A014085 -id:A014085 - cp's OEIS frontend

A061235 Number of primes between n^4 and (n+1)^4.

0, 6, 16, 32, 60, 96, 147, 207, 283, 382, 486, 619, 773, 945, 1139, 1351, 1610, 1870, 2165, 2496, 2848, 3237, 3653, 4125, 4572, 5118, 5698, 6269, 6894, 7586, 8309, 9033, 9907, 10656, 11616, 12522, 13509, 14552, 15639, 16708, 18009, 19140, 20527

Offset: 0

Amarnath Murthy, Apr 23 2001

nonn

			a(3) = 32, as the number of primes between 3^4 = 81 and 4^4 = 256 is 32.

Amiram Eldar, Table of n, a(n) for n = 0..10000 (terms 0..300 from Vincenzo Librandi)

Cf. A014085, A060199, A062517.

[0] cat [#PrimesInInterval(n^4, (n+1)^4): n in [1..50]]; // Vincenzo Librandi, Apr 30 2017

Table[PrimePi[(w+1)^4]-PrimePi[w^4], {w, 0, 100}]

a(n) = primepi((n+1)^4) - primepi(n^4); \\ Michel Marcus, Apr 29 2017

More terms from Labos Elemer, Jul 10 2001

Edited for consistency by Peter Munn, Apr 28 2017

A062517 Number of primes between n^5 and (n+1)^5.

Original entry on oeis.org

0, 11, 42, 119, 273, 540, 954, 1573, 2456, 3624, 5181, 7177, 9666, 12797, 16514, 21098, 26454, 32836, 40134, 48760, 58508, 69714, 82277, 96723, 112702, 130639, 150488, 172617, 197039, 223915, 253318, 285540, 320450, 358839, 400159, 445011, 493504

Offset: 0

Labos Elemer, Jul 10 2001

nonn

			a(1) = 11 the number of primes between 1 = 1^5 and 32 = 2^5.

Amiram Eldar, Table of n, a(n) for n = 0..4000 (terms 0..83 from Harry J. Smith)

Cf. A014085, A060199, A061235.

Table[PrimePi[(w+1)^5]-PrimePi[w^5], {w, 0, 50}]

a(n) = { primepi((n + 1)^5) - primepi((n)^5) } \\ Harry J. Smith, Aug 08 2009

Edited for consistency by Peter Munn, Apr 30 2017

A076957 Smallest k such that there are exactly n primes strictly between k^2 and (k+1)^2.

Original entry on oeis.org

1, 4, 6, 10, 15, 16, 25, 24, 31, 39, 38, 45, 64, 48, 52, 57, 75, 82, 81, 70, 76, 79, 106, 112, 145, 111, 121, 117, 123, 134, 144, 139, 146, 154, 163, 192, 169, 176, 179, 193, 202, 218, 204, 226, 223, 240, 233, 238, 243, 259, 291, 256, 286, 309, 278

Offset: 2

Amarnath Murthy, Oct 20 2002

nonn

From David W. Wilson, Jan 08 2017: (Start)
a(n)^2 = A076956(n).
A014085(a(n)) = n.
Conjecturally, a(n) is undefined for n = 1 and defined for all n >= 2. (End)

T. D. Noe and David W. Wilson, Table of n, a(n) for n = 2..10000 (a(n) for n = 2..1000 from T. D. Noe).

Cf. A014085, A076956, A084597.

a := proc(n) local k, h, SEARCHLIMIT; SEARCHLIMIT := 10000; h := proc(k) option remember; nops(select(j->isprime(j), [$k^2+1..(k+1)^2])) end: k := 1; while h(k) <> n and k < SEARCHLIMIT do k := k+1 od; `if`(k=SEARCHLIMIT, print("Search limit reached!"), k) end: seq(a(n), n=2..56); # Peter Luschny, Jan 10 2017

Table[k = 1; While[Count[Map[PrimeQ, Range[k^2 + 1, (k + 1)^2]], True] != n, k++]; k, {n, 2, 56}] (* Michael De Vlieger, Jan 10 2017 *)
With[{pp=Table[Count[Range[n^2+1,(n+1)^2-1],?(PrimeQ[#]&)],{n,500}]},Table[ Position[pp,k,1,1],{k,60}]]//Flatten (* _Harvey P. Dale, Aug 01 2021 *)

More terms from Ralf Stephan, Oct 31 2002

A091591 Number of pairs of twin primes between n^2 and (n+1)^2.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 2, 1, 2, 2, 1, 1, 0, 2, 1, 1, 1, 2, 2, 0, 0, 3, 2, 0, 1, 3, 2, 0, 3, 2, 1, 3, 0, 3, 2, 1, 3, 2, 4, 2, 2, 3, 0, 2, 2, 4, 0, 2, 1, 1, 5, 4, 4, 1, 2, 3, 4, 3, 5, 2, 2, 3, 2, 4, 1, 2, 2, 3, 4, 3, 0, 3, 3, 2, 4, 5, 2, 2, 3, 4, 1, 2, 3, 2, 3, 3, 1, 5, 1, 3, 4, 4, 2, 5, 3, 4, 1, 3, 5, 1, 2

Offset: 3

Hugo Pfoertner, Jan 22 2004

a(1) and a(2) are omitted because they are dependent on the treatment of the twin pair (3,5). It is conjectured that a(n)>0 for all n>122. Proving this would also prove the twin prime conjecture.

Proving a(n)>0 for n>122 would also prove Legendre's conjecture that there is a prime between n^2 and (n+1)^2. - T. D. Noe, Feb 28 2007

			a(3)=1 because the interval [3^2,4^2] contains one pair of twins (11,13).
a(9)=0 because the interval [9^2,10^2] is one of the few known intervals (given in A091592) not containing twin primes.

T. D. Noe, Table of n, a(n) for n=3..10000
Eric Weisstein's World of Mathematics, Twin Prime Conjecture.

Cf. A000290, A001359, A006512, A091592.

Cf. A014085 (number of primes between n^2 and (n+1)^2)

a[n_] := (k = 0; For[p = NextPrime[n^2], p <= NextPrime[(n + 1)^2, -2], q = NextPrime[p]; If[q - p == 2, k++; p = NextPrime[q], p = q]]; k); Table[a[n], {n, 3, 107}] (* Jean-François Alcover, Jun 13 2012 *)
With[{tps=Select[Partition[Prime[Range[2000]],2,1],Last[#]-First[#] == 2&]},Table[ Count[tps,?(#[[1]]>n^2&&#[[2]]<(n+1)^2&)],{n,3,110}]] (* _Harvey P. Dale, Feb 19 2013 *)

A117896 Number of perfect powers between consecutive squares n^2 and (n+1)^2.

Original entry on oeis.org

0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0

Offset: 1

T. D. Noe, Mar 31 2006, Feb 15 2010

nonn

a(n)=2 only 14 times for n^2 < 2^63. What is the least n such that a(n)=3? Is a(n) bounded?

			a(5)=2 because powers 27 and 32 are between 25 and 36.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
J. H. Loxton, Some problems involving powers of integers, Acta Arith., 46 (1986), pp. 113-123.
C. L. Stewart, On heights of multiplicatively dependent algebraic numbers, Acta Arith. 133 (2008), pp. 97-108.
J. Turk, Multiplicative properties of integers in short intervals, Proc. Kon. Ned. Akad. Wet. (A) 83 (1980), pp. 429-436.

Cf. A001597 (perfect powers), A014085 (primes between squares), A097055, A097056, A117934.

nn=151^2; powers=Join[{1}, Union[Flatten[Table[n^i, {i,Prime[Range[PrimePi[Log[2,nn]]]]}, {n,2,nn^(1/i)}]]]]; t=Table[0,{Sqrt[nn]-1}]; Do[n=Floor[Sqrt[i]]; If[i>n^2, t[[n]]++], {i,powers}]; t (* revised, T. D. Noe, Apr 19 2011 *)

a(n)=my(k);-sum(e=3,2*log(n+1)\log(2),k=round((n+1/2)^(2/e))^e;if(n^2Charles R Greathouse IV, Dec 19 2011

Trivially, a(n) << log n/log log n. Turk gives a(n) << sqrt(log n) and Loxton improves this to a(n) <= exp(40 sqrt(log log n log log log n)). Stewart improves the constant from 40 to 30 and conjectures that a(n) < 3 for all but finitely many n. - Charles R Greathouse IV, Dec 11 2012

A144831 (n+1)^2 - (smallest prime > n^2).

Original entry on oeis.org

2, 4, 5, 8, 7, 12, 11, 14, 17, 20, 17, 20, 23, 28, 29, 32, 31, 30, 33, 40, 41, 42, 35, 48, 45, 52, 51, 54, 47, 54, 57, 58, 65, 62, 67, 72, 71, 74, 77, 80, 71, 72, 75, 76, 89, 80, 91, 92, 89, 98, 95, 102, 97, 108, 99, 112, 113, 110, 109, 114, 117, 122, 107, 126, 127, 132, 131

Offset: 1

Enoch Haga, Sep 21 2008

Suggested by Conjecture 60 in Carlos Rivera's The Prime Puzzles & Problems Connection.

Legendre's conjecture that there is always a prime between n^2 and (n+1)^2 is equivalent to a(n) >= 0 for all n. As the conjecture is still opened, it is not proved that a(n) is nonn, although the keyword is automatically added. - Jean-Christophe Hervé, Oct 26 2013

			a(2)=4 because n=2, 2^2=4 and (2+1)^2=9. The gap in which primes are to be found is 4 - 9. Next prime=5 and 9-5=4. For a(3)=5, 3^2=9 and (3+1)^2=16. Next prime=11 and 16-11=5.

Jean-Christophe Hervé, Table of n, a(n) for n = 1..10000

Cf. A007491, A053000, A144832, A144256, A014085.

Table[n^2-NextPrime[(n-1)^2],{n,2,70}] (* Harvey P. Dale, Jan 22 2019 *)

a(n) = (n+1)^2 - nextprime(n^2); \\ Michel Marcus, Jun 08 2014

Calculate n^2 and (n+1)^2, e.g. 4 - 9. Find the next prime following n^2 and subtract from (n+1)^2. Next prime is 5 so 9-5=4, the distance from next prime to (n+1)^2.

a(n) = (n+1)^2 - A007491(n).

Definition rewritten by N. J. A. Sloane, Sep 28 2008

Definition rewritten by Jean-Christophe Hervé, Oct 26 2013

A161622 Denominators of the ratios (in lowest terms) of numbers of primes in one square interval to that of the interval and its successor.

Original entry on oeis.org

2, 2, 5, 5, 3, 7, 7, 7, 8, 9, 9, 2, 9, 5, 13, 12, 11, 2, 13, 2, 2, 13, 5, 17, 15, 15, 17, 17, 2, 9, 19, 19, 19, 19, 19, 2, 7, 23, 23, 23, 20, 7, 23, 24, 23, 23, 28, 5, 21, 26, 31, 7, 25, 24, 23, 29, 30, 29, 2, 29, 30, 32, 29, 15, 31, 2, 32, 30, 34, 12, 2, 32, 2, 35, 20, 18, 16, 41, 36, 33

Offset: 1

Daniel Tisdale, Jun 14 2009

The numerators are derived from sequence A014085.
The expression is: R(n) = (PrimePi((n+1)^2) - PrimePi(n^2))/(PrimePi((n+2)^2) - PrimePi(n^2)).
The first few ratios are 1/2, 2/5, 3/5, 1/3, 4/7, ...
Conjecture: lim_{n->infinity} R(n) = 1/2. See also more extensive comment entered with sequence of numerators. This conjecture implies Legendre's conjecture.

			R(3) = (PrimePi(4^2)-PrimePi(3^2)) / (PrimePi(5^2)-PrimePi(3^2)) = (PrimePi(16)-PrimePi(9)) / (PrimePi(25)-PrimePi(9)) = (6-4)/(9-4) = 2/5. Hence a(3) = 5. - _Klaus Brockhaus_, Jun 15 2009

Cf. A014085.

Cf. A161621 (numerators). - Klaus Brockhaus, Jun 15 2009

[ Denominator((#PrimesUpTo((n+1)^2) - a) / (#PrimesUpTo((n+2)^2) - a)) where a is #PrimesUpTo(n^2): n in [1..80] ]; // Klaus Brockhaus, Jun 15 2009

a(1) inserted and extended beyond a(11) by Klaus Brockhaus, Jun 15 2009

A185150 Number of odd primes p between n^2 and (n+1)^2 with (n/p) = 1, where (-) is the Legendre symbol.

Original entry on oeis.org

1, 1, 2, 3, 2, 2, 2, 3, 3, 1, 4, 2, 4, 3, 5, 7, 2, 3, 4, 6, 5, 3, 3, 4, 8, 5, 4, 5, 4, 4, 6, 6, 6, 4, 9, 9, 7, 7, 5, 6, 7, 5, 9, 5, 7, 3, 9, 6, 10, 6, 10, 6, 8, 8, 7, 7, 10, 3, 12, 8, 7, 10, 8, 14, 11, 7, 10, 10, 5, 9, 11, 8, 7, 9, 9, 18, 11, 11, 12, 9, 20, 6, 13, 6, 10, 9, 13, 9, 8, 10, 10, 12, 12, 6, 13, 9, 12, 12, 8, 23

Offset: 1

Zhi-Wei Sun, Dec 29 2012

nonn

Conjecture: a(n)>0 for all n>0.
This is a refinement of Legendre's conjecture that for each n=1,2,3,... the interval (n^2,(n+1)^2) contains a prime.
We have verified the conjecture for n up to 10^9.
Zhi-Wei Sun also made some similar conjectures involving primes and Legendre symbols, below are few examples:
(1) If n>10 then there is a prime p between n^2 and (n+1)^2 with (n/p) = ((1-n)/p) = 1. If n>2 is different from 7 and 17, then there is a prime p between n^2 and (n+1)^2 with (n/p) = ((n+1)/p) = 1. If n>1 is not equal to 27, then there is a prime p between n^2 and (n+1)^2 with (n/p) = ((n+2)/p) = 1.
(2) If n>2 is different from 6, 12, 58, then there is a prime p between n^2 and n^2+n such that (n/p) = 1. If n>20 is not a square, and different from 37 and 77, then there is a prime p between n^2 and n^2+n such that (n/p) = -1.
(3) For each n=15,16,... there is a prime p between n and 2n such that (n/p) = 1. If n>0 is not a square, then
there is a prime p between n and 2n such that (n/p) = -1.

			a(10)=1 since 107 is the only prime p between 10^2 and 11^2 with (10/p) = 1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv:1211.1588.

Cf. A014085, A185636.

a[n_]:=a[n]=Sum[If[n^2+k>2&&PrimeQ[n^2+k]==True&&JacobiSymbol[n,n^2+k]==1,1,0],{k,1,2n}]
Do[Print[n," ",a[n]],{n,1,100}]

A191225 Number of Ramanujan primes R_k between triangular numbers T(n-1) < R_k <= T(n).

Original entry on oeis.org

0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 2, 0, 2, 1, 1, 2, 1, 2, 0, 2, 3, 2, 1, 2, 2, 2, 1, 4, 2, 2, 1, 0, 4, 3, 5, 1, 3, 2, 1, 5, 1, 2, 3, 4, 4, 4, 2, 2, 2, 4, 2, 3, 4, 3, 5, 4, 3, 2, 5, 4, 2, 5, 1, 6, 1, 5, 5, 7, 2, 2, 1, 10, 6, 6, 2, 2, 5, 0, 3, 7, 5, 4, 6, 7, 4

Offset: 1

John W. Nicholson, May 27 2011

nonn

The function eta(x), A191228, returns the greatest value of k of R_k <= x, and where R_k is the k-th Ramanujan prime (A104272).

			Write the numbers 1, 2, ... in a triangle with n terms in the n-th row; a(n) = number of Ramanujan primes in n-th row.
Triangle begins
1                 (0 Ramanujan primes, eta(1) = 0)
2  3              (1 Ramanujan primes, eta(3) - eta(1) = 1)
4  5  6           (0 Ramanujan primes, eta(6) - eta(3) = 0)
7  8  9  10       (0 Ramanujan primes, eta(10) - eta(6) = 0)
11 12 13 14 15    (1 Ramanujan primes, eta(15) - eta(10) = 1)
16 17 18 19 20 21 (1 Ramanujan primes, eta(21) - eta(15) = 1)

T. D. Noe, Table of n, a(n) for n = 1..10000
J. Sondow, J. W. Nicholson, and T. D. Noe, Ramanujan Primes: Bounds, Runs, Twins, and Gaps, J. Integer Seq. 14 (2011) Article 11.6.2

Cf. A000217, A000040, A104272, A191228, A014085, A190661, A083382, A191226, A191227.

terms = 100; nn = terms^2; R = Table[0, {nn}]; s = 0;
Do[If[PrimeQ[k], s++]; If[PrimeQ[k/2], s--]; If[s < nn, R[[s+1]] = k], {k, Prime[3 nn]}];
A104272 = R + 1;
eta = Table[Boole[MemberQ[A104272, k]], {k, 1, nn}] // Accumulate;
T[n_] := n(n+1)/2;
a[1] = 0; a[n_] := eta[[T[n]]] - eta[[T[n-1]]];
Array[a, terms] (* Jean-François Alcover, Nov 07 2018, using T. D. Noe's code for A104272 *)

use ntheory ":all"; sub a191225 { my $n = shift; ramanujan_prime_count( (($n-1)*$n)/2+1, ($n*($n+1))/2 ); } say a191225($) for 1..10; # _Dana Jacobsen, Dec 30 2015

a(n) = eta(T(n))- eta(T(n-1)).

A333846 Numbers k such that the number of primes between k^2 and (k+1)^2 increases to a new record.

Original entry on oeis.org

0, 1, 4, 6, 10, 15, 16, 24, 31, 38, 45, 48, 52, 57, 70, 76, 79, 106, 111, 117, 123, 134, 139, 146, 154, 163, 169, 176, 179, 193, 202, 204, 223, 233, 238, 243, 256, 278, 284, 318, 326, 336, 359, 369, 412, 419, 430, 456, 458, 468, 479, 517, 550, 564, 595, 601, 612

Offset: 1

Bernard Schott, Apr 08 2020

nonn

Legendre's conjecture (still open) states that for n > 0 there is always a prime between n^2 and (n+1)^2. The number of primes between n^2 and (n+1)^2 is equal to A014085(n), so, the corresponding records are given by A014085(a(n)) = 0, 2, 3, 4, 5, 6, 7, 9, 10, 12, 13, ... (A349996).
m = 25 is the smallest number such that there are exactly 8 primes between m^2 = 625 et (m+1)^2 = 676, namely {631, 641, 643, 647, 653, 659, 661, 673} but there are 9 primes between 24^2 = 576 et 25^2 = 625, namely {577, 587, 593, 599, 601, 607, 613, 617, 619} so 24 is a term but not 25; hence, 25 is the first term of A076957 that is not a record.
This sequence is infinite. Suppose for contradiction that a(n) = k was the last term, with s primes between k^2 and (k+1)^2. Then there are at most s primes between (k+1)^2 and (k+2)^2, at most s primes between (k+2)^2 and (k+3)^3, and at most s*sqrt(x) + pi(k^2) primes up to x. But there are ~ x/log x primes up to x by the Prime Number Theorem, a contradiction. This can be made sharp with various explicit estimates. - Charles R Greathouse IV, Apr 10 2020

			There are 7 primes between 16^2 and 17^2, i.e., 256 and 289, which are 257, 263, 269, 271, 277, 281, 283, and there does not exist k < 16 with 7 or more primes between k^2 and (k+1)^2, hence, 16 is in the sequence.

Hugo Pfoertner, Table of n, a(n) for n = 1..2533
Mac Tutor History of Mathematics, Adrien-Marie Legendre
Wikipedia, Legendre's conjecture

Cf. A014085, A076957, A084597.
Cf. A333241 (Similar records between k and (9/8)*k).
Cf. A349996, A349997, A349998, A349999.

primeCount[n_] := PrimePi[(n + 1)^2] - PrimePi[n^2]; pmax = -1; seq = {}; Do[p = primeCount[n]; If[p > pmax, pmax = p; AppendTo[seq, n]], {n, 0, 612}]; seq (* Amiram Eldar, Apr 08 2020 *)

print1(pr=0,", ");pp=0;for(k=1,650,my(pc=primepi(k*k));if(pc-pp>pr,print1(k-1,", ");pr=pc-pp);pp=pc) \\ Hugo Pfoertner, Apr 10 2020

More terms from Michel Marcus, Apr 08 2020

cp's OEIS Frontend

A061235 Number of primes between n^4 and (n+1)^4.

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A062517 Number of primes between n^5 and (n+1)^5.

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A076957 Smallest k such that there are exactly n primes strictly between k^2 and (k+1)^2.

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A091591 Number of pairs of twin primes between n^2 and (n+1)^2.

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A117896 Number of perfect powers between consecutive squares n^2 and (n+1)^2.

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A144831 (n+1)^2 - (smallest prime > n^2).

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A161622 Denominators of the ratios (in lowest terms) of numbers of primes in one square interval to that of the interval and its successor.

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