A328274
Numbers m such that phi(m) = rad(m)^3, where phi is the Euler totient function (A000010) and rad is the squarefree kernel function (A007947).
Original entry on oeis.org
1, 16, 648, 2500, 101250, 259308, 1542294, 3660250, 4691556, 33734898, 846948966, 1877404326, 21107677374, 39691260010, 535377671178, 178726991395974
Offset: 1
16 is in the sequence since phi(16) = 8, rad(16) = 2 and 8 = 2^3.
A328275
Numbers m such that phi(m) = rad(m)^4, where phi is the Euler totient function (A000010) and rad is the squarefree kernel function (A007947).
Original entry on oeis.org
1, 32, 3888, 25000, 2839714, 3037500, 10890936, 120298932, 402627500, 534837384, 7489147356, 8508543750, 48919241250, 111945866022, 336977358354, 417841706250, 553904623764, 1498168652148, 2627525125250, 2761526809032, 2898701538750, 7978057537338, 16548448068126, 20978349935382
Offset: 1
32 is in the sequence since phi(32) = 16, rad(32) = 2 and 16 = 2^4.
-
rad[n_] := Times @@ First /@ FactorInteger[n]; aQ[n_] := EulerPhi[n] == rad[n]^4; Select[Range[3*10^6], aQ]
-
rad(n) = factorback(factorint(n)[, 1]); \\ A007947
isok(m) = eulerphi(m) == rad(m)^4; \\ Michel Marcus, Oct 15 2019
A328276
The number of solutions to phi(x) = rad(x)^n, where phi is the Euler totient function (A000010) and rad is the squarefree kernel function (A007947).
Original entry on oeis.org
3, 6, 16, 85, 969
Offset: 1
a(1) = 3 since there are only 3 solutions to phi(x) = rad(x): x = 1, 4, and 18.
a(2) = 6 since there are only 6 solutions to phi(x) = rad(x)^2: x = 1, 8, 108, 250, 6174, and 41154 (the terms of A211413).
- Jean-Marie De Konick and Stefan Gubo, When the totient is the product of the squared prime divisors: problem 10966, American Mathematical Monthly, Vol. 111, No. 6 (2004), p. 536.
- Jean-Marie De Koninck, Florian Luca and A. Sankaranarayanan, Positive integers whose Euler function is a power of their kernel function, Rocky Mountain Journal of Mathematics, Vol. 36, No. 1 (2006), pp. 81-96, alternative link.
A337775
a(n) is the least natural k which is a multiple of prime(n) such that for some m >= 0, phi(k) = rad(k)^m, where phi(k) = A000010(k) and rad(k) = A007947(k).
Original entry on oeis.org
2, 18, 250, 6174, 3660250, 1542294, 2839714, 41154, 117793122328750, 7978057537338, 2898701538750, 33734898, 29688151506250, 21107677374, 69834458642125879757481250, 3999523458421521342
Offset: 1
For n=12 the initial prime is prime(12) = 37 and a(12) = 33734898 because phi(33734898) = 10941048, rad(33734898) = 222 and 222^3 = 10941048 and there is no smaller number satisfying the requirements. The order of a(12) is 3.
- J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 108, p. 38, Ellipses, Paris 2008.
- J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique Des Nombres, Problème 745 ; pp 95; 317-8, Ellipses Paris 2004.
-
nn = 16;
Sar = Table[0, {nn}]; Sar[[1]] = 2;
(*It is a list oh the sequence A337775*)
OrdSar = Table[0, {nn}]; OrdSar[[1]] = 0;
(*It is a sequence A337776 - the orders of members in sequence A337775*) For[Index = 2, Index <= nn, Index++,
InitialPrime = Prime[Index];
InitialInteger = InitialPrime - 1;
InitialArray = FactorInteger[InitialInteger];
For[i = 1, i <= Length[InitialArray], i++,
CurrentArray =
FactorInteger[InitialArray[[-i, 1]] - 1] ~Join~ InitialArray;
InitialInterger =
Product[CurrentArray[[k, 1]] ^ CurrentArray[[k, 2]], {k, 1,
Length[CurrentArray]}];
InitialArray = FactorInteger[InitialInterger];
];
InitialArray = InitialArray ~Join~ {{InitialPrime, 0}};
Ord = Max[InitialArray[[All, 2]]];
Lint = Product[
Power[InitialArray[[k, 1]], Ord - InitialArray[[k, 2]] + 1], {k,
1, Length[InitialArray]}];
radn = Product[InitialArray[[k, 1]], {k, 1, Length[InitialArray]}];
Sar[[Index]] = Lint;
OrdSar[[Index]] = Ord;
];
Print["Sar= ", Sar]
Print["OrdSar= ", OrdSar]
-
rad(n) = factorback(factorint(n)[, 1]);
isok(k) = my(phik=eulerphi(k), radk=rad(k), x=logint(phik, radk)); radk^x == phik;
a(n) = {my(p=prime(n), k=p); while (!isok(k), k+=p); k;} \\ Michel Marcus, Sep 23 2020
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