

A000006


Integer part of square root of nth prime.
(Formerly M0259 N0092)


19



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
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OFFSET

1,3


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.  JeanChristophe HervĂ©, 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 (7Zip compressed file)


FORMULA

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


MATHEMATICA

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


PROG

(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
(Haskell) a000006 = a000196 . a000040  Reinhard Zumkeller, Mar 24 2012
(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


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).
Sequence in context: A084557 A024417 A060021 * A061017 A331535 A248170
Adjacent sequences: A000003 A000004 A000005 * A000007 A000008 A000009


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane


STATUS

approved



