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%I M0259 N0092 #51 Jun 25 2020 15:04:53
%S 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,
%T 11,11,11,12,12,12,12,12,13,13,13,13,13,14,14,14,14,15,15,15,15,15,15,
%U 16,16,16,16,16,16,16,17,17,17,17,17,18,18,18,18,18
%N Integer part of square root of n-th prime.
%C 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
%C Equals the number of squares less than prime(n). Cf. A014689. - _Zak Seidov_ Nov 04 2007
%C 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
%D 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.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe, <a href="/A000006/b000006.txt">Table of n, a(n) for n = 1..10000</a>
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H Matthew Parker, <a href="https://oeis.org/A000006/a000006_100M.7z">The first 100 million terms (7-Zip compressed file)</a>
%F a(n) = A000196(A000040(n)). - _Reinhard Zumkeller_, Mar 24 2012
%t a[n_] := IntegerPart[Sqrt[Prime[n]]]
%t IntegerPart[Sqrt[#]]&/@Prime[Range[80]] (* _Harvey P. Dale_, Mar 06 2012 *)
%o (PARI) (a(n)=sqrtint(prime(n))); vector(100,n,a(n)) \\ Edited by _M. F. Hasler_, Oct 19 2018
%o (PARI) apply(sqrtint,primes(100)) \\ _Charles R Greathouse IV_, Apr 26 2012
%o (PARI) apply( A000006=n->sqrtint(prime(n)), [1..100]) \\ _M. F. Hasler_, Oct 19 2018
%o (Haskell) a000006 = a000196 . a000040 -- _Reinhard Zumkeller_, Mar 24 2012
%o (Python)
%o from sympy import sieve
%o A000006 = lambda n: int(sieve[n]**.5)
%o print([A000006(n) for n in range(1,100+1)])
%o # _Albert Lahat_, Jun 25 2020
%Y Cf. A014085.
%Y See also A263846 (floor of cube root of prime(n)), A000196 (floor of sqrt(n)), A048766 (floor of cube root of n).
%K nonn,easy,nice
%O 1,3
%A _N. J. A. Sloane_
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