A219930 n such that phi(n) represents a new lower bound for the phi function.

1, 3, 8, 14, 20, 36, 48, 66, 70, 96, 126, 132, 156, 240, 252, 300, 336, 450, 480, 540, 660, 690, 714, 870, 900, 1080, 1320, 1470, 1530, 1710, 1950, 2340, 2940, 2970, 3360, 3780, 4200, 4830, 5040, 5610, 5670, 5880, 6270, 7140, 7350, 7410, 8400, 9660, 9870

Offset: 1

Author

Jon Perry, Dec 01 2012

Keywords

• nonn

Comments

Conjecture: If n is in the sequence, then the sequence contains an infinite number of multiples of n.

Conjecture: Except for 1 and 3, all members of the sequence are even. If n is odd, it cannot be squarefree.

Conjecture: There does not exist N such that for all n > N, a(n) is divisible by 30.

A036912 gives the values of the phi function at these n.

Examples

			phi(1)=1, and for n>=1, phi(n)>=1.
phi(3)=2, and for n>=3, phi(n)>=2.
phi(8)=4, and for n>=8, phi(n)>=4.
phi(14)=6, and for n>=14, phi(n)>=6.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..200

Crossrefs

Cf. A000010, A036912, A057635, A014197.

Programs

JavaScript

p = new Array();
p[0] = NaN;
p[1] = 2;
p[2] = 3;
mj = 2;
for (k = 3; k < 50000; k += 2) makeprimes(k);
function makeprimes(i) {
for (j = 2; j <= mj; j++)
if (i%p[j] == 0) return false;
p[++mj] = i;
return true;
}
function primeFactorize(n) {
var pf = new Array(), pc, pfc;
pf[0] = new Array();
pf[1] = new Array();
pc = 1;
pfc = -1;
while (n != 1) {
if (n%p[pc] == 0) {pfc++; pf[0][pfc] = p[pc]; pf[1][pfc] = 0;}
while (n%p[pc] == 0) {n /= p[pc]; pf[1][pfc]++;}
pc++;
}
return pf;
}
function phi(n) {
var f, i, v;
v = 1;
f = primeFactorize(n);
for (i = 0; i < f[0].length; i++) v *= Math.pow(f[0][i], f[1][i] - 1)*(f[0][i] - 1);
return v;
}
function isMin(arr, ik, k) {
var i, im;
im = true;
for (i = ik; i < arr.length; i++) if (arr[i] < k) {im = false; break;}
return im;
}
phiV = new Array();
for (k = 1; k < 50000; k++) phiV[k] = phi(k);
cm = 1;
for (n = 1; n < 3000; n++) if (phiV[n] > cm && isMin(phiV, n, phiV[n])) {cm = phiV[n]; document.write(n + ", ");}

Mathematica

nn = 8!; t = Table[EulerPhi[n], {n, nn}]; min = Infinity; t2 = {}; Do[If[t[[n]] <= min, AppendTo[t2, {n, t[[n]]}]; min = t[[n]]], {n, Length[t], 1, -1}]; t2 = Reverse[t2]; t3 = {}; mx = 0; Do[If[i[[2]] > mx, mx = i[[2]]; AppendTo[t3, i[[1]]]], {i, t2}]; t3 (* T. D. Noe, Dec 04 2012 *)