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 1-4 of 4 results.

A259916 Least positive integer k such that sigma(k) and phi(k*n) are both squares, where sigma(k) is the sum of all positive divisors of k, and phi(.) is Euler's totient function.

Original entry on oeis.org

1, 1, 210, 3, 1, 170, 81, 1, 70, 1, 400, 1, 210, 81, 357, 3, 1, 119, 3, 3, 3, 651, 1990, 170, 66, 70, 210, 884, 3810, 357, 1066, 1, 217, 1, 81, 3, 1, 3, 70, 1, 22, 3, 1624, 217, 119, 3383, 11510, 1, 364, 22, 210, 81, 8743, 170, 510, 81, 1, 1270, 2902, 1, 385, 1155, 1, 3, 357, 217, 966, 3, 4179, 81
Offset: 1

Views

Author

Zhi-Wei Sun, Jul 08 2015

Keywords

Comments

The conjecture in A259915 implies that a(n) exists for any n > 0.

Examples

			a(3) = 210 since sigma(210) = 576 =24^2 and phi(210*3) = 144 = 12^2.
a(719) = 42862647 since sigma(42862647) = 58003456 = 7616^2 and phi(42862627*719) = phi(30818243193) = 20210602896 = 142164^2.
		

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
    SQ[n_]:=IntegerQ[Sqrt[n]]
    sigma[n_]:=DivisorSigma[1, n]
    Do[k=0; Label[aa]; k=k+1; If[SQ[sigma[k]]&&SQ[EulerPhi[k*n]], Goto[bb], Goto[aa]]; Label[bb]; Print[n, " ", k]; Continue, {n, 1, 70}]
    lpi[n_]:=Module[{k=1},While[!IntegerQ[Sqrt[DivisorSigma[1,k]]]|| !IntegerQ[ Sqrt[ EulerPhi[ n*k]]],k++];k]; Array[lpi,70] (* Harvey P. Dale, Jul 17 2020 *)

A334337 Least positive integer m such that phi(m*n) is a cube, where phi is Euler's totient function (A000010).

Original entry on oeis.org

1, 1, 5, 4, 3, 4, 37, 2, 37, 2, 101, 2, 19, 37, 1, 1, 5, 36, 13, 1, 19, 101, 13333, 1, 55, 19, 13, 19, 985, 1, 1057, 4, 401, 4, 73, 18, 7, 13, 9, 4, 275, 18, 2649, 401, 9, 13333, 169285, 4, 1813, 50, 4, 73, 3385, 12, 25, 73, 7, 788, 40371, 4, 3737, 1057, 12, 2, 37, 401, 4357, 2, 6537, 73, 5401, 9, 35, 7, 25, 7, 3737, 9, 48673, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 23 2020

Keywords

Comments

Conjecture: a(n) exists for any n > 0. In other words, for any positive integer n, there is a positive integer m with phi(m*n) equal to a cube.
We note that there is no positive integer m <= 10^8 with phi(107*m) equal to a fourth power.
The author's conjecture in A259915 implies that for any positive integer n there is a positive integer m with phi(m*n) equal to a square.
See also A334339 for a similar conjecture.

Examples

			a(3) = 5 with phi(3*5) = 2^3.
a(7) = 37 with phi(7*37) = 6^3.
a(863) = 21176773 with phi(863*21176773) = 17293606056 = 2586^3.
		

Crossrefs

Programs

  • Mathematica
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
    phi[n_]:=phi[n]=EulerPhi[n];
    tab={};Do[m=0;Label[aa];m=m+1;If[CQ[phi[m*n]],tab=Append[tab,m],Goto[aa]],{n,1,80}];tab
  • PARI
    a(n) = my(m=1); while (!ispower(eulerphi(n*m), 3), m++); m; \\ Michel Marcus, Apr 23 2020

A334339 Least positive integer m such that sigma(m * n) is a cube, where sigma(k) is the sum of the divisors of k.

Original entry on oeis.org

1, 51, 34, 291, 22, 17, 1, 1347, 597, 11, 10, 97, 892, 51, 46, 1758, 6, 3540, 343, 1649, 34, 5, 30, 449, 2928, 446, 199, 291, 472, 23, 34, 879, 235, 3, 22, 1770, 8661, 356, 3007, 1593, 884, 17, 241, 298, 1416, 15, 22, 586, 133, 1464, 2, 223, 3, 1180, 2, 1347, 711, 236, 232, 1062, 1200, 17, 597, 96771, 586, 265, 577, 485, 10, 11
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 23 2020

Keywords

Comments

Conjecture: a(n) exists for any n > 0. In other words, for any positive integer n, there is a positive integer m with sigma(m * n) equal to a cube.
The author's conjecture in A259915 implies that for each n = 1, 2, 3, ... there is a positive integer m with sigma(m * n) equal to a square.
See also A334337 for a similar conjecture.

Examples

			a(2) = 51 with sigma(2*51) = 216 = 6^3.
a(4) = 291 with sigma(4*291) = 2744 = 14^3.
a(578) = 34312749 with sigma(578*34312749) = 42144192000 = 3480^3.
a(673) = 49061802 with sigma(673*49061802) = 66135317184 = 4044^3.
		

Crossrefs

Programs

  • Mathematica
    cubeQ[n_] := cubeQ[n] = IntegerQ[n^(1/3)];
    sigma[n_] := sigma[n] = DivisorSigma[1, n];
    tab = {}; Do[m = 0; Label[aa]; m = m + 1; If[cubeQ[sigma[m * n]], tab = Append[tab, m], Goto[aa]], {n, 70}]; tab
    lpi[n_]:=Module[{k=1},While[!IntegerQ[Surd[DivisorSigma[1,n*k],3]],k++]; k]; Array[lpi,70] (* Harvey P. Dale, Nov 05 2020 *)
  • PARI
    a(n) = my(m=1); while (!ispower(sigma(n*m), 3), m++); m; \\ Michel Marcus, Apr 23 2020

Extensions

Corrected and extended by Harvey P. Dale, Nov 05 2020

A334350 Least positive integer m relatively prime to n such that phi(m*n) = phi(m)*phi(n) is a fourth power, where phi is Euler's totient function (A000010).

Original entry on oeis.org

1, 1, 16, 15, 8, 703, 247, 5, 247, 489, 1255, 5, 109, 247, 4, 3, 1, 247, 73, 3, 109, 1255, 13315, 163, 753, 109, 73, 109, 1373, 163, 27331, 1, 625, 1, 81, 109, 57, 73, 1295, 1, 251, 109, 74663, 625, 949, 13315, 1557377, 1, 74663, 753, 16, 81, 175765, 73, 251, 81, 37, 1373, 243895, 1
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 24 2020

Keywords

Comments

Conjecture: For any positive integers k and m, there is a positive integer n relatively prime to m such that phi(m*n) = phi(m)*phi(n) is a k-th power.
This conjecture implies that a(n) exists for every n = 1,2,3,....
See also A334353 for a similar conjecture involving the sigma function (A000203).
a(n) = 1 if and only if n is in A078164. - Charles R Greathouse IV, Apr 24 2020

Examples

			a(3) = 16 with gcd(3,16) = 1 and phi(3*16) = phi(3)*phi(16) = 2*8 = 2^4.
a(167) = 370517977 with gcd(167, 370517977) = 1 and phi(167*370517977) = phi(167)*phi(370517977) = 166*370517976 = 61505984016 = 498^4.
		

Crossrefs

Programs

  • Mathematica
    QQ[n_]:=QQ[n]=IntegerQ[n^(1/4)];
    phi[n_]:=phi[n]=EulerPhi[n];
    tab={};Do[m=0;Label[aa];m=m+1;If[GCD[m,n]==1&&QQ[phi[m]*phi[n]],tab=Append[tab,m],Goto[aa]],{n,1,60}];tab
  • PARI
    a(n) = my(m=1,e=eulerphi(n)); while (!((gcd(n, m) == 1) && ispower(e*eulerphi(m), 4)), m++); m; \\ Michel Marcus, Apr 25 2020
Showing 1-4 of 4 results.