A057057 -id:A057057 - cp's OEIS frontend

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

Clark Kimberling, Jul 30 2000

nonn

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

			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.

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

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

Cf. A010051.

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

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

a(n)=(sqrtint(8*prime(n))+1)\2 \\ Charles R Greathouse IV, Jul 26 2012

from math import isqrt
from sympy import prime
def A057062(n): return isqrt(prime(n)<<3)+1>>1 # Chai Wah Wu, Jun 19 2024

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

A057045 Let R(i,j) be the rectangle with antidiagonals 1; 2,3; 4,5,6; ...; the n-th Lucas number is in antidiagonal a(n).

Original entry on oeis.org

2, 1, 2, 3, 4, 5, 6, 8, 10, 12, 16, 20, 25, 32, 41, 52, 66, 85, 107, 137, 174, 221, 281, 358, 455, 579, 737, 937, 1192, 1516, 1929, 2454, 3121, 3970, 5050, 6424, 8171, 10394, 13221, 16818, 21393, 27212

Offset: 1

Clark Kimberling, Jul 30 2000

nonn

Chai Wah Wu, Table of n, a(n) for n = 1..1000

Cf. A057062, A057048, A022846, A057057, A057054.

from gmpy2 import isqrt_rem, lucas
def A057045(n):
    i, j = isqrt_rem(2*lucas(n-1))
    return int(i + int(4*(j-i) >= 1)) # Chai Wah Wu, Aug 16 2016

Round(sqrt(2*A000032(n-1))). - Vladeta Jovovic, Jun 14 2003

A057054 Let R(i,j) be the rectangle with antidiagonals 1; 2,3; 4,5,6; ...; n^3 is in antidiagonal a(n).

Original entry on oeis.org

1, 4, 7, 11, 16, 21, 26, 32, 38, 45, 52, 59, 66, 74, 82, 91, 99, 108, 117, 126, 136, 146, 156, 166, 177, 187, 198, 210, 221, 232, 244, 256, 268, 280, 293, 305, 318, 331, 344, 358, 371, 385, 399, 413, 427, 441, 456, 470, 485, 500

Offset: 1

Clark Kimberling, Jul 30 2000

nonn

Cf. A057045, A057062, A057048, A022846, A057057.

Table[Round[n Sqrt[2n]],{n,50}] (* Harvey P. Dale, Feb 10 2020 *)

Round(n*(sqrt(2*n))). - Vladeta Jovovic, Jun 14 2003

cp's OEIS Frontend

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).

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A057045 Let R(i,j) be the rectangle with antidiagonals 1; 2,3; 4,5,6; ...; the n-th Lucas number is in antidiagonal a(n).

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A057054 Let R(i,j) be the rectangle with antidiagonals 1; 2,3; 4,5,6; ...; n^3 is in antidiagonal a(n).

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