A014085 -id:A014085 - cp's OEIS frontend

A224363 Primes p such that there are no squares between p and the prime following p.

2, 5, 11, 17, 19, 29, 37, 41, 43, 53, 59, 67, 71, 73, 83, 89, 101, 103, 107, 109, 127, 131, 137, 149, 151, 157, 163, 173, 179, 181, 191, 197, 199, 211, 227, 229, 233, 239, 241, 257, 263, 269, 271, 277, 281, 293, 307, 311, 313, 331, 337, 347, 349, 353, 367, 373

Offset: 1

César Aguilera, Apr 04 2013

nonn

Legendre's Conjecture states that there is a prime between n^2 and (n+1)^2 for every integer n > 0 and thus that between two adjacent primes there can be at most one square. As of April 2013, the conjecture is still unproved.

a(n) = A000040(A221056(n)). - Reinhard Zumkeller, Apr 15 2013

			5 is a term because there are no squares between the adjacent primes 5 and 7.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Legendre's Conjecture
Wikipedia, Legendre's conjecture

Cf. A061265, A014085.

a224363 = a000040 . a221056  -- Reinhard Zumkeller, Apr 15 2013

Select[Prime[Range[60]], Floor[Sqrt[NextPrime[#]]] == Floor[Sqrt[#]] &] (* Giovanni Resta, Apr 10 2013 *)

Corrected and edited by Giovanni Resta, Apr 10 2013

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

Original entry on oeis.org

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

N. J. A. Sloane

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

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).

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)

Cf. A014085.

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

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

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

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

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

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

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

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

A050216 Number of primes between (prime(n))^2 and (prime(n+1))^2, with a(0) = 2 by convention.

Original entry on oeis.org

2, 2, 5, 6, 15, 9, 22, 11, 27, 47, 16, 57, 44, 20, 46, 80, 78, 32, 90, 66, 30, 106, 75, 114, 163, 89, 42, 87, 42, 100, 354, 99, 165, 49, 299, 58, 182, 186, 128, 198, 195, 76, 356, 77, 144, 75, 463, 479, 168, 82, 166, 270, 90, 438, 275, 274, 292, 91, 292, 199, 99

Offset: 0

Eric W. Weisstein

The function in Brocard's Conjecture, which states that for n >= 2, a(n) >= 4.
The lines in the graph correspond to prime gaps of 2, 4, 6, ... . - T. D. Noe, Feb 04 2008
Lengths of blocks of consecutive primes in A000430 (union of primes and squares of primes). - Reinhard Zumkeller, Sep 23 2011
In the n-th step of the sieve of Eratosthenes, all multiples of prime(n) are removed. Then a(n) gives the number of new primes obtained after the n-th step. - Jean-Christophe Hervé, Oct 27 2013
More precisely, after the n-th step, one is sure to have eliminated all composites less than prime(n+1)^2, since any composite N has a prime factor <= sqrt(N). It is in exactly this (restricted) sense that a(n) yields the number of "new primes" (additional numbers known to be prime) after the n-th step. But one knows after the n-th step also that all remaining numbers between prime(n+1)^2 and prime(n+1)*(prime(n+1)+2) are prime: By construction they don't have a factor less than prime(n+1) and they don't have a factor prime(n+1) so the least prime factor could be prime(n+2) >= prime(n+1)+2. For example, after eliminating multiples of 3 in the 2nd step, one has (2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 31, 35, ...) and one knows that all remaining numbers strictly in between 5^2=25 and 5*(5+2)=35 are prime, too. - M. F. Hasler, Dec 31 2014
Numerically, the slope of the lowest "ray" m(n) = min {a(k); k>n}, seems to converge to a value somewhere in the range 1.75 < m(n)/n < 1.8; with m(n)/n > 1.7 for n > 900, m(n)/n > 1.75 for n > 2700. - M. F. Hasler, Dec 31 2014
Legendre's conjecture (see A014085) would imply that a(n) >= 2 for all n and that sequences A054272, A250473 and A250474 were thus strictly increasing (see the Wikipedia article about Brocard's conjecture). - Antti Karttunen, Jan 01 2015
a(n) >= 4 up to at least n = 4*10^5. - Eric W. Weisstein, Jan 13 2025

			There are 2 primes less than 2^2, there are 2 primes between 2^2 and 3^2, 5 primes between 3^2 and 5^2, etc. [corrected by Jonathan Sperry, Aug 30 2013]

Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 183.

T. D. Noe, Table of n, a(n) for n = 0..10000
Joel E. Cohen, Conjectures about Primes and Cyclic Numbers, arXiv:2508.08335 [math.NT], 2025. See p. 16.
Anthony G. Shannon and Jean V. Leyendekkers, On Legendre's Conjecture, Notes on Number Theory and Discrete Mathematics, Vol. 23, No. 2 (2017): 117-125.
Nicolas Vaillant, Graph of A050216(n) / (n * A001223(n))
Nicolas Vaillant, Average density of primes between prime(n)^2 and prime(n+1)^2
Eric Weisstein's World of Mathematics, Brocard's Conjecture.
Wikipedia, Brocard's Conjecture

First differences of A000879.
One more than A251723.
Cf. A010051, A001248, A014085, A089609, A251719, A256468, A256469, A256470, A029707, A137859.
Cf. A380135 (High water marks for number of primes between prime(n)^2 and prime(n+1)^2).
Cf. A380136 (Positions of the high water marks for number of primes between prime(n)^2 and prime(n+1)^2).

import Data.List (group)
a050216 n = a050216_list !! (n-1)
a050216_list =
   map length $ filter (/= [0]) $ group $ map a010051 a000430_list
-- Reinhard Zumkeller, Sep 23 2011

A050216 := proc(n)
    local p,pn ;
    if n = 0 then
        2;
    else
        p := ithprime(n) ;
        pn := nextprime(p) ;
        numtheory[pi](pn^2)-numtheory[pi](p^2) ;
    end if;
end proc:
seq(A050216(n),n=0..40) ; # R. J. Mathar, Jan 27 2025

-Subtract @@@ Partition[PrimePi[Prime[Range[20]]^2], 2, 1] (* Eric W. Weisstein, Jan 10 2025 *)

a(n)={n||return(2);primepi(prime(n+1)^2)-primepi(prime(n)^2)} \\ M. F. Hasler, Dec 31 2014

For all n >= 1, a(n) = A256468(n) + A256469(n). - Antti Karttunen, Mar 30 2015

Limit_{N->oo} (Sum_{n=1..N} a(n)) / (Sum_{n=1..N} prime(n)) = 1. - Alain Rocchelli, Sep 30 2023

Edited by N. J. A. Sloane, Nov 15 2009

A053000 a(n) = (smallest prime > n^2) - n^2.

Original entry on oeis.org

2, 1, 1, 2, 1, 4, 1, 4, 3, 2, 1, 6, 5, 4, 1, 2, 1, 4, 7, 6, 1, 2, 3, 12, 1, 6, 1, 4, 3, 12, 7, 6, 7, 2, 7, 4, 1, 4, 3, 2, 1, 12, 13, 12, 13, 2, 13, 4, 5, 10, 3, 8, 3, 10, 1, 12, 1, 2, 7, 10, 7, 6, 3, 20, 3, 4, 1, 4, 13, 22, 3, 10, 5, 4, 1, 14, 3, 10, 5, 6, 21, 2, 9, 10, 1, 4, 15, 4, 9, 6, 1, 6, 3, 14

Offset: 0

N. J. A. Sloane, Feb 21 2000

Suggested by Legendre's conjecture (still open) that there is always a prime between n^2 and (n+1)^2.
Record values are listed in A070317, their indices in A070316. - M. F. Hasler, Mar 23 2013
Conjecture: a(n) <= 1+phi(n) = 1+A000010(n), for n>0. This improves on Oppermann's conjecture, which says a(n) < n. - Jianglin Luo, Sep 22 2023

J. R. Goldman, The Queen of Mathematics, 1998, p. 82.
R. K. Guy, Unsolved Problems in Number Theory, Section A1.

T. D. Noe, Table of n, a(n) for n = 0..10000

Cf. A007491, A013632, A053001, A014085, A070316, A085099, A058055, A069003.

[NextPrime(n^2) - n^2: n in [0..100]]; // Vincenzo Librandi, Jul 06 2015

Maple
```
A053000 := n->nextprime(n^2)-n^2;
```

nxt[n_]:=Module[{n2=n^2},NextPrime[n2]-n2]
nxt/@Range[0,100]  (* Harvey P. Dale, Dec 20 2010 *)

A053000(n)=nextprime(n^2)-n^2  \\ M. F. Hasler, Mar 23 2013

from sympy import nextprime
def a(n): nn = n*n; return nextprime(nn) - nn
print([a(n) for n in range(94)]) # Michael S. Branicky, Feb 17 2022

a(n) = A013632(n^2). - Robert Israel, Jul 06 2015

More terms from James Sellers, Feb 22 2000

A056892 a(n) = square excess of the n-th prime.

Original entry on oeis.org

1, 2, 1, 3, 2, 4, 1, 3, 7, 4, 6, 1, 5, 7, 11, 4, 10, 12, 3, 7, 9, 15, 2, 8, 16, 1, 3, 7, 9, 13, 6, 10, 16, 18, 5, 7, 13, 19, 23, 4, 10, 12, 22, 24, 1, 3, 15, 27, 2, 4, 8, 14, 16, 26, 1, 7, 13, 15, 21, 25, 27, 4, 18, 22, 24, 28, 7, 13, 23, 25, 29, 35, 6, 12, 18, 22, 28, 36, 1, 9, 19, 21

Offset: 1

Henry Bottomley, Jul 05 2000

nonn

			a(5) = 2 since the 5th prime is 11 = 3^2 + 2.
From _M. F. Hasler_, Oct 19 2018: (Start)
Written as a table, starting a new row when a square is reached, the sequence reads:
  1, 2,    // = 2 - 1, 3 - 1 = {primes between 1^2 = 1 and 2^2 = 4} - 1
  1, 3,     // = 5 - 4, 7 - 4 = {primes between 2^2 = 4 and 3^2 = 9} - 4
  2, 4,      // = 11 - 9, 13 - 9 = {primes between 3^2 = 9 and 4^2 = 16} - 9
  1, 3, 7,    // = 17 - 16, 19 - 16, 23 - 16 = {primes between 16 and 25} - 16
  4, 6,        // = 29 - 25, 31 - 25 = {primes between 5^2 = 25 and 6^2 = 36} - 25
  1, 5, 7, 11,  // = {37, 41, 43, 47: primes between 6^2 = 36 and 7^2 = 49} - 36
  4, 10, 12,    // = {53, 59, 61: primes between 7^2 = 49 and 8^2 = 64} - 49
  3, 7, 9, 15,  // = {67, 71, 73, 79: primes between 8^2 = 64 and 9^2 = 81} - 64
  2, 8, 16,     // = {83, 89, 97: primes between 9^2 = 81 and 10^2 = 100} - 81
  etc. (End)

M. F. Hasler, Table of n, a(n) for n = 1..10000

Cf. A000040, A000196, A002496, A048760, A053186, A056893 .. A056898.

When written as a table, row lengths are A014085, and row sums are A108314 - A014085 * A000290 = A320688.

lst={};Do[p=Prime[n];s=p^(1/2);f=Floor[s];a=f^2;d=p-a;AppendTo[lst,d],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, Mar 11 2009 *)
#-Floor[Sqrt[#]]^2&/@Prime[Range[90]] (* Harvey P. Dale, Jul 06 2014 *)

A056892(n)={my(p=prime(n));p-sqrtint(p)^2} \\ M. F. Hasler, Oct 04 2009

a(n) = A053186(A000040(n)).

a(n) = A000040(n) - A000006(n)^2. - M. F. Hasler, Oct 04 2009

A060199 Number of primes between n^3 and (n+1)^3.

Original entry on oeis.org

0, 4, 5, 9, 12, 17, 21, 29, 32, 39, 49, 52, 58, 73, 76, 88, 92, 109, 117, 125, 140, 151, 159, 176, 188, 199, 207, 233, 247, 254, 267, 284, 305, 320, 346, 338, 373, 385, 416, 418, 437, 458, 481, 504, 517, 551, 555, 583, 599, 636, 648, 678, 686, 733, 723, 753, 810

Offset: 0

Labos Elemer, Mar 19 2001

nonn

Ingham showed that for n large enough and k=5/8, prime(n+1)-prime(n) = O(prime(n)^k). Ingham's result implies that there is a prime between sufficiently large consecutive cubes. Therefore a(n) is nonzero for n sufficiently large. Using the Riemann Hypothesis, Caldwell and Cheng prove there is a prime between all consecutive cubes. The question is undecided for squares. Many authors have reduced the value of k. The best value of k is 21/40, proved by Baker, Harman and Pintz in 2001. - corrected by Jonathan Sondow, May 19 2013
Conjecture: There are always more than 3 primes between two consecutive nonzero cubes. - Cino Hilliard, Jan 05 2003
Dudek (2014), correcting a claim of Cheng, shows that a(n) > 0 for n > exp(exp(33.217)) = 3.06144... * 10^115809481360808. - Charles R Greathouse IV, Jun 27 2014
Cully-Hugill shows the above for n > exp(exp(32.892)) = 6.92619... * 10^83675518094285. - Charles R Greathouse IV, Aug 02 2021
Mossinghoff, Trudgian, & Yang improve this to n > exp(exp(32.76)) = 3.62275 * 10^73328286790528. - Charles R Greathouse IV, Jul 31 2024

			n = 2: there are 5 primes between 8 and 27, 11,13,17,19,23.
n = 9, n+1 = 10: PrimePi(1000)-PrimePi(729) = 168-129 = a(9) = 39.

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
R. C. Baker, G. Harman and J. Pintz, The difference between consecutive primes, II, Proc. London Math. Soc. (3) 83 (2001), no. 3, 532-562.
Chris K. Caldwell and Yuanyou Cheng, Determining Mills's Constant and a Note on Honaker's Problem, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.1.
Y.-Y. F.-R. Cheng, Explicit Estimate on Primes between consecutive cubes, Rocky Mountain Journal of Mathematics 40:1 (2010), pp. 117-153. arXiv:0810.2113 [math.NT], 2008-2013.
Michaela Cully-Hugill, Primes between consecutive powers, arXiv:2107.14468 [math.NT]
Adrian Dudek, An explicit result for primes between cubes arXiv:1401.4233 [math.NT], 2014.
Adrian Dudek, An explicit result for primes between cubes, Functiones et Approximatio Commentarii Mathematici Vol. 55, Issue 2 (Dec 2016), pp. 177-197. See also Explicit Estimates in the Theory of Prime Numbers, arXiv:1611.07251 [math.NT], 2016; PhD thesis, Australian National University, 2016.
A. E. Ingham, On the difference between consecutive primes, Quart. J. Math. Oxford 8 (1937), 255-266.
MacTutor, A. E. Ingham Biography
Michael J. Mossinghoff, Timothy S. Trudgian, and Andrew Yang, Explicit zero-free regions for the Riemann zeta-function, arXiv preprint (2022). arXiv:2212.06867 [math.NT]

First differences of A038098.

Cf. A000720, A014085, A014220, A061235, A062517.

[0] cat [#PrimesInInterval(n^3, (n+1)^3): n in [1..70]]; // Vincenzo Librandi, Feb 13 2016

PrimePi[(#+1)^3]-PrimePi[#^3]&/@Range[0,60] (* Harvey P. Dale, Feb 08 2013 *)
Last[#]-First[#]&/@Partition[PrimePi[Range[0,60]^3],2,1] (* Harvey P. Dale, Feb 02 2015 *)

cubespr(n)= for(x=0,n, ct=0; for(y=x^3,(x+1)^3, if(isprime(y), ct++; )); if(ct>=0,print1(ct, ", ")))  \\ Cino Hilliard, Jan 05 2003

from sympy import primepi
def a(n): return primepi((n+1)**3) - primepi(n**3)
print([a(n) for n in range(57)]) # Michael S. Branicky, Jun 22 2021

Table[PrimePi[(j+1)^3]-PrimePi[j^3], {j, 1, 100}]

Corrected and added more detail to the Ingham references. - T. D. Noe, Sep 23 2008
Combined two comments, correcting a bad error in the first comment. - T. D. Noe, Sep 27 2008
Edited by N. J. A. Sloane, Jan 17 2009 at the suggestion of R. J. Mathar

A066888 Number of primes p between triangular numbers T(n) < p <= T(n+1).

Original entry on oeis.org

0, 2, 1, 1, 2, 2, 1, 2, 3, 2, 2, 3, 3, 3, 3, 2, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 6, 4, 5, 3, 6, 6, 7, 5, 5, 6, 4, 8, 5, 6, 6, 8, 6, 8, 5, 7, 5, 11, 4, 6, 9, 7, 8, 9, 8, 7, 7, 9, 7, 8, 7, 12, 5, 9, 9, 11, 9, 7, 7, 12, 10, 10, 9, 9, 9, 6, 11, 10, 11, 9, 12, 11, 12, 9, 10, 11, 12, 10, 13, 9, 11, 10, 12

Offset: 0

N. J. A. Sloane, Jun 06 2003

It is conjectured that for n > 0, a(n) > 0. See also A190661. - John W. Nicholson, May 18 2011

If the above conjecture is true, then for any k > 1 there is a prime p > k such that p <= (n+1)(n+2)/2, where n = floor(sqrt(2k)+1/2). Ignoring the floor function we can obtain a looser (but nicer) lower bound of p <= 1 + k + 2*sqrt(2k). - Dmitry Kamenetsky, Nov 26 2016

			Write the numbers 1, 2, ... in a triangle with n terms in the n-th row; a(n) = number of primes in n-th row.
Triangle begins
   1              (0 primes)
   2  3           (2 primes)
   4  5  6        (1 prime)
   7  8  9 10     (1 prime)
  11 12 13 14 15  (2 primes)

T. D. Noe, Table of n, a(n) for n = 0..10000

Cf. A083382.
Essentially the same as A065382 and A090970.
Cf. A000217, A000040, A014085, A190661.

Table[PrimePi[(n^2 + n)/2] - PrimePi[(n^2 - n)/2], {n, 96}] (* Alonso del Arte, Sep 03 2011 *)
PrimePi[#[[2]]]-PrimePi[#[[1]]]&/@Partition[Accumulate[Range[0,100]],2,1] (* Harvey P. Dale, Jun 04 2019 *)

{ tp(m)=local(r, t); r=1; for(n=1,m,t=0; for(k=r,n+r-1,if(isprime(k),t++)); print1(t","); r=n+r; ) }

{tpf(m)=local(r, t); r=1; for(n=1,m,t=0; for(k=r,n+r-1,if(isprime(k),t++); print1(k" ")); print1(" ("t" prime)"); print(); r=n+r;) }

from sympy import primerange
def A066888(n): return sum(1 for p in primerange((n*(n+1)>>1)+1,((n+2)*(n+1)>>1)+1)) # Chai Wah Wu, May 22 2025

a(n) = pi(n*(n+1)/2) - pi(n*(n-1)/2).

a(n) equals the number of occurrences of n+1 in A057062. - Esko Ranta, Jul 29 2011

More terms from Vladeta Jovovic and Jason Earls, Jun 06 2003

Offset corrected by Daniel Forgues, Sep 05 2012

A083382 Write the numbers from 1 to n^2 consecutively in n rows of length n; a(n) = minimal number of primes in a row.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 3, 2, 2, 2, 2, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 4, 4, 3, 3, 4, 5, 4, 3, 4, 5, 4, 5, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 7, 6, 4, 5, 6, 6, 5, 6, 6, 6, 6, 6, 7, 7, 6, 6, 6, 7, 7, 7, 7, 6, 6, 7, 7, 7

Offset: 1

James Propp, Jun 05 2003

Conjectured by Schinzel (Hypothesis H2) to be always positive for n > 1.
The conjecture has been verified for n = prime < 790000 by Aguilar.
If this is true, then Legendre's conjecture is true as well. (See A014085). - Antti Karttunen, Jan 01 2019

			For n = 3 the array is
1 2 3 (2 primes)
4 5 6 (1 prime)
7 8 9 (1 prime)
so a(3) = 1

P. Ribenboim, The New Book of Prime Number Records, Chapter 6.
P. Ribenboim, The Little Book Of Big Primes, Springer-Verlag, NY 1991, page 185.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Carlos Rivera, The calendar-like square conjecture
A. Schinzel and W. Sierpinski, Sur certaines hypothèses concernant les nombres premiers, Acta Arithmetica 4 (1958), 185-208; erratum 5 (1958) p. 259.

A084927 generalizes this to three dimensions.
Cf. A083415, A083383, A066888, A092556, A092557. See A083414 for primes in columns.
Cf. A139326.
Cf. A000720, A010051, A014085.

a083382 n = f n n a010051_list where
   f m 0 _     = m
   f m k chips = f (min m $ sum chin) (k - 1) chips' where
     (chin,chips') = splitAt n chips
-- Reinhard Zumkeller, Jun 10 2012

A083382 := proc(n) local t1,t2,at; t1 := n; at := 0; for i from 1 to n do t2 := 0; for j from 1 to n do at := at+1; if isprime(at) then t2 := t2+1; fi; od; if t2 < t1 then t1 := t2; fi; od; t1; end;

Table[minP=n; Do[s=0; Do[If[PrimeQ[c+(r-1)*n], s++ ], {c, n}]; minP=Min[s, minP], {r, n}]; minP, {n, 100}]
Table[Min[Count[#,?PrimeQ]&/@Partition[Range[n^2],n]],{n,110}] (* _Harvey P. Dale, May 29 2013 *)

A083382(n) = { my(m=-1); for(i=0,n-1,my(s=sum(j=(i*n),((i+1)*n)-1,isprime(1+j))); if((m<0) || (s < m), m = s)); (m); }; \\ Antti Karttunen, Jan 01 2019

Edited by Charles R Greathouse IV, Jul 07 2010

A106044 Difference between n-th prime and next larger perfect square.

Original entry on oeis.org

2, 1, 4, 2, 5, 3, 8, 6, 2, 7, 5, 12, 8, 6, 2, 11, 5, 3, 14, 10, 8, 2, 17, 11, 3, 20, 18, 14, 12, 8, 17, 13, 7, 5, 20, 18, 12, 6, 2, 23, 17, 15, 5, 3, 28, 26, 14, 2, 29, 27, 23, 17, 15, 5, 32, 26, 20, 18, 12, 8, 6, 31, 17, 13, 11, 7, 30, 24, 14, 12, 8, 2, 33, 27, 21, 17, 11, 3, 40, 32, 22

Offset: 1

Zak Seidov, May 06 2005

Can be read as a table, since there are always several primes between two squares, although this is the yet unproved Legendre's conjecture, cf. A014085. Whenever a(n+1) > a(n), the n-th prime is the largest one below a given square and prime(n+1) is the smallest prime larger than that square. For n > 1, these are also the indices where the parity of the terms changes. - M. F. Hasler, Oct 19 2018

			From _M. F. Hasler_, Oct 19 2018: (Start)
Written as a table, starting a new row when a square is reached, the sequence reads:
  2, 1,  // 4 - {2, 3: primes between 1^2 = 1 and 2^2 = 4}
  4, 2,   // 9 - {5, 7: primes between 2^2 = 4 and 3^2 = 9}
  5, 3,    // 16 - {11, 13: primes between 3^2 = 9 and 4^2 = 16}
  8, 6, 2,  // 25 - {17, 19, 23: primes between 4^2 = 16 and 5^2 = 25}
  7, 5,      // 36 - {29, 31: primes between 5^2 = 25 and 6^2 = 36}
  12, 8, 6, 2,// 49 - {37, 41, 43, 47: primes between 6^2 = 36 and 7^2 = 49}
  11, 5, 3,    // 64 - {53, 59, 61: primes between 7^2 = 49 and 8^2 = 64}
  14, 10, 8, 2, // 81 - {67, 71, 73, 79: primes between 8^2 = 64 and 9^2 = 81}
  17, 11, 3,     // 100 - {83, 89, 97: primes between 9^2 = 81 and 10^2 = 100}
  etc. (End)

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A158038 (analog for cubes).
Read as a table, row lengths are A014085 (number of primes between squares).
Row sums are A014085 * A000290(.+1) - A108314.

lst={};Do[p=Prime[n];s=p^(1/2);f=Floor[s];a=(f+1)^2;d=a-p;AppendTo[lst,d],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, Mar 11 2009 *)
(Floor[Sqrt[#]]+1)^2-#&/@Prime[Range[90]] (* Harvey P. Dale, Feb 08 2013 *)

A106044(n)=(sqrtint(n=prime(n))+1)^2-n \\ M. F. Hasler, Oct 19 2018

Edited by M. F. Hasler, Oct 19 2018

A143227 (Number of primes between n and 2n) - (number of primes between n^2 and (n+1)^2), if > 0.

Original entry on oeis.org

1, 2, 1, 1, 1, 1, 2, 2, 3, 3, 3, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 2, 1, 1, 2, 2, 6, 3, 3, 1, 1, 1, 2, 1, 1, 1, 1, 6, 3, 8, 3, 2, 3, 2, 3, 1, 1, 4, 3, 10, 2, 1, 1, 2, 3, 1, 3, 4, 2, 2, 9, 7, 2, 2, 4, 3, 3, 1, 2, 3, 5, 1, 2, 3, 2, 11, 3, 1, 2, 4, 7, 1, 1, 1, 1, 1, 5, 1, 2, 3, 3, 4, 2, 2, 9, 5, 1, 4, 2, 2

Offset: 1

Jonathan Sondow, Aug 02 2008

nonn

If the sequence is bounded (e.g., if it is finite), then Legendre's conjecture is true: there is always a prime between n^2 and (n+1)^2, at least for all sufficiently large n. This follows from the strong form of Bertrand's postulate proved by Ramanujan (see A104272 Ramanujan primes).

			The first positive value of ((pi(2n) - pi(n)) - (pi((n+1)^2) - pi(n^2))) is 1 (at n = 42), the 2nd is 2 (at n = 55) and the 3rd is 1 (at n = 56), so a(1) = 1, a(2) = 2, a(3) = 1.

M. Aigner and C. M. Ziegler, Proofs from The Book, Chapter 2, Springer, NY, 2001.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1989, p. 19.
S. Ramanujan, Collected Papers of Srinivasa Ramanujan (G. H. Hardy, S. Aiyar, P. Venkatesvara and B. M. Wilson, eds.), Amer. Math. Soc., Providence, 2000, pp. 208-209.

T. D. Noe, Table of n, a(n) for n=1..413
T. Hashimoto, On a certain relation between Legendre's conjecture and Bertrand's postulate, arXiv:0807.3690 [math.GM], 2008.
M. Hassani, Counting primes in the interval (n^2,(n+1)^2), arXiv:math/0607096 [math.NT], 2006.
T. D. Noe, Plot of the points (A143226(n), A143227(n))
J. Pintz, Landau's problems on primes
S. Ramanujan, A proof of Bertrand's postulate, J. Indian Math. Soc., 11 (1919), 181-182.
J. Sondow, Ramanujan Prime in MathWorld.
J. Sondow and E. W. Weisstein, Bertrand's Postulate in MathWorld.
E. W. Weisstein, Legendre's Conjecture in MathWorld.

Cf. A000720, A014085, A060715, A104272, A143223, A143224, A143225, A143226 = corresponding values of n.

L={}; Do[ With[ {d=(PrimePi[2n]-PrimePi[n])-(PrimePi[(n+1)^2]-PrimePi[n^2])}, If[d>0, L=Append[L,d]]], {n,0,1000}]; L
Select[Table[(PrimePi[2n]-PrimePi[n])-(PrimePi[(n+1)^2]-PrimePi[n^2]),{n,1000}],#>0&] (* Harvey P. Dale, Jun 19 2019 *)

a(n) = |A143223(A143226(n))|.

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