A256468 -id:A256468 - cp's OEIS frontend

A256470 a(n) = A256469(n) - A256468(n).

0, 1, 2, 3, 1, 6, 1, 3, 3, 0, 3, 2, -2, 0, 4, 6, 0, 4, 4, 0, 2, 3, 10, 13, -1, -2, 3, 4, 4, 34, 5, 3, 5, 17, 6, 2, 8, -2, -6, 3, -4, -4, -3, -2, -1, 9, 25, -2, -6, -4, 12, 4, 6, 9, -6, 18, 1, -2, -11, 7, -8, 27, -10, 3, -1, 12, 11, 13, -3, 5, 3, 5, -13, -8, 10, 16, -4, 14, 3, 12, -3, 23, 5, 4, 6, -8, 19, -13, 1, 0

Offset: 1

Antti Karttunen, Mar 30 2015

sign

a(n) = Difference between the number of primes occurring in range [prime(n)*prime(n+1), prime(n+1)^2] and the number of primes occurring in range [prime(n)^2, prime(n)*prime(n+1)].

In other words, a(n) tells how many more primes there are in the latter part of the range prime(n)^2 .. prime(n+1)^2 (after the geometric mean of its limits), than in its first part (before the geometric mean of its limits).

Antti Karttunen, Table of n, a(n) for n = 1..6541
A. Karttunen, Sequences A256470 and A050216 compared with OEIS Plot2-script (See also the ratio-plots linked in A256468.)

Positions of zeros: A256471. Cf. also A256472, A256473.
Positions of nonnegative terms: A256474, negative terms: A256475.
Positions of strictly positive terms: A256476, terms less than or equal to zero: A256477.
Cf. A256449, A256468, A256469.

(define (A256470 n) (- (A256469 n) (A256468 n)))

a(n) = A256469(n) - A256468(n).

a(n) = 3 - A256449(n).

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

A256447 Number of integers in range (prime(n)^2)+1 .. (prime(n)*prime(n+1)) whose smallest prime factor is at least prime(n): a(n) = A250477(n) - A250474(n).

Original entry on oeis.org

2, 3, 3, 7, 5, 9, 6, 13, 23, 9, 28, 22, 12, 24, 39, 37, 17, 44, 32, 16, 53, 37, 53, 76, 46, 23, 43, 20, 49, 161, 48, 82, 23, 142, 27, 91, 90, 66, 103, 97, 41, 181, 41, 74, 39, 228, 228, 86, 45, 86, 130, 44, 217, 134, 141, 138, 46, 148, 106, 47, 261, 355, 116, 53, 109, 387, 166, 284, 65, 119, 181, 243, 198, 195, 122, 190, 268, 125, 265, 330, 78

Offset: 1

Antti Karttunen, Mar 29 2015

nonn

a(n) = number of integers in range [(prime(n)^2)+1, (prime(n) * prime(n+1))] whose smallest prime factor is at least prime(n).
All the terms are strictly positive, because at least for the last number in the range we have A020639(prime(n)*prime(n+1)) = prime(n).
See the conjectures in A256448.

			For n=1, we have in range [(prime(1)^2)+1, (prime(1) * prime(2))], that is, in range [5,6], two numbers, 5 and 6, whose smallest prime factor (A020639) is at least 2, thus a(1) = 2.
For n=2, we have in range [10, 15] three numbers, {11, 13, 15}, whose smallest prime factor is at least 3, thus a(2) = 3.
For n=3, we have in range [26, 35] three numbers, {29, 31, 35}, whose smallest prime factor is at least prime(3) = 5, thus a(3) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..564
A. Karttunen, Ratio a(n)/A256448(n) plotted with OEIS Plot2-script
A. Karttunen, Ratio a(n)/A251723(n) plotted with OEIS Plot2-script

One more than A256468.

Cf. A020639, A250474, A250477, A251723, A256446, A256448, A256449.

f[n_] := Count[Range[Prime[n]^2 + 1, Prime[n] Prime[n + 1]],
  x_ /; Min[First /@ FactorInteger[x]] >=
Prime@n]; Array[f, 81] (* Michael De Vlieger, Mar 30 2015 *)

(define (A256447 n) (- (A250477 n) (A250474 n)))

a(n) = A250477(n) - A250474(n).
a(n) = A251723(n) - A256448(n).
a(n) = A256448(n) + A256449(n).
a(n) = A256468(n) + 1.
Other identities. For all n >= 1:
a(n+1) = A256446(n) - A256448(n).

A256448 a(n) = A250474(n+1) - A250477(n).

Original entry on oeis.org

-1, 1, 2, 7, 3, 12, 4, 13, 23, 6, 28, 21, 7, 21, 40, 40, 14, 45, 33, 13, 52, 37, 60, 86, 42, 18, 43, 21, 50, 192, 50, 82, 25, 156, 30, 90, 95, 61, 94, 97, 34, 174, 35, 69, 35, 234, 250, 81, 36, 79, 139, 45, 220, 140, 132, 153, 44, 143, 92, 51, 250, 379, 103, 53, 105, 396, 174, 294, 59, 121, 181, 245, 182, 184, 129, 203, 261, 136, 265, 339, 72

Offset: 1

Antti Karttunen, Mar 29 2015

sign

a(n) tells how many more positive integers there are <= prime(n+1)^2 whose smallest prime factor is at least prime(n+1), as compared to how many positive integers there are <= (prime(n) * prime(n+1)) whose smallest prime factor is at least prime(n).
Conjecture 1: for n >= 2, a(n) > 0.
Conjecture 2: ratio a(n)/A256447 converges towards 1. See the associated plots in A256447 and A256449 and comments in A050216.
As what comes to the second conjecture, it's not necessarily true. See the plots linked into A256468. - Antti Karttunen, Mar 30 2015

			For n=1, the respective primes are prime(1) = 2 and prime(2) = 3, and the ranges in question are [1, 9] and [1, 6]. The former range contains 4 such numbers whose lpf (A020639) is at least 3, namely {3, 5, 7, 9}, while the latter range contains 5 such numbers whose lpf is at least 2, namely {2, 3, 4, 5, 6}, thus a(1) = 4 - 5 = -1.
For n=2, the respective primes are prime(2) = 3 and prime(3) = 5, and the ranges in question are [1, 25] and [1, 15]. The former range contains 8 such numbers whose lpf is at least 5, namely {5, 7, 11, 13, 17, 19, 23, 25}, while the latter range contains 7 such numbers whose lpf is at least 3, namely {3, 5, 7, 9, 11, 13, 15}, thus a(2) = 8 - 7 = 1.
For n=3, the respective primes are prime(3) = 5 and prime(4) = 7, and the ranges in question are [1, 49] and [1, 35]. The former range contains 13 such numbers whose lpf is at least 7, namely {7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49}, while the latter range contains 11 such numbers whose lpf is at least 5, namely {5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35}, thus a(3) = 13 - 11 = 2.

Antti Karttunen, Table of n, a(n) for n = 1..564

Two less than A256469.

Cf. A050216, A250474, A250477, A251723, A256446, A256447, A256449.

(define (A256448 n) (- (A250474 (+ n 1)) (A250477 n)))

a(n) = A256469(n) - 2.
a(n) = A250474(n+1) - A250477(n).
a(n) = A251723(n) - A256447(n).
a(n) = A256446(n) - A256447(n+1).
a(n) = A256447(n) - A256449(n).

A256469 Number of primes between prime(n)*prime(n+1) and prime(n+1)^2.

Original entry on oeis.org

1, 3, 4, 9, 5, 14, 6, 15, 25, 8, 30, 23, 9, 23, 42, 42, 16, 47, 35, 15, 54, 39, 62, 88, 44, 20, 45, 23, 52, 194, 52, 84, 27, 158, 32, 92, 97, 63, 96, 99, 36, 176, 37, 71, 37, 236, 252, 83, 38, 81, 141, 47, 222, 142, 134, 155, 46, 145, 94, 53, 252, 381, 105, 55, 107, 398, 176, 296, 61

Offset: 1

Antti Karttunen, Mar 30 2015

nonn

			For n=1, there is only one prime in range prime(1)*prime(2) .. prime(2)^2, [6 .. 9], namely 7, thus a(1) = 1.
For n=2, the primes in range prime(2)*prime(3) .. prime(3)^2, [15 .. 25] are {17, 19, 23}, thus a(2) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..6541

Cf. A050216, A256448, A256468, A256470.

Table[Count[Range[Prime[n] Prime[n + 1], Prime[n + 1]^2], ?PrimeQ], {n, 69}] (* _Michael De Vlieger, Mar 30 2015 *)
Table[PrimePi[Prime[n+1]^2]-PrimePi[Prime[n]Prime[n+1]],{n,70}] (* Harvey P. Dale, Jul 31 2021 *)

allocatemem(234567890);
default(primelimit,4294965247);
A256469(n) = (primepi(prime(n+1)^2) - primepi(prime(n)*prime(n+1)));
for(n=1, 6541, write("b256469.txt", n, " ", A256469(n)));

(define (A256469 n) (let* ((p (A000040 n)) (q (A000040 (+ 1 n))) (q2 (* q q))) (let loop ((s 0) (k (* p q))) (cond ((= k q2) s) (else (loop (+ s (if (prime? k) 1 0)) (+ k 1)))))))

a(n) = A256448(n)+2.
a(n) = A050216(n) - A256468(n).
a(n) = A256468(n) + A256470(n).

cp's OEIS Frontend

A256470 a(n) = A256469(n) - A256468(n).

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

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A256447 Number of integers in range (prime(n)^2)+1 .. (prime(n)*prime(n+1)) whose smallest prime factor is at least prime(n): a(n) = A250477(n) - A250474(n).

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A256448 a(n) = A250474(n+1) - A250477(n).

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A256469 Number of primes between prime(n)*prime(n+1) and prime(n+1)^2.

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