A006511 -id:A006511 - cp's OEIS frontend

A206942 Numbers of the form Phi_k(m) with k > 2 and |m| > 1.

3, 5, 7, 10, 11, 13, 17, 21, 26, 31, 37, 43, 50, 57, 61, 65, 73, 82, 91, 101, 111, 121, 122, 127, 133, 145, 151, 157, 170, 183, 197, 205, 211, 226, 241, 257, 273, 290, 307, 325, 331, 341, 343, 362, 381, 401, 421, 442, 463, 485, 507, 521, 530, 547, 553

Offset: 1

Lei Zhou, Feb 13 2012

nonn

Phi_k(m) denotes the k-th cyclotomic polynomial evaluated at m.
We can see that for any integer b, b = Phi_2(b-1). However, if we make k>2 and |m|>1, Phi(k,m) are always positive integers that do not traverse the positive integer set.
The Mathematica program can generate this sequence to arbitrary upper bound maxdata without user's chosen of parameters. The parameter determination part of this program is explained in A206864.

			a(1) = 3 = Phi_6(2) = Cyclotomic(6,2).
a(2) = 5 = Phi_4(2) = Cyclotomic(4,2).
...
a(15) = 61 = Phi_5(-3) = Cyclotomic(5,-3).

Étienne Fouvry, Claude Levesque, Michel Waldschmidt, Representation of integers by cyclotomic binary forms, arXiv:1712.09019 [math.NT], 2017.

Cf. A206225, A206710, A194712, A206292, A206864.

Cf. A006511 for phiinv function in the Mathematica program.

using Nemo
function isA206942(n)
    if n < 3 return false end
    R, x = PolynomialRing(ZZ, "x")
    K = Int(floor(5.383*log(n)^1.161)) # Bounds from
    M = Int(floor(2*sqrt(n/3)))        # Fouvry & Levesque & Waldschmidt
    for k in 3:K
        c = cyclotomic(k, x)
        for m in 2:M
            n == subst(c, m) && return true
        end
    end
    return false
end
L = [n for n in 1:553 if isA206942(n)]; print(L) # Peter Luschny, Feb 21 2018

phiinv[n_, pl_] :=  Module[{i, p, e, pe, val}, If[pl == {}, Return[If[n == 1, {1}, {}]]]; val = {}; p = Last[pl]; For[e = 0; pe = 1, e == 0 || Mod[n, (p - 1) pe/p] == 0, e++; pe *= p, val = Join[val, pe*phiinv[If[e == 0, n, n*p/pe/(p - 1)], Drop[pl, -1]]]]; Sort[val]]; phiinv[n_] := phiinv[n, Select[1 + Divisors[n], PrimeQ]]; maxdata = 560; max =  Ceiling[(1 + Sqrt[1 + 4*(maxdata - 1)])/4]*2; eb =  2*Floor[(Log[2, maxdata])/2 + 0.5]; While[eg = phiinv[eb]; lu = Length[eg]; lu == 0, eb = eb + 2]; t = Select[Range[eg[[Length[eg]]]], EulerPhi[#] <= eb &]; ap = SortBy[t, Cyclotomic[#, 2] &]; an =  SortBy[t, Cyclotomic[#, -2] &]; a = {}; Do[i = 2; While[i++; cc = Cyclotomic[ap[[i]], m]; cc <= maxdata, a = Append[a, cc]]; i = 2;  While[i++; cc = Cyclotomic[an[[i]], -m]; cc <= maxdata, a = Append[a, cc]], {m, 2, max}]; Union[a]
(* Alternatively: *)
isA206942[n_] := If[n < 3, Return[False],
    K = Floor[5.383 Log[n]^1.161]; M = Floor[2 Sqrt[n/3]];
    For[k = 3, k <= K, k++, For[x = 2, x <= M, x++,
        If[n == Cyclotomic[k, x], Return[True]]]];
    Return[False]
]; Select[Range[555], isA206942] (* Peter Luschny, Feb 21 2018 *)

A215240 Sum of the numbers p such that phi(p) = n, where phi is Euler's totient function.

Original entry on oeis.org

3, 13, 0, 35, 0, 48, 0, 105, 0, 33, 0, 166, 0, 0, 0, 231, 0, 138, 0, 218, 0, 69, 0, 621, 0, 0, 0, 87, 0, 93, 0, 581, 0, 0, 0, 655, 0, 0, 0, 833, 0, 276, 0, 299, 0, 141, 0, 1514, 0, 0, 0, 159, 0, 243, 0, 377, 0, 177, 0, 1114, 0, 0, 0, 1315, 0, 201, 0, 0, 0, 213, 0

Offset: 1

T. D. Noe, Oct 12 2012

nonn

These terms (greater than 0) are not unique. The first duplicate appears at a(256) = a(2236) = 6711.

T. D. Noe, Table of n, a(n) for n = 1..10000
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).
David M. Bressoud, A Course in Computational Number Theory (web page), CNT.m, Computational Number Theory Mathematica package.

Cf. A002181 (smallest inverse), A006511 (largest inverse), A217842 (product of inverses).

Cf. A007617, A032447 (inverse of phi).

Needs["CNT`"]; Table[Total[PhiInverse[n]], {n, 100}]

a(n) = vecsum(invphi(n)); \\ Amiram Eldar, Nov 15 2024, using Max Alekseyev's invphi.gp

a(n) = 0 if and only if n is in A007617. - Amiram Eldar, Nov 15 2024

A293928 Totients phi(m) having one or more solutions m to phi(m)^2 = phi(phi(m)*m).

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 32, 36, 40, 48, 54, 64, 72, 80, 84, 96, 100, 108, 120, 128, 144, 160, 162, 168, 192, 200, 216, 240, 252, 256, 272, 288, 312, 320, 324, 336, 360, 384, 400, 432, 440, 480, 486, 500, 504, 512, 544, 576, 588, 600, 624, 640, 648, 672, 684

Offset: 1

Torlach Rush, Oct 19 2017

nonn

"Totients" are terms of A000010. - N. J. A. Sloane, Oct 22 2017
The smallest totient absent from the list is 10. This is because the totient inverses of 10, 11 and 22 are not solutions to phi(m)^2 = phi(phi(m)*m).
The formula is recursive. For example, taking a(22) we get the following: 11664 = phi(108*324), 1259712 = phi(11664*324), 136048896 = phi(1259712*324), ...
Where (if ever) does this first differ from A068997? - R. J. Mathar, Oct 30 2017
Apparently the set of the m is A151999. - R. J. Mathar, Mar 25 2024
If m satisfies phi(m)^2 = phi(phi(m)*m), then it satisfies phi(m)^(k+1) = phi(phi(m)^k*m) for all k >= 1. - Max Alekseyev, Dec 03 2024

			96 is a term since 96^2 = phi(96*288), with m=288 where phi(288) = 96.

Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Cf. A000010, A006511, A032447, A007366.

Subsequence of A002202.

isok(n) = {my(iv = invphi(n)); if (#iv, for (m = 1, #iv, if (n^2 == eulerphi(n*iv[m]), return (1)););); return (0);} \\ using the invphi script by Max Alekseyev; Michel Marcus, Nov 01 2017

More terms from Michel Marcus, Oct 24 2017

Definition simplified by Max Alekseyev, Dec 03 2024

A139795 Least m such that k>=m implies phi(k)>=n (where phi is the Euler totient function, sequence A000010).

Original entry on oeis.org

1, 3, 7, 7, 13, 13, 19, 19, 31, 31, 31, 31, 43, 43, 43, 43, 61, 61, 61, 61, 67, 67, 67, 67, 91, 91, 91, 91, 91, 91, 91, 91, 121, 121, 121, 121, 127, 127, 127, 127, 151, 151, 151, 151, 151, 151, 151, 151, 211, 211, 211, 211, 211, 211, 211, 211, 211, 211, 211, 211, 211

Offset: 1

Benoit Jubin, May 21 2008

nonn

Define b(n)=A006511(m)+1 where m is the unique integer such that A002202(m)A002202(m+1) (with the convention A002202(0)=A006511(0)=0). Then a(1)=b(1) and a(n+1)=max(a(n),b(n+1)).

The sequence a(n) without the repetitions is 1+A036913(n).

			a(5)=13 because if k>=13, then phi(k)>=5, but phi(12)=4.

Max Alekseyev, Table of n, a(n) for n = 1..10000
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Different from A137315 (see Comments in that entry).

{m=0;for(n=1,100,print1(m+1,",");trap(,0,m=max(m,vecmax(invphi(n)))))}

A217842 Product of the numbers p such that phi(p) = n, where phi is Euler's totient function.

Original entry on oeis.org

2, 72, 1, 4800, 1, 15876, 1, 3456000, 1, 242, 1, 300500928, 1, 1, 1, 2130739200, 1, 1052676, 1, 119790000, 1, 1058, 1, 531598161669120000, 1, 1, 1, 1682, 1, 1922, 1, 20864198246400, 1, 1, 1, 1159208596538496, 1, 1, 1, 265804426800000000, 1, 17757796, 1

Offset: 1

T. D. Noe, Oct 12 2012

nonn

It appears that all terms greater than 1 are distinct. This is true for all n <= 10^6.

T. D. Noe, Table of n, a(n) for n = 1..1000
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).
David M. Bressoud, A Course in Computational Number Theory (web page), CNT.m, Computational Number Theory Mathematica package.

Cf. A002181 (smallest inverse), A006511 (largest inverse), A215240 (sum of inverses).

Cf. A032447 (inverse of phi).

Needs["CNT`"]; Table[Times @@ PhiInverse[n], {n, 100}]

a(n) = vecprod(invphi(n)); \\ Amiram Eldar, Nov 15 2024, using Max Alekseyev's invphi.gp

A297475 Numbers n such that phi(x) = n for more than one value of x, and the smallest such x divides the largest.

Original entry on oeis.org

1, 2, 8, 10, 22, 28, 30, 44, 46, 52, 54, 56, 58, 66, 70, 78, 82, 92, 102, 104, 106, 110, 116, 126, 128, 130, 136, 138, 140, 148, 150, 164, 166, 172, 178, 190, 196, 198, 204, 210, 212, 222, 226, 228, 238, 250, 260, 262, 268, 270, 282, 292, 294, 296, 306, 310, 316, 330, 332, 342, 344, 346, 356, 358, 366, 368, 372

Offset: 1

Torlach Rush, Dec 30 2017

nonn

The larger endpoint is always twice the value of the smaller endpoint.
Conjecture 1: The number of solutions, excluding endpoints is always 0, or an odd number. (known to n = 2 * 10^5)
Conjecture 2: If both endpoints are divisible by 5, then the number of solutions (excluding terms of A007366) is of the form 4k + 1. (known to n = 2 * 10^5)
A007366 is contained in this sequence and the number of solutions, excluding endpoints is always 0.
Terms of this sequence are totients with a single odd totient inverse.

			2 is in the sequence because {phi^-1(2)} = {3,4,6}, and 2 = 6 / 3.
8 is in the sequence because {phi^-1(8)} = {15,...,30}, and 2 = 30 / 15.
10 is in the sequence because {phi^-1(10)} = {11,22}, and 2 = 22 / 11.

Michel Marcus, Table of n, a(n) for n = 1..2441
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).

Cf. A002181, A002202, A006511, A007366, A032447.

With[{nn = 67}, Take[#, nn] &@ Keys@ Select[KeySort@ PositionIndex@ Array[EulerPhi, nn^2], IntegerQ[#2/#1] & @@ {First@ #, Last@ #} &]] (* Michael De Vlieger, Dec 31 2017 *)

isok(n) = my(vx = invphi(n)); (#vx > 1) && ((vecmax(vx) % vecmin(vx)) == 0); \\ Michel Marcus, Jul 18 2018

2 = max({phi^-1(n)}) / min({phi^-1(n)}).

0 = A006511(n) mod A002181(n).

A143423 Least even number k such that phi(k) = n, where n runs through the values (A002202) taken by phi.

Original entry on oeis.org

2, 4, 8, 14, 16, 22, 26, 32, 38, 44, 46, 52, 58, 62, 64, 74, 82, 86, 92, 94, 104, 106, 162, 116, 118, 122, 128, 134, 142, 146, 158, 164, 166, 172, 178, 188, 194, 202, 206, 212, 214, 218, 242, 226, 236, 244, 254, 256, 262, 268, 274, 278, 284, 292, 298, 302, 314

Offset: 1

T. D. Noe, Aug 14 2008

nonn

Such an even number always exists.

T. D. Noe, Table of n, a(n) for n=1..10000
Maxim Rytin, Finding the Inverse of Euler Totient Function (1999).

Cf. A002181 (least k such that phi(k)=n), A006511 (largest k such that phi(k)=n).

f:= proc(m) local L;
      L:= numtheory:-invphi(m);
      if L = [] then NULL
      else min(select(type,L,even))
      fi
end proc:
map(f, [1,seq(2*k,k=1..1000)]); # Robert Israel, Oct 07 2015

f[m_] := Module[{L}, L = invphi[m]; If[L == {}, Nothing, Min[Select[L, EvenQ]]]];
f /@ Join [{1}, 2 Range[1000]] (* Jean-François Alcover, Aug 28 2020, using Maxim Rytin's invphi function *)

A196079 Difference between the largest and smallest inverse of totient function.

Original entry on oeis.org

1, 3, 7, 11, 15, 11, 29, 43, 35, 41, 23, 55, 29, 31, 69, 89, 109, 55, 69, 47, 145, 53, 81, 87, 59, 137, 155, 67, 71, 197, 79, 207, 83, 165, 187, 141, 323, 149, 103, 159, 107, 269, 121, 235, 177, 319, 127, 255, 131, 253, 137, 139, 213, 445, 149, 151

Offset: 1

Franz Vrabec, Sep 27 2011

nonn

No terms are zero if Carmichael's conjecture is true.

Even terms are rare: e.g., all inverses of 257*2^16 are even [Foster], so the difference between the largest and smallest inverse is even.

			Let n=3. The largest inverse of A002202(3)=4 is A006511(3)=12, the smallest inverse is A002181(3)=5, so a(3)=12-5=7.

T. D. Noe, Table of n, a(n) for n = 1..10000
William P. Wardlaw, L. L. Foster and R. J. Simpson, Problem E3361, Amer. Math. Monthly, Vol. 98, No. 5 (May, 1991), 443-444.

Cf. A002181, A002202, A006511.

max = 300; inversePhi[?OddQ] = {}; inversePhi[1] = {1, 2}; inversePhi[m] := Module[{p, nmax, n, nn}, p = Select[Divisors[m] + 1, PrimeQ]; nmax = m*Times @@ (p/(p - 1)); n = m; nn = Reap[While[n <= nmax, If[EulerPhi[n] == m, Sow[n]]; n++]] // Last; If[nn == {}, {}, First[nn]]]; Join[{2}, Reap[For[n = 2, n <= max, n = n + 2, nn = inversePhi[n] ; If[ nn != {} , Sow[Max[nn] - Min[nn]]]]] // Last // First] (* Jean-François Alcover, Nov 21 2013 *)

a(n) = A006511(n) - A002181(n).

a(1) corrected by the editors, Nov 23 2013

a(1) in b-file corrected by Andrew Howroyd, Feb 22 2018

A294618 a(n) is the number of solutions of x^2 = eulerphi(x * m) where x is A293928(n).

Original entry on oeis.org

2, 2, 3, 1, 4, 2, 5, 1, 1, 4, 6, 3, 3, 5, 1, 7, 6, 4, 1, 7, 1, 3, 1, 8, 10, 5, 1, 1, 9, 3, 8, 4, 1, 9, 1, 13, 1, 7, 4, 3, 1, 12, 5, 14, 1, 7, 1, 1, 2, 10, 2, 18, 1, 1, 1, 9, 9, 3, 1, 5, 1, 14, 7, 22, 3, 1

Offset: 1

Torlach Rush, Nov 05 2017

The valid values of m in the equation are the terms of the sequence A151999 in order.
m is a solution if all squarefree divisors of x also divide m.
The formula is recursive. For example, taking A151999(68) we get the following: 11664=phi(108*324), 1259712=phi(11664*324), 136048896=phi(1259712*324), ...
If a solution exists then x^(k+1) = phi(x^k * m) for a fixed m, and the smallest value of k must be 1. This follows from a|b implies phi(a)|phi(b), and for k >= 1 a^(k-1)|a^k.
The smallest solution where solutions exist are the terms of the sequence A055744 not in order.
The values of phi(m) are the terms of the sequence A068997 not in order.

			The first 1 is a term since there is only 1 solution when phi(m)=6. The solution is m=18.
The first 5 is a term since there are 5 solutions when phi(m)=16. These are 32, 34, 40, 48, and 60.
From _Michel Marcus_, Nov 08 2017: (Start)
Illustration of first few terms:
   1: [1, 2],
   2: [4, 6],
   4: [8, 10, 12],
   6: [18],
   8: [16, 20, 24, 30],
  12: [36, 42],
  16: [32, 34, 40, 48, 60],
  18: [54],
  20: [50],
  24: [72, 78, 84, 90],
  32: [64, 68, 80, 96, 102, 120],
  ... (End)

Max Alekseyev, PARI scripts for various problems

Cf. A000010, A006511, A032447, A151999, A055744, A068997, A293928.

isok(n) = {iv = invphi(n); if (#iv, return (sum(m=1, #iv, n^2 == eulerphi(n*iv[m])))); return (0);}
lista(nn) = {for (n=1, nn, if (v = isok(n), print1(v, ", ")););} \\ \\ using the invphi script by Max Alekseyev; Michel Marcus, Nov 07 2017

0 < (phi(m)^(k+1) = phi(phi(m)^k*m)), k >= 1, m >= 1.

cp's OEIS Frontend

A206942 Numbers of the form Phi_k(m) with k > 2 and |m| > 1.

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Julia

Mathematica

A215240 Sum of the numbers p such that phi(p) = n, where phi is Euler's totient function.

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A293928 Totients phi(m) having one or more solutions m to phi(m)^2 = phi(phi(m)*m).

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A139795 Least m such that k>=m implies phi(k)>=n (where phi is the Euler totient function, sequence A000010).

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PARI

A217842 Product of the numbers p such that phi(p) = n, where phi is Euler's totient function.

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Mathematica

PARI

A297475 Numbers n such that phi(x) = n for more than one value of x, and the smallest such x divides the largest.

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Mathematica

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A143423 Least even number k such that phi(k) = n, where n runs through the values (A002202) taken by phi.

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Maple

Mathematica

A196079 Difference between the largest and smallest inverse of totient function.

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