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

Author

Clark Kimberling, Jul 30 2000

Keywords

• nonn

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

Examples

			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.
		

Links

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

Crossrefs

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

Cf. A010051.

Programs

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

Mathematica

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

PARI

a(n)=(sqrtint(8*prime(n))+1)\2 \\ Charles R Greathouse IV, 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

Formula

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