cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-8 of 8 results.

A038107 Number of primes < n^2.

Original entry on oeis.org

0, 0, 2, 4, 6, 9, 11, 15, 18, 22, 25, 30, 34, 39, 44, 48, 54, 61, 66, 72, 78, 85, 92, 99, 105, 114, 122, 129, 137, 146, 154, 162, 172, 181, 191, 200, 210, 219, 228, 240, 251, 263, 274, 283, 295, 306, 319, 329, 342, 357, 367, 378, 393, 409, 421, 434, 445, 457, 474
Offset: 0

Views

Author

Joe K. Crump (joecr(AT)carolina.rr.com)

Keywords

Comments

Also number of primes <= n^2 since n^2 is not prime.
Also the number of primes contained within an n X n square spiral. - William A. Tedeschi, Mar 03 2008
For large n, these numbers closely approximate the sum of primes less than n. For example, n = 10^10, sum of primes < n = 2220822432581729238. The number of primes < (10^10)^2 = 10^20 = 2220819602560918840. The error is 0.0000012743... The derivation of this is in the link Sum of Primes. - Cino Hilliard, Jun 09 2008
a(n) - A000720(n) = A073882(n) - A010051(n) = A117490(n). - Reinhard Zumkeller, May 20 2010
A061265(a(n)) = 1 for n > 1. - Reinhard Zumkeller, Apr 15 2013
From Zhi-Wei Sun, Feb 17 2014: (Start)
Conjecture:
(i) The sequence a(n)^(1/n) (n = 3, 4, ...) is strictly decreasing (to the limit 1).
(ii) If n > 0 is not among 25, 35, 44, 46, 105, then the interval [a(n), a(n+1)] contains at least one prime. (End)
A classical conjecture of Legendre asserts that a(n) < a(n+1) for all n > 0.
Conjecture: All the numbers Sum_{i=j,...,k} 1/a(i) with 1 < j <= k have pairwise distinct fractional parts. - Zhi-Wei Sun, Sep 24 2015

Examples

			a(2)=2 because the only primes < 4 are 2 and 3.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187. (See Conjectures 2.14-2.16.)

Crossrefs

Cf. A014085 (first differences), A111208, A194189, A262408, A262443, A262447, A262462.

Programs

  • Haskell
    a038107 0 = 0
    a038107 n = a000720 $ a000290 n
    -- Reinhard Zumkeller, Apr 15 2013, Nov 01 2011
    
  • Maple
    A038107 := proc(n) numtheory[pi]( n^2) ; end: seq(A038107(n),n=0..100) ; # R. J. Mathar, Jun 22 2009
  • Mathematica
    Table[PrimePi[n^2], {n, 0, 100}] (* Ray Chandler, Oct 22 2005 *)
  • PARI
    a(n)=primepi(n^2) \\ Charles R Greathouse IV, Apr 26 2012
  • Sage
    [prime_pi(n^2) for n in range(0, 59)] # Zerinvary Lajos, Jun 06 2009
    

Formula

a(n) = A000720(A000290(n)).
a(n) ~ 1/2 * n^2/log n. - Charles R Greathouse IV, Apr 26 2012

Extensions

Extended by Ray Chandler, Oct 22 2005

A262409 Positive integers m such that pi(m^3) = pi(j^3) + pi(k^3) for some 0 < j <= k < m.

Original entry on oeis.org

4, 89, 97, 101, 110, 196, 237, 372, 410, 1457, 2522, 3327, 4244, 4437, 5684, 5777, 7647, 8827, 9608, 9680, 9807, 10744, 17563, 19146, 21208, 23188, 27153, 28286, 34086, 35443, 40057, 49338, 49613, 54425, 55360, 56906, 61304, 69147, 69515, 73694, 84508, 95674
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 22 2015

Keywords

Comments

Conjecture: The Diophantine equation pi(x^3) + pi(y^3) = pi(z^3) with 0 < x <= y < z has infinitely many solutions.
The 25 terms we have found yield the following 25 solutions to the equation: (x,y,z) = (3,3,4), (54,80,89), (63,85,97), (27,100,101), (47,106,110), (80,190,196), (122,223,237), (229,335,372), (151,401,410), (263,1453,1457), (1302,2382,2522), (879,3301,3327), (2190,4011,4244), (498,4434,4437), (3792,4991,5684), (4496,4584,5777), (3113,7442,7647), (5239,8090,8827), (6904,8116,9608), (5659,8910,9680), (5323,9187,9807), (5527,10168,10744), (7395,17050,17563), (11637,17438,19146), (4486,21125,21208).
See also the conjecture in A262408 involving the n-th powers with n = 2,4,5,....
Solution triples (x,y,z) corresponding to a(n) for n = 26..42: (16440, 19774, 23188), (4775, 27091, 27153), (10708, 27687, 28286), (25272, 28248, 34086), (6302, 35360, 35443), (3941, 40040, 40057), (16336, 48639, 49338), (33631, 43365, 49613), (6206, 54390, 54425), (6741, 55317, 55360), (28160, 54247, 56906), (25339, 59637, 61304), (41473, 63300, 69147), (27684, 67825, 69515), (29690, 71841, 73694), (65989, 67172, 84508), (55781, 88294, 95674) - Chai Wah Wu, May 24 2018

Examples

			a(1) = 4 since pi(4^3) = pi(64) = 18 = 9 + 9 = pi(27) + pi(27) = pi(3^3) + pi(3^3).
a(2) = 89 since pi(89^3) = 56924 = 14479 + 42445 = pi(157464) + pi(512000) = pi(54^3) + pi(80^3).
a(22) = 10744 since pi(10744^3) = pi(1240217910784) = 46266787130 = 6805722064 + 39461065066 = pi(168837298183) + pi(1051251461632) = pi(5527^3) + pi(10168^3).
a(23) = 17563 since pi(17563^3) = pi(5417464872547) = 191548794617 = 15745791385 + 175803003232 = pi(404403154875) + pi(4956477625000) = pi(7395^3) + pi(17050^3).
a(24) = 19146 since pi(19146^3) = pi(7018336124136) = 245897610272 = 58267274193 + 187630336079 = pi(1575879851853) + pi(5302614071672) = pi(11637^3) + pi(17438^3).
a(25) = 21208 since pi(21208^3) = pi(9538918630912) = 330649999352 = 3733416265 + 326916583087 = pi(90277143256) + pi(9427361328125) = pi(4486^3) + pi(21125^3).
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[n_]:=PrimePi[n^3]
    T[1]:={0}
    T[n_]:=Union[T[n-1],{f[n]}]
    Do[n=0;Do[If[MemberQ[T[m-1],f[m]-f[k]],n=n+1;Print[n," ",m];Goto[aa]],{k,1,m-1}];Label[aa];Continue,{m,1,21350}]

Extensions

a(26)-a(42) from Chai Wah Wu, May 24 2018

A262447 Primes p such that pi(p^2) = pi(q^2) + pi(r^2) for some distinct primes q and r.

Original entry on oeis.org

13, 53, 73, 131, 199, 277, 281, 283, 313, 353, 641, 643, 647, 701, 773, 839, 887, 977, 1033, 1103, 1117, 1163, 1187, 1223, 1259, 1409, 1433, 1439, 1487, 1489, 1583, 1721, 1913, 1931, 2239, 2243, 2269, 2309, 2371, 2441, 2473, 2477, 2621, 2683, 2707, 2797, 2843, 2851, 2953, 3049, 3137, 3257, 3307, 3499, 3511, 3613, 3659, 3769, 3779, 3911
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 23 2015

Keywords

Comments

Conjecture: The sequence has infinitely many terms.
See also A262408 and A262443 for related conjectures.

Examples

			a(1) = 13 since pi(13^2) = pi(169) = 39 = 9 + 30 = pi(5^2) + pi(11^2) with 13, 5 and 11 distinct primes.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[n_]:=PrimePi[Prime[n]^2]
    T[n_]:=Table[f[k],{k,1,n}]
    n=0;Do[Do[If[2*f[k]>=f[m],Goto[aa]];If[MemberQ[T[m-1],f[m]-f[k]],n=n+1;Print[n," ",Prime[m]];Goto[aa]];Continue,{k,1,m-1}];Label[aa];Continue,{m,1,541}]

A262462 Positive integers k with pi(k^3) a square, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 2, 3, 14, 1122
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 23 2015

Keywords

Comments

Conjecture: (i) The Diophantine equation pi(x^2) = y^2 with x > 0 and y > 0 has infinitely many solutions.
(ii) The only solutions to the Diophantine equation pi(x^m) = y^n with {m,n} = {2,3}, x > 0 and y > 0 are as follows:
pi(89^2) = 10^3, pi(2^3) = 2^2, pi(3^3) = 3^2, pi(14^3) = 20^2 and pi(1122^3) = 8401^2.
(iii) For m > 1 and n > 1 with m + n > 5, the equation pi(x^m) = y^n with x > 0 and y > 0 has no integral solution.
The conjecture seems reasonable in view of the heuristic arguments.
Part (ii) of the conjecture implies that the only terms of the current sequence are 1, 2, 3, 14 and 1122.

Examples

			a(1) = 1 since pi(1^3) = 0^2.
a(2) = 2 since pi(2^3) = 2^2.
a(3) = 3 since pi(3^3) = 3^2.
a(4) = 14 since pi(14^3) = pi(2744) = 400 = 20^2.
a(5) = 1122 since pi(1122^3) = pi(1412467848) = 70576801 = 8401^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]]
    f[n_]:=PrimePi[n^3]
    n=0;Do[If[SQ[f[k]],n=n+1;Print[n," ",k]],{k,1,1200}]
    Select[Range[1200],IntegerQ[Sqrt[PrimePi[#^3]]]&] (* Harvey P. Dale, Aug 21 2024 *)

A262536 Positive integers z such that pi(x^3+y^3) = pi(z^3) for some 0 < x <= y < z, where pi(m) denotes the number of primes not exceeding m.

Original entry on oeis.org

7, 9, 11, 12, 34, 46, 49, 65, 89, 95, 103, 127, 144, 150, 163, 172, 206, 236, 249, 258, 275, 288, 300, 309, 312, 385, 492, 495, 505, 577, 641, 683, 729, 738, 751, 796, 835, 873, 904, 990, 995, 1010, 1154, 1210, 1297, 1312, 1403, 1458, 1476, 1502, 1544, 1626, 1661, 1731, 1808, 1852, 1985, 1988, 2020, 2059, 2107, 2214, 2304, 2316, 2370, 2448, 2594, 2833, 2840, 2883, 2920, 3073, 3088, 3097
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 24 2015

Keywords

Comments

The sequence has infinitely many terms. In fact, for any integer t > 0, we have (6*t^2)^3 + (6*t^3-1)^3 = (6*t^3+1)^3 - 2 and hence pi((6*t^2)^3+(6*t^3-1)^3) = pi((6*t^3+1)^3) since neither (6*t^3+1)^3 nor (6*t^3+1)^3-1 is prime.
Concerning the equation pi(x^3+y^3) = pi(z^3) with 0 < x <= y < z, there are exactly 70 solutions with z <= 2700. They are (x,y,z) = (5,6,7),(6,8,9),(7,10,11),(9,10,12),(15,33,34),(23,44,46),(24,47,49),(43,58,65),(41,86,89),(47,91,95),(64,94,103),(95,106,127),(71,138,144),(73,144,150),(54,161,163),(135,138,172),(128,188,206),(55,235,236),(135,235,249),(197,212,258),(159,256,275),(142,276,288),(146,288,300),(192,282,309),(161,297,312),(96,383,385),(252,345,385),(390,391,492),(334,438,495),(372,426,505),(426,486,577),(297,619,641),(353,650,683),(242,720,729),(244,729,738),(150,749,751),(602,659,796),(161,833,835),(470,825,873),(566,823,904),(668,876,990),(514,947,995),(744,852,1010),(791,812,1010),(509,1120,1154),(852,972,1154),(236,1207,1210),(216,1295,1297),(459,1293,1312),(915,1259,1403),(484,1440,1458),(488,1458,1476),(300,1498,1502),(368,1537,1544),(511,1609,1626),(420,1652,1661),(1278,1458,1731),(1132,1646,1808),(1033,1738,1852),(1241,1808,1985),(1010,1897,1988),(1582,1624,2020),(294,2057,2059),(237,2106,2107),(732,2187,2214),(575,2292,2304),(577,2304,2316),(1518,2141,2370),(1611,2189,2448),(432,2590,2594).
Recall Fermat's Last Theorem, which asserts that the Diophantine equation x^n + y^n = z^n with n > 2 and x,y,z > 0 has no solution. In 1936 K. Mahler discovered that
(9*t^3+1)^3 + (9*t^4)^3 - (9*t^4+3*t)^3 = 1.
Conjecture: (i) For any integers n > 3 and x,y,z > 0 with {x,y} not equal to {1,z}, we have |x^n+y^n-z^n| >= 2^n-2, unless n = 5, {x,y} = {13,16} and z = 17.
(ii) For any integer n > 3 and x,y,z > 0 with {x,y} not containing z, there is a prime p with x^n+y^n < p < z^n or z^n < p < x^n+y^n, unless n = 5, {x,y} = {13,16} and z = 17.
(iii) For any integers n > 3, x > y >= 0 and z > 0 with x not equal to z, there always exists a prime p with x^n-y^n < p < z^n or z^n < p < x^n-y^n.
We have verified part (i) of the conjecture for n = 4..10 and 0 < x,y,z <= 1700.

Examples

			a(1) = 7 since pi(5^3+6^3) = pi(125+216) = pi(341) = 68 = pi(343) = pi(7^3).
a(2) = 9 since pi(6^3+8^3) = pi(216+512) = pi(728) = 129 = pi(729) = pi(9^3).
a(50) = 1502 since pi(300^3+1498^3) = pi(27000000+3361517992) = pi(3388517992) = 162202081 = pi(3388518008) = pi(1502^3).
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    pi[n_]:=PrimePi[n]
    n=0;Do[Do[If[pi[x^3+y^3]==pi[z^3],n=n+1;Print[n," ",z];Goto[aa]],{x,1,z-1},{y,x,z-1}];Label[aa];Continue,{z,1,2700}]

A262707 Positive integers m such that pi(k^2)*pi(m^2) is a square for some 1 < k < m, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

5, 8, 10, 14, 16, 19, 23, 31, 35, 39, 45, 63, 65, 66, 68, 71, 74, 82, 87, 92, 94, 115, 130, 145, 151, 162, 172, 204, 250, 279, 292, 304, 334, 391, 413, 415, 418, 449, 451, 454, 461, 499, 514, 524, 552, 557, 626, 664, 676, 683, 691, 706, 708, 724, 763, 766, 846, 848, 858, 866
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 27 2015

Keywords

Comments

Conjecture: The sequence has infinitely many terms.
See also A262700 for a related conjecture.

Examples

			a(2) = 8 since pi(8^2)*pi(2^2) = 18*2 = 6^2.
a(3) = 10 since pi(10^2)*pi(3^2) = 25*4 = 10^2.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[n_]:=PrimePi[n^2]
    SQ[n_]:=IntegerQ[Sqrt[n]]
    n=0;Do[Do[If[SQ[f[x]*f[y]],n=n+1;Print[n," ",y];Goto[aa]],{x,2,y-1}];Label[aa];Continue,{y,1,870}]

A262700 Primes p such that pi(p^2)*pi(q^2) is a square for some prime q < p, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

5, 19, 31, 151, 691, 1181, 1489, 1511, 1601, 2579, 3037, 7297, 9661, 10993, 11699, 20407, 25657, 33937, 65099, 96419, 102911, 133157, 251789, 411841, 417271, 670729, 808211, 1179907, 1671277
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 27 2015

Keywords

Comments

Conjecture: (i) The sequence has infinitely many terms.
(ii) The Diophantine equation pi(x^n)*pi(y^n) = z^n with n > 2 and x,y,z > 0 has no solution.

Examples

			a(1) = 5 since pi(5^2)*pi(3^2) = 9*4 = 6^2 with 5 and 3 both prime.
a(2) = 19 since pi(19^2)*pi(2^2) = 72*2 = 12^2 with 19 and 2 both prime.
a(21) = 102911 since pi(102911^2)*pi(919^2) = pi(10590673921)*pi(844561) = 480670430*67230 = 32315473008900 = 5684670^2 with 102911 and 919 both prime.
a(22) = 133157 since pi(133157^2)*pi(19^2) = pi(17730786649)*pi(361) = 786299168*72 = 56613540096 = 237936^2 with 133157 and 19 both prime.
a(23) = 251789 since pi(251789^2)*pi(10513^2) = pi(63397700521)*pi(110523169) = 2660789341*6331444 = 16846638708338404 = 129794602^2 with 251789 and 10513 both prime.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[n_]:=PrimePi[Prime[n]^2]
    SQ[n_]:=IntegerQ[Sqrt[n]]
    n=0;Do[Do[If[SQ[f[k]*f[m]],n=n+1;Print[n, " ", Prime[m]];Goto[aa]],{k,1,m-1}];Label[aa];Continue,{m,2,22200}]

Extensions

a(24)-a(29) from Chai Wah Wu, Aug 21 2019

A262722 Positive integers m such that pi(k^3+m^3) is a cube for some k = 1..m, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 41, 56, 74, 103, 157, 384, 491, 537, 868, 1490, 1710, 4322, 4523, 4877, 4942, 5147, 5407, 7564, 17576, 67722, 131455, 220641, 438895, 443475, 553878, 571473, 625611
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 28 2015

Keywords

Comments

Conjecture: (i) There are infinitely many distinct primes p,q,r such that pi(p^2+q^2) = r^2.
(ii) The Diophantine equation pi(x^3+y^3) = z^3 with 0 < x <= y and z > 0 only has the following 13 solutions: (x,y,z) = (1,1,1), (5,41,19), (47,56,29), (28,74,33), (2,103,44), (3,103,44), (6,157,65), (235,384,160), (266,491,198), (91,537,206), (359,868,331), (783,1490,565), (1192,1710,677).
(iii) The Diophantine equation pi(x^n+y^n) = z^n with n > 3 and x,y,z > 0 has no solution.
Part (ii) of the conjecture implies that the current sequence only has 12 terms as shown here.
Conjecture (ii) is false as there are more terms beyond 1710. It is likely the sequence has an infinite number of terms. (x,y,z) for 1710 < y <= 7564: (1429, 4322, 1514), (1974, 4523, 1604), (3361, 4877, 1840), (3992, 4942, 1949), (3253, 5147, 1902), (971, 5407, 1859), (935, 7564, 2563). - Chai Wah Wu, Apr 12 2021
Solutions (x,y,z) for 7564 < y <= 67722: (3484, 17576, 5783), (25184, 67722, 21604). - Chai Wah Wu, Apr 17 2021
Solutions (x,y,z) for 67722 < y <= 625611: (61021, 131455, 41715), (93577, 220641, 68507), (394510, 438895, 155930), (3086, 443475, 131933), (338485, 553878, 175133), (239982, 571473, 172855), (610794, 625611, 228409). - Chai Wah Wu, Apr 26 2021

Examples

			a(2) = 41 since pi(5^3+41^3) = pi(125+68921) = pi(69046) = 6859 = 19^3.
		

References

  • Zhi-Wei Sun, Problems on combinatorial properties of primes, in: M. Kaneko, S. Kanemitsu and J. Liu (eds.), Number Theory: Plowing and Starring through High Wave Forms, Proc. 7th China-Japan Seminar (Fukuoka, Oct. 28 - Nov. 1, 2013), Ser. Number Theory Appl., Vol. 11, World Sci., Singapore, 2015, pp. 169-187.

Crossrefs

Programs

  • Mathematica
    f[x_,y_]:=PrimePi[x^3+y^3]
    CQ[n_]:=IntegerQ[n^(1/3)]
    n=0;Do[Do[If[CQ[f[x,y]],n=n+1;Print[n," ",y];Goto[aa]],{x,1,y}];Label[aa];Continue,{y,1,1800}]

Extensions

a(13)-a(19) from Chai Wah Wu, Apr 12 2021
a(20)-a(21) from Chai Wah Wu, Apr 17 2021
a(22)-a(28) from Chai Wah Wu, Apr 26 2021
Showing 1-8 of 8 results.