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 1-7 of 7 results.

A248004 Least positive integer m with prime(m*n) == 1 (mod m+n).

Original entry on oeis.org

3, 4, 1, 2, 2, 15, 1, 1, 5, 10, 2, 3, 4, 18, 6, 27, 4, 7, 35, 4, 45, 2, 47, 9, 5, 10, 16, 11, 3, 3, 9, 61, 1, 52, 3, 60, 53, 74, 8, 47, 7, 60, 128, 5, 21, 12, 2, 29, 15, 127, 53, 28, 17, 21, 303, 80, 72, 8, 61, 36
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 29 2014

Keywords

Comments

Conjecture: (i) a(n) exists for any n > 0. Moreover, a(n) does not exceed n*(n-1)/2 if n > 2.
(ii) For each positive integer n, there is an integer m > 0 with prime(m*n) == -1 (mod m+n). Moreover, we may require that m does not exceed n*(n-1)/2 if n > 2.

Examples

			a(2) = 4 since prime(2*4) = 19 is congruent to 1 modulo 2 + 4 = 6.
a(5146) = 593626 since prime(5146*593626) = prime(3054799396) = 73226821741 is congruent to 1 modulo 5146 + 593626 = 598772.

Links

Crossrefs

Cf. A000040, A247824, A247975.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[Prime[m*n],m+n]==1,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
lpim[n_]:=Module[{m=1},While[Mod[Prime[m*n],m+n]!=1,m++];m]; Array[lpim,60] (* Harvey P. Dale, Oct 01 2017 *)

A248036 Least positive integer m such that m + n divides sigma(m)^2 + sigma(n)^2, where sigma(k) denotes the number of positive divisors of k.

Original entry on oeis.org

1, 3, 2, 1, 10, 6, 3, 50, 1, 5, 34, 28, 7, 6, 10, 18, 3, 16, 33, 5, 20, 14, 83, 24, 1, 10, 10, 12, 56, 6, 33, 2, 15, 11, 93, 13, 204, 27, 52, 38, 17, 6, 7, 6, 15, 14, 5, 944, 1, 8, 17, 39, 32, 33, 5, 24, 7, 59, 58, 15
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 29 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 0.

Examples

			a(5) = 10 since 10 + 5 = 15 divides sigma(10)^2 + sigma(5)^2 = 18^2 + 6^2 = 360.
a(1024) = 2098177 since 2098177 + 1024 = 2099201 divides sigma(2098177)^2 + sigma(1024)^2 = 2103300^2 + 2047^2 = 4423875080209 = 2099201*2107409.

Links

Crossrefs

Cf. A000203, A247975, A248008, A248035.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[DivisorSigma[1,m]^2+DivisorSigma[1,n]^2,m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
lpi[n_]:=Module[{m=1,dsn=DivisorSigma[1,n]^2},While[ !Divisible[ DivisorSigma[ 1,m]^2+ dsn, m+n], m++];m]; Array[lpi, 60] (* Harvey P. Dale, May 07 2016 *)

A248035 Least positive integer m such that m + n divides phi(m)^2 + phi(n)^2, where phi(.) is Euler's totient function.

Original entry on oeis.org

1, 3, 2, 1, 15, 14, 3, 8, 9, 30, 30, 14, 7, 6, 5, 9, 3, 8, 55, 60, 9, 4, 83, 28, 25, 71, 9, 1, 24, 4, 43, 32, 1523, 30, 13, 9, 35, 3, 21, 24, 17, 18, 7, 8, 99, 166, 5, 4, 3, 32, 205, 6, 36, 18, 19, 19, 40, 78, 9, 8
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 29 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 0. Moreover, a(n) <= n^2 except for n = 33.

Examples

			a(5) = 15 since 15 + 5 = 20 divides phi(15)^2 + phi(5)^2 = 8^2 + 4^2 = 80.
a(33) = 1523 since 1523 + 33 = 1556 divides phi(1523)^2 + phi(33)^2 = 1522^2 + 20^2 = 2316884 = 1489*1556.

Links

Crossrefs

Cf. A000010, A247975, A248007, A248036.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[EulerPhi[m]^2+EulerPhi[n]^2,m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
lpim[n_]:=Module[{m=1,p2=EulerPhi[n]^2},While[Mod[p2+EulerPhi[m]^2,m+n]!=0,m++];m]; Array[lpim,60] (* Harvey P. Dale, Nov 19 2020 *)

A248052 Least positive integer m such that m + n divides pi(m^2) + pi(n^2), where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 2, 1, 8, 7, 7, 5, 1, 25, 35, 15, 14, 13, 12, 1, 4, 23, 532, 22, 385, 113, 1, 17, 138, 8, 92, 80, 44, 116, 128, 586, 165, 5, 464, 10, 39, 80, 38, 1, 52, 33, 118, 6, 28, 11, 1239, 47, 92, 517, 3, 145, 40, 8, 184, 104, 104, 16, 73, 53, 52, 5, 145, 172, 68, 11
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 30 2014

Keywords

Comments

Conjecture: (i) a(n) exists for any n > 0.
(ii) For each integer m > 0, there is a positive integer n such that m + n divides prime(m^2) + prime(n^2).

Examples

			a(4) = 8 since 8 + 4 = 12 divides pi(8^2) + pi(4^2) = 18 + 6 = 24.

Links

Crossrefs

Cf. A000040, A000720, A247824, A247975, A248044, A248054.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[PrimePi[m^2]+PrimePi[n^2],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,70}]
a[ n_] := If[ n<1, 0, Module[ {m=1}, While[ Not @ Divisible[ PrimePi[m^2] + PrimePi[n^2], m + n], m++]; m]]; (* Michael Somos, Sep 30 2014 *)

A248044 Least positive integer m such that m + n divides pi(m)^2 + pi(n)^2, where pi(x) denotes the number of primes not exceeding x.

Original entry on oeis.org

1, 3, 1, 4, 12, 11, 1, 8, 7, 16, 2, 5, 26, 25, 24, 4, 228, 227, 46, 45, 44, 43, 16, 6, 5, 1, 27, 26, 45, 44, 12526, 12525, 12524, 12523, 2970, 502, 351, 350, 46, 45, 236, 235, 10, 9, 8, 4, 1078, 1077, 576, 575, 574, 198, 63, 62, 61, 176, 16, 10, 362, 70
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 30 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 0.

Examples

			a(5) = 12 since 12 + 5 = 17 divides pi(12)^2 + pi(5)^2 = 5^2 + 3^2 = 34.

Links

Crossrefs

Cf. A000720, A247824, A247975, A248035, A248036.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[PrimePi[m]^2+PrimePi[n]^2,m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]

A248197 Least positive integer m such that m + n divides prime(prime(m)) + prime(prime(n)).

Original entry on oeis.org

1, 9, 4, 1, 17, 12, 3, 4, 2, 4, 15, 6, 1, 20, 4, 74, 4, 3, 2, 8, 9, 5, 3, 17, 5, 9, 8, 26, 8, 1, 14, 4, 17, 35, 33, 52, 29, 46, 35, 95, 4, 4, 23, 24, 23, 38, 135, 64, 11, 62, 222, 36, 92, 41, 1, 39, 6, 37, 3, 18
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 03 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 0. Moreover, a(n) < n*(n-1) if n > 2.

Examples

			a(3) = 4 since 3 + 4 = 7 divides prime(prime(3)) + prime(prime(4)) = prime(5) + prime(7) = 11 + 17 = 28.
a(2479) = 3386154 since 2479 + 3386154 = 3388633 divides prime(prime(2479)) + prime(prime(3386154)) = prime(22111) + prime(56851657) = 250963 + 1124775193 = 1125026156 = 332*3388633.

Links

Crossrefs

Cf. A000040, A247824, A247975.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[Prime[Prime[m]]+Prime[Prime[n]],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]

A248354 Least positive integer m such that m + n divides prime(m^2) + prime(n^2).

Original entry on oeis.org

1, 1, 2, 1, 3, 8, 2, 6, 6, 45, 9, 4, 15, 2, 13, 17, 4, 12, 9, 8, 11, 6, 101, 20, 2, 15, 7, 50, 4, 183, 48, 15, 9, 5, 4, 4, 157, 1, 123, 4, 13, 112, 76, 4, 7, 13, 44, 2, 16, 28, 83, 202, 114, 50, 85, 31, 14, 62, 19, 25
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 05 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 0. Moreover, a(n) <= n*(n-1)/2 for all n > 1.
See also the comments in A248052.

Examples

			a(3) = 2 since 2 + 3 = 5 divides prime(2^2) + prime(3^2) = 7 + 23 = 30.

Links

Crossrefs

Cf. A000040, A000290, A011757, A247824, A247975, A248052.

Programs

  • Mathematica
    Do[m = 1; Label[aa]; If[Mod[Prime[m^2] + Prime[n^2], m + n] == 0, Print[n, " ", m]; Goto[bb]]; m = m + 1; Goto[aa]; Label[bb]; Continue, {n, 1, 60}]
  • PARI
    a(n)=my(N=prime(n^2),m); while((prime(m++^2)+N)%(m+n), ); m \\ Charles R Greathouse IV, Oct 05 2014
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