A057062 Let R(i,j) be the infinite square array with antidiagonals 1; 2,3; 4,5,6; ...; the n-th prime is in antidiagonal a(n).

2, 2, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 9, 9, 10, 10, 11, 11, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 17, 17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23, 24, 24, 24, 24, 25, 25

Offset

1,1

Comments

The smallest integer in the j-th antidiagonal is A000124(j-1). So a(n) is the index j such that A000124(j-1) <= prime(n) < A000124(j). - R. J. Mathar, Dec 02 2011

Links

T. D. Noe, Table of n, a(n) for n = 1..1000

Formula

a(n) = round(sqrt(2*prime(n))). - Vladeta Jovovic, Jun 14 2003

Example

The array begins
   1  3  6 10 15 ...
   2  5  9 14 ...
   4  8 13 ...
   7 12 ...
  11 ...
  ...
The third prime, 5, is in the 3rd antidiagonal, so a(3) = 3.

Mathematica

Table[Round[Sqrt[2*Prime[n]]], {n, 100}] (* T. D. Noe, Dec 03 2011 *)

Prog

(PARI) a(n)=(sqrtint(8*prime(n))+1)\2 \\ Charles R Greathouse IV, Jul 26 2012
(Haskell)
a057062 n = a057062_list !! (n-1)
a057062_list = f 1 [1..] where
   f j xs = (replicate (sum $ map a010051 dia) j) ++ f (j + 1) xs'
     where (dia, xs') = splitAt j xs
-- Reinhard Zumkeller, Jul 26 2012
(Python)
from math import isqrt
from sympy import prime
def A057062(n): return isqrt(prime(n)<<3)+1>>1 # Chai Wah Wu, Jun 19 2024

Crossrefs

Cf. A057045, A057048, A022846, A057057, A057054. A066888 counts how many times each positive integer appears in this sequence.

Cf. A010051.

Sequence in context: A305557 A099249 A050296 * A283993 A255572 A065855

Adjacent sequences: A057059 A057060 A057061 * A057063 A057064 A057065

Keyword

nonn

Author

Clark Kimberling, Jul 30 2000

Status

approved