars Blomberg, Jan 09 2014.) This conjecture now follows from a proof given by David Applegate, Jan 10 2014 (see [Applegate]).
a(n) = min(A048761(n), A259225(n)). - Michel Marcus, Jun 22 2015
Sum_{n>=1} 1/a(n)^2 = 2 - Pi^2/6 + zeta(3). - Amiram Eldar, Aug 16 2022
MATHEMATICA
Join[{0}, Table[Ceiling[n/Ceiling[Sqrt[n]]] Ceiling[Sqrt[n]], {n, 100}]] (* Alonso del Arte, Nov 18 2013 *)
PROG
(PARI) a(n)=my(t=sqrtint(n-1)+1); t*((n-1)\t+1) \\ Charles R Greathouse IV, Nov 18 2013
(Magma) [(Ceiling(n /Ceiling(Sqrt(n)))*Ceiling(Sqrt(n))): n in [1..80]]; // Vincenzo Librandi, Jun 22 2015
CROSSREFS
Cf. A000290 (squares), A002378 (promic or oblong numbers), A002620 (A000290 union A002378).
Sequence in context: A211514 A111457 A161765 * A364066 A164798 A087554
KEYWORD
nonn,easy,changed
AUTHOR
L. Edson Jeffery, Nov 18 2013
EXTENSIONS
Extended by Charles R Greathouse IV, Nov 18 2013
a(0)=0 prepended by Michel Marcus, Jun 22 2015
STATUS
approved