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 1-5 of 5 results.

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

Views

Author

Zhi-Wei Sun, Oct 07 2015

Keywords

Comments

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.

Examples

			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).
		

Crossrefs

Programs

  • Mathematica
    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}]

A263001 Number of ordered pairs (k, m) with k > 0 and m > 0 such that n = pi(k*(k+1)) + pi(m*(m+1)/2), where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 0, 2, 1, 3, 1, 3, 2, 3, 3, 3, 4, 3, 4, 2, 5, 4, 2, 7, 2, 4, 5, 2, 7, 2, 5, 4, 4, 5, 3, 5, 6, 4, 5, 6, 3, 6, 6, 2, 9, 3, 5, 5, 5, 6, 5, 6, 5, 4, 7, 4, 7, 4, 5, 6, 7, 3, 5, 6, 7, 4, 7, 7, 5, 3, 9, 5, 7, 3, 8, 7, 5, 4, 8, 6, 6, 3, 10, 7, 3, 3, 11, 5, 7, 4, 8, 5, 4, 7, 7, 5, 8, 3, 8, 7, 4, 5, 9, 6, 9
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 07 2015

Keywords

Comments

Conjecture: a(n) > 0 for all n > 2, and a(n) = 1 only for n = 1, 4, 6.
We have verified this for n up to 10^5.
See also A262995, A262999 and A263020 for similar conjectures.

Examples

			a(1) = 1 since 1 = pi(1*2) + pi(1*2/2).
a(4) = 1 since 4 = pi(1*2) + pi(3*4/2).
a(6) = 1 since 6 = pi(2*3) + pi(3*4/2).
		

Crossrefs

Programs

  • Mathematica
    s[n_]:=s[n]=PrimePi[n(n+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

Views

Author

Zhi-Wei Sun, Oct 07 2015

Keywords

Comments

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.

Examples

			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).
		

Crossrefs

Programs

  • Mathematica
    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}]

A263100 Number of ordered pairs (k, m) with k > 0 and m > 0 such that n = pi(k^2) + pi(m^2/2), where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 2, 1, 3, 2, 3, 3, 2, 4, 2, 6, 2, 5, 2, 5, 4, 4, 4, 4, 5, 3, 5, 5, 4, 4, 6, 6, 1, 7, 4, 6, 4, 4, 7, 6, 4, 5, 5, 5, 6, 6, 4, 6, 3, 7, 6, 5, 6, 6, 6, 5, 5, 6, 4, 7, 8, 4, 3, 10, 2, 6, 6, 6, 6, 7, 5, 5, 9, 3, 6, 8, 6, 7, 5, 5, 6, 7, 7, 8, 3, 9, 3, 10, 2, 7, 9, 7, 2, 7, 8, 5, 8, 4, 6, 9, 5, 7, 6, 5, 7
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 09 2015

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 28.
(ii) Any integer n > 0 can be written as pi(k^2) + pi((m^2+1)/2) with k and m positive integers.
(iii) Each n = 1,2,3,... can be written as pi(k^2/2) + pi((m^2+1)/2) with k and m positive integers.
See also A262995, A262999, A263001 and A263020 for similar conjectures.

Examples

			a(1) = 1 since 1 = 0 + 1 = pi(1^2) + pi(2^2/2).
a(3) = 1 since 3 = 2 + 1 = pi(2^2) + pi(2^2/2).
a(28) = 1 since 28 = 11 + 17 = pi(6^2) + pi(11^2/2).
		

Crossrefs

Programs

  • Mathematica
    s[n_]:=s[n]=PrimePi[n^2]
    t[n_]:=t[n]=PrimePi[n^2/2]
    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}]

A263107 Number of ordered pairs (k, m) with k > 0 and m > 0 such that n = pi(k^2/2) + pi(3*m^2/2), where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 1, 4, 5, 1, 6, 3, 4, 5, 2, 4, 5, 2, 5, 5, 3, 4, 5, 4, 6, 2, 4, 4, 6, 3, 4, 5, 2, 6, 8, 1, 5, 6, 2, 4, 7, 3, 6, 5, 2, 6, 7, 1, 4, 7, 3, 6, 5, 3, 7, 4, 5, 5, 6, 5, 5, 4, 2, 6, 10, 3, 4, 6, 5, 3, 6, 5, 7, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 09 2015

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
We have verified this for n up to 3*10^5.
It seems that a(n) = 1 only for n = 1, 2, 3, 6, 19, 22, 48, 60, 396, 1076, 3033, 3889, 4741, 6804, 7919, 9604, 16938, 19169, 32533, 59903, 100407.

Examples

			a(1) = 1 since 1 = 1 + 0 = pi(2^2/2) + pi(3*1^2/2).
a(2) = 1 since 2 = 2 + 0 = pi(3^2/2) + pi(3*1^2/2).
a(3) = 1 since 3 = 0 + 3 = pi(1^2/2) + pi(3*2^2/2).
a(6) = 1 since 6 = 0 + 6 = pi(1^2/2) + pi(3*3^2/2).
a(19) = 1 since 19 = 7 + 12 = pi(6^2/2) + pi(3*5^2/2).
a(22) = 1 since 22 = 1 + 21 = pi(2^2/2) + pi(3*7^2/2).
a(48) = 1 since 48 = 1 + 47 = pi(2^2/2) + pi(3*12^2/2).
a(60) = 1 since 60 = 25 + 35 = pi(14^2/2) + pi(3*10^2/2).
a(396) = 1 since 396 = 334 + 62 = pi(67^2/2) + pi(3*14^2/2).
a(1076) = 1 since 1076 = 47 + 1029 = pi(21^2/2) + pi(3*74^2/2).
a(3033) = 1 since 3033 = 7 + 3026 = pi(6^2/2) + pi(3*136^2/2).
a(3889) = 1 since 3889 = 1808 + 2081 = pi(176^2/2) + pi(3*110^2/2).
a(4741) = 1 since 4741 = 4699 + 42 = pi(301^2/2) + pi(3*11^2/2).
a(6804) = 1 since 6804 = 6047 + 757 = pi(346^2/2) + pi(3*62^2/2).
a(7919) = 1 since 7919 = 4049 + 3870 = pi(277^2/2) + pi(3*156^2/2).
a(9604) = 1 since 9604 = 4754 + 4850 = pi(303^2/2) + pi(3*177^2/2).
a(16938) = 1 since 16938 = 2223 + 14715 = pi(198^2/2) + pi(3*327^2/2).
a(19169) = 1 since 19169 = 6510 + 12659 = pi(361^2/2) + pi(3*301^2/2).
a(32533) = 1 since 32533 = 1768 + 30765 = pi(174^2/2) + pi(3*490^2/2).
a(59903) = 1 since 59903 = 59210 + 693 = pi(1213^2/2) + pi(3*59^2/2).
a(100407) = 1 since 100407 = 7554 + 92853 = pi(392^2/2) + pi(3*894^2/2).
		

Crossrefs

Programs

  • Mathematica
    s[n_]:=s[n]=PrimePi[3n^2/2]
    t[n_]:=t[n]=PrimePi[n^2/2]
    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}]
Showing 1-5 of 5 results.