A262403 -id:A262403 - cp's OEIS frontend

A262408 Positive integers m such that pi(m^2) = pi(j^2) + pi(k^2) for no 0 < j <= k < m.

1, 2, 5, 10, 21, 23, 46, 103, 105, 193, 222, 232, 285, 309, 345, 392, 404, 476, 587, 670, 779, 912, 1086, 1162, 1249, 2508, 2592, 2852, 2964, 3362, 3673, 3895, 4218, 4732, 5452, 6417, 7667, 7759, 8430, 8796, 9606, 11096, 11953, 12014, 12125, 13956, 14474, 15018, 17854, 18861, 18879, 19307, 22843, 28106, 29423, 31576, 37182

Offset: 1

Zhi-Wei Sun, Sep 21 2015

nonn

Conjecture: (i) There are infinitely many positive integers m such that pi(m^2) = pi(x^2) + pi(y^2) for some 0 < x <= y < m. Also, the current sequence has infinitely many terms.
(ii) For every n = 4,5,... the equation pi(x^n) + pi(y^n) = pi(z^n) has no integral solution with 0 < x <= y < z.
It is interesting to compare this conjecture with Fermat's Last Theorem. See also A262409 for the equation pi(x^3) + pi(y^3) = pi(z^3).

			a(3) = 5 since pi(5^2) = 9 cannot be written as pi(j^2) + pi(k^2) with 0 < j <= k < 5. Note that pi(1^2) = 0, pi(2^2) = 2, pi(3^2) = 4 and pi(4^2) = 6 are all even.

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.

Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.

Cf. A000720, A038107, A262403, A262409.

f[n_]:=PrimePi[n^2]
T[n_]:=Table[f[k],{k,1,n}]
n=0;Do[Do[If[MemberQ[T[m-1],f[m]-f[k]],Goto[aa]],{k,1,m-1}];n=n+1;Print[n," ",m];Label[aa];Continue,{m,1,32000}]

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

Zhi-Wei Sun, Sep 22 2015

nonn

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

			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).

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.

Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.

Cf. A000720, A000578, A262403, A262408, A262443.

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}]

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

A262995 Number of ordered pairs (k,m) with k > 0 and m > 0 such that n = pi(k(k+1)/2) + pi(1+m(m+1)/2), where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Oct 07 2015

nonn

Conjecture: a(n) > 0 for all n > 0.
This has been verified for n up to 10^5. See also A262999 for a similar conjecture.
By Chebyshev's inequality, pi(n*(n+1)/2) > n-1 for all n > 1.
In A262403 and A262439, the author conjectured that the sequences pi(n*(n+1)/2) (n = 1,2,3,...) and pi(1+n*(n+1)/2) (n = 1,2,3,...) are both strictly increasing.

			a(1) = 1 since 1 = pi(1*2/2) + pi(1+1*2/2).
a(2) = 1 since 2 = pi(1*2/2) + pi(1+2*3/2).
a(3) = 1 since 3 = pi(2*3/2) + pi(1+1*2/2).
a(4) = 3 since 4 = pi(1*2/2) + pi(1+3*4/2) = pi(2*3/2) + pi(1+2*3/2) = pi(3*4/2) + pi(1+1*2/2).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000217, A000720, A111208, A262403, A262439, A262446, A262999.

s[k_]:=s[k]=PrimePi[k(k+1)/2+1]
t[n_]:=t[n]=PrimePi[n(n+1)/2]
Do[r=0;Do[If[s[k]>n,Goto[bb]];Do[If[t[j]>n-s[k],Goto[aa]];If[t[j]==n-s[k],r=r+1];Continue,{j,1,n-s[k]+1}];Label[aa];Continue,{k,1,n}];Label[bb];Print[n," ",r];Continue,{n,1,100}]

A263020 Number of ordered pairs (k, m) with k > 0 and m > 0 such that n = pi(k(k+1)/2) + pi(m(3*m-1)/2), where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

0, 1, 2, 1, 2, 2, 2, 3, 3, 1, 5, 2, 2, 5, 2, 3, 4, 2, 6, 1, 5, 3, 3, 5, 2, 4, 5, 2, 4, 5, 1, 6, 5, 2, 6, 4, 3, 5, 4, 5, 3, 6, 4, 4, 4, 5, 4, 5, 4, 5, 6, 2, 3, 7, 5, 3, 6, 5, 2, 3, 8, 5, 3, 5, 5, 6, 5, 1, 8, 8, 2, 4, 6, 6, 3, 5, 8, 4, 4, 5, 3, 9, 2, 6, 8, 3, 3, 6, 4, 7, 3, 6, 6, 5, 5, 5, 3, 7, 6, 6

Offset: 1

Zhi-Wei Sun, Oct 07 2015

nonn

Conjecture: a(n) > 0 for all n > 1.
We have verified this for n up to 3*10^5. It seems that a(n) = 1 only for n = 2, 4, 10, 20, 31, 68, 147, 252, 580, 600, 772, 1326, 1381, 2779, 3136, 3422, 3729, 7151, 9518, 13481, 18070, 18673, 36965, 48181, 69250, 91130, 93580, 99868.
Note that n*(n+1)/2 (n = 0,1,2,...) are the triangular numbers while n*(3n-1)/2 (n = 0,1,2,...) are the pentagonal numbers.

			a(2) = 1 since 2 = 2 + 0 = pi(2*3/2) + pi(1*(3*1-1)/2).
a(4) = 1 since 4 = 4 + 0 = pi(4*5/2) + pi(1*(3*1-1)/2).
a(10) = 1 since 10 = 2 + 8 = pi(2*3/2) + pi(4*(3*4-1)/2).
a(20) = 1 since 20 = 9 + 11 = pi(7*8/2) + pi(5*(3*5-1)/2).
a(31) = 1 since 31 = 16 + 15 = pi(10*11/2) + pi(6*(3*6-1)/2).
a(68) = 1 since 68 = 2 + 66 = pi(2*3/2) + pi(15*(3*15-1)/2).
a(147) = 1 since 147 = pi(31*32/2) + pi(13*(3*13-1)/2).
a(252) = 1 since 252 = pi(29*30/2) + pi(26*(3*26-1)/2).
a(580) = 1 since 580 = pi(5*6/2) + pi(53*(3*53-1)/2).
a(600) = 1 since 600 = pi(42*43/2) + pi(46*(3*46-1)/2).
a(772) = 1 since 772 = pi(107*108/2) + pi(6*(3*6-1)/2).
a(1326) = 1 since 1326 = pi(139*140/2) + pi(22*(3*22-1)/2).
a(1381) = 1 since 1381 = pi(145*146/2) + pi(18*(3*18-1)/2).
a(2779) = 1 since 2779 = pi(212*213/2) + pi(33*(3*33-1)/2).
a(3136) = 1 since 3136 = pi(147*148/2) + pi(102*(3*102-1)/2).
a(3422) = 1 since 3422 = pi(151*152/2) + pi(109*(3*109-1)/2).
a(3729) = 1 since 3729 = pi(29*30/2) + pi(151*(3*151-1)/2).
a(7151) = 1 since 7151 = pi(100*101/2) + pi(208*(3*208-1)/2).
a(9518) = 1 since 9518 = pi(82*83/2) + pi(250*(3*250-1)/2).
a(13481) = 1 since 13481 = pi(539*540/2) + pi(6*(3*6-1)/2).
a(18070) = 1 since 18070 = pi(632*633/2) + pi(17*(3*17-1)/2).
a(18673) = 1 since 18673 = 14493 + 4180 = pi(561*562/2) + pi(163*(3*163-1)/2).
a(36965) = 1 since 36965 = 3780 + 33185 = pi(266*267/2) + pi(511*(3*511-1)/2).
a(48181) = 1 since 48181 = 30755 + 17426 = pi(848*849/2) + pi(359*(3*359-1)/2).
a(69250) = 1 since 69250 = 20669 + 48581 = pi(682*683/2) + pi(629*(3*629-1)/2).
a(91130) = 1 since 91130 = 81433 + 9697 = pi(1442*1443/2) + pi(260*(3*260-1)/2).
a(93580) = 1 since 93580 = 91865 + 1715 = pi(1539*1540/2) + pi(99*(3*99-1)/2).
a(99868) = 1 since 99868 = 66079 + 33789 = pi(1287*1288/2) + pi(516*(3*516-1)/2).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Conjectures on the prime-counting function, a message to Number Theory Mailing List, Oct. 19, 2015.

Cf. A000217, A000326, A000720, A111208, A262403, A262995, A262999, A263001, A263107.

s[n_]:=s[n]=PrimePi[n(3n-1)/2]
t[n_]:=t[n]=PrimePi[n(n+1)/2]
Do[r=0;Do[If[s[k]>n,Goto[bb]];Do[If[t[j]>n-s[k],Goto[aa]];If[t[j]==n-s[k],r=r+1];Continue,{j,1,n-s[k]+1}];Label[aa];Continue,{k, 1, n}];Label[bb];Print[n," ",r];Continue,{n,1,100}]

A262439 Number of primes not exceeding 1+n*(n+1)/2.

Original entry on oeis.org

1, 2, 4, 5, 6, 8, 10, 12, 14, 16, 19, 22, 24, 27, 30, 33, 36, 39, 43, 47, 50, 54, 59, 62, 66, 70, 75, 79, 84, 90, 94, 99, 102, 108, 115, 121, 126, 131, 137, 142, 149, 154, 161, 167, 174, 180, 189, 193, 200, 205, 217, 220, 226, 235, 242, 251, 259, 267, 274, 282, 290, 297, 306, 313, 324, 329, 338, 348, 358, 367

Offset: 1

Zhi-Wei Sun, Sep 22 2015

nonn

Conjecture: (i) The sequence is strictly increasing, and also a(n)^(1/n) >  a(n+1)^(1/(n+1)) for all n = 3,4,....
(ii) The sequence is an addition chain. In other words, for each n = 2,3,... we have a(n) = a(k) + a(m) for some 0 < k <= m < n.
(iii) All the numbers Sum_{i=j..k} 1/a(i) with 0 < min{2,k} <= j <= k have pairwise distinct fractional parts.
See also A262446 related to part (ii) of this conjecture.
Concerning part (ii) of the conjecture, Neill Clift verified in 2024 that for all 1 < n <= 2^24 = 16777216 we have a(n) = a(k) + a(m) for some 0 < k <= m < n. - Zhi-Wei Sun, Jan 29 2024

			a(3) = 4 since there are exactly four primes (namely, 2, 3, 5, 7) not exceeding 1 + 3*4/2 = 7.

R. K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004. (Cf. Section C6 on addition chains.)
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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Neill Clift, Prime Count and Addition Chains, 2024.
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.

Cf. A000040, A000217, A000720, A262403, A262446.

a[n_]:=PrimePi[1+n(n+1)/2]
Do[Print[n," ",a[n]],{n,1,70}]

A262446 Number of ways to write A262439(n) = A262439(k) + A262439(m) with 0 < k < m < n.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Sep 23 2015

nonn

Conjecture: a(n) > 0 for all n > 3, and a(n) = 1 only for n = 4, 6, 11, 21, 54, 253, 325.
This is slightly stronger than part (ii) of the conjecture in A262439.
I have verified the conjecture for n up to 10^5. - Zhi-Wei Sun, Sep 27 2015

			a(4) = 1 since pi(4*5/2+1) = pi(11) = 5 = 1 + 4 = pi(2) + pi(7) = pi(1*2/2+1) + pi(3*4/2+1).
a(6) = 1 since pi(6*7/2+1) = pi(22) = 8 = 2 + 6 = pi(4) + pi(16) = pi(2*3/2+1) + pi(5*6/2+1).
a(11) = 1 since pi(11*12/2+1) = pi(67) = 19 = 5 + 14 = pi(11) + pi(46) = pi(4*5/2+1) + pi(9*10/2+1).
a(21) = 1 since pi(21*22/2+1) = pi(232) = 50 = 14 + 36 = pi(46) + pi(154) = pi(9*10/2+1) + pi(17*18/2+1).
a(54) = 1 since pi(54*55/2+1) = pi(1486) = 235 = 30 + 205 = pi(121) + pi(1276) = pi(15*16/2+1) + pi(50*51/2+1).
a(253) = 1 since pi(253*254/2+1) = pi(32132) = 3447 = 747 + 2700 = pi(5672) + pi(24311) = pi(106*107/2+1) + pi(220*221/2+1).
a(325) = 1 since pi(325*326/2+1) = pi(52976) = 5406 = 1446 + 3960 = pi(12091) + pi(37402) = pi(155*156/2+1) + pi(37402*37403/2+1).

R. K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, 2004. (Cf. Section C6 on addition chains.)
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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Neil Clift, Prime Count and Addition Chains, 2024.
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641 [math.NT], 2014.

Cf. A000217, A000720, A262403, A262439.

f[n_]:=PrimePi[n(n+1)/2+1]
T[n_]:=Table[f[k],{k,1,n}]
Do[r=0;Do[If[2*f[k]>=f[n],Goto[aa]];If[MemberQ[T[n],f[n]-f[k]],r=r+1];Continue,{k,1,n-1}];Label[aa];Print[n," ",r];Continue,{n,1,100}]

cp's OEIS Frontend

A262408 Positive integers m such that pi(m^2) = pi(j^2) + pi(k^2) for no 0 < j <= k < m.

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

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A262995 Number of ordered pairs (k,m) with k > 0 and m > 0 such that n = pi(k*(k+1)/2) + pi(1+m*(m+1)/2), where pi(x) denotes the number of primes not exceeding x.

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A263020 Number of ordered pairs (k, m) with k > 0 and m > 0 such that n = pi(k*(k+1)/2) + pi(m*(3*m-1)/2), where pi(x) denotes the number of primes not exceeding x.

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A262439 Number of primes not exceeding 1+n*(n+1)/2.

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A262446 Number of ways to write A262439(n) = A262439(k) + A262439(m) with 0 < k < m < n.

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A262995 Number of ordered pairs (k,m) with k > 0 and m > 0 such that n = pi(k(k+1)/2) + pi(1+m(m+1)/2), where pi(x) denotes the number of primes not exceeding x.

A263020 Number of ordered pairs (k, m) with k > 0 and m > 0 such that n = pi(k(k+1)/2) + pi(m(3*m-1)/2), where pi(x) denotes the number of primes not exceeding x.