A000006 Integer part of square root of n-th prime.

1, 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Comments

Conjecture: No two successive terms in the sequence differ by more than 1. Proof of this would prove the converse of the theorem that every prime is surrounded by two consecutive squares, namely |sqrt(p)|^2 and (|sqrt(p)|+1)^2. - Cino Hilliard, Jan 22 2003

Equals the number of squares less than prime(n). Cf. A014689. - Zak Seidov Nov 04 2007

The above conjecture is Legendre's conjecture that for n > 0 there is always a prime between n^2 and (n+1)^2. See A014085, number of primes between two consecutive squares, which is also the number of times n is repeated in the present sequence. - Jean-Christophe Hervé, Oct 25 2013

References

• M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 2.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
• Matthew Parker, The first 100 million terms (7-Zip compressed file)

Crossrefs

Cf. A014085.

See also A263846 (floor of cube root of prime(n)), A000196 (floor of sqrt(n)), A048766 (floor of cube root of n).

Programs

Haskell

a000006 = a000196 . a000040  -- Reinhard Zumkeller, Mar 24 2012

Mathematica

a[n_] := IntegerPart[Sqrt[Prime[n]]]
IntegerPart[Sqrt[#]]&/@Prime[Range[80]] (* Harvey P. Dale, Mar 06 2012 *)

PARI

(a(n)=sqrtint(prime(n))); vector(100,n,a(n)) \\ Edited by M. F. Hasler, Oct 19 2018

PARI

apply(sqrtint,primes(100)) \\ Charles R Greathouse IV, Apr 26 2012

PARI

apply( A000006=n->sqrtint(prime(n)), [1..100]) \\ M. F. Hasler, Oct 19 2018

Python

from sympy import sieve
A000006 = lambda n: int(sieve[n]**.5)
print([A000006(n) for n in range(1,100+1)])
# Albert Lahat, Jun 25 2020

Formula

a(n) = A000196(A000040(n)). - Reinhard Zumkeller, Mar 24 2012