

A057062


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


6



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
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OFFSET

1,1


COMMENTS

The smallest integer in the jth antidiagonal is A000124(j1). So a(n) is the index j such that A000124(j1) <= 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 !! (n1)
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


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



