A050518 An arithmetic progression of at least 6 terms having the same value of phi starts at these numbers.

583200, 1166400, 1749600, 2332800, 2916000, 3499200, 4082400, 4665600, 5248800, 5832000, 6415200, 6998400, 7581600, 8164800, 8748000, 9331200, 9914400, 10497600, 11080800, 11664000, 12247200, 12830400, 13413600, 13996800, 14580000, 15163200, 15746400

Offset: 1

Author

Jud McCranie, Dec 28 1999

Keywords

• nonn

Comments

From Mauro Fiorentini, Apr 12 2015 (Start):

The following are all the terms between 13413600 and 10^9 with increment <= 1000:

13996800, 14580000, 15163200, 15746400, 16329600, 16912800, 17496000, 18079200, 18662400, 19245600, 65621220, 85731240, 131242440, 165488430, 171462480, 196863660, 257193720, 262484880, 330976860, 342924960, 496465290, 504932430, 544924830, 661953720, 827442150, 892306830, 992930580.

(End)

If phi is constant on the arithmetic progression A = [x, x+d, ..., x+m*d], and k is an integer such that each prime factor of k divides either all members of A or no members of A, then phi is also constant on the arithmetic progression k*A = [x*k, x*k+d*k, ..., x*k+m*(d*k)]. - Robert Israel, Apr 12 2015

The a.p. of 7 terms starting at 1158419010 with increment 210 have the same value of phi. - Robert Israel, Apr 15 2015

a(n) = 583200*n for n <= 112, but a(113) = 65621220. - Robert Israel, May 10 2015

Links

• Robert Israel, Table of n, a(n) for n = 1..114 (all the terms <= 6.6*10^7).
• Tanya Khovanova, Non Recursions
• Eric Weisstein's World of Mathematics, Totient function.

Crossrefs

Cf. A000010, A050495, A050496, A050497, A050515-A050520.

The increments are in A050519. The values of phi are in A050520.

Programs

Maple

N:= 10^7: # to get all terms <= N
with(numtheory):
Res:= NULL:
phis:= {seq(phi(i),i=2..N)}:
for m in phis do
   S:= convert(invphi(m),set);
   if nops(S) < 6 then next fi;
   for d from 0 to 4 do
     Sd[d]:= select(t-> (t mod 5 = d),S, d);
     nd:= nops(Sd[d]);
     for i0 from 1 to nd-1 do
       s0:= Sd[d][i0];
       if s0 > N then break fi;
       for i5 from i0+1 to nd do
         s5:= Sd[d][i5];
         incr:= (s5 - s0)/5;
         if {s0+incr,s0+2*incr,s0+3*incr,s0+4*incr} subset S then
           Res:= Res, [s0, incr];
         fi
       od
     od;
   od;
od:
sort([Res],(s,t)->s[1]A050518 and A050519 entries
map2(op,1,%); # Robert Israel, Apr 16 2015