# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a050518 Showing 1-1 of 1 %I A050518 #57 May 12 2015 10:16:09 %S A050518 583200,1166400,1749600,2332800,2916000,3499200,4082400,4665600, %T A050518 5248800,5832000,6415200,6998400,7581600,8164800,8748000,9331200, %U A050518 9914400,10497600,11080800,11664000,12247200,12830400,13413600,13996800,14580000,15163200,15746400 %N A050518 An arithmetic progression of at least 6 terms having the same value of phi starts at these numbers. %C A050518 From _Mauro Fiorentini_, Apr 12 2015 (Start): %C A050518 The following are all the terms between 13413600 and 10^9 with increment <= 1000: %C A050518 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. %C A050518 (End) %C A050518 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 %C A050518 The a.p. of 7 terms starting at 1158419010 with increment 210 have the same value of phi. - _Robert Israel_, Apr 15 2015 %C A050518 a(n) = 583200*n for n <= 112, but a(113) = 65621220. - _Robert Israel_, May 10 2015 %H A050518 Robert Israel, Table of n, a(n) for n = 1..114 (all the terms <= 6.6*10^7). %H A050518 Tanya Khovanova, Non Recursions %H A050518 Eric Weisstein's World of Mathematics, Totient function. %p A050518 N:= 10^7: # to get all terms <= N %p A050518 with(numtheory): %p A050518 Res:= NULL: %p A050518 phis:= {seq(phi(i),i=2..N)}: %p A050518 for m in phis do %p A050518 S:= convert(invphi(m),set); %p A050518 if nops(S) < 6 then next fi; %p A050518 for d from 0 to 4 do %p A050518 Sd[d]:= select(t-> (t mod 5 = d),S, d); %p A050518 nd:= nops(Sd[d]); %p A050518 for i0 from 1 to nd-1 do %p A050518 s0:= Sd[d][i0]; %p A050518 if s0 > N then break fi; %p A050518 for i5 from i0+1 to nd do %p A050518 s5:= Sd[d][i5]; %p A050518 incr:= (s5 - s0)/5; %p A050518 if {s0+incr,s0+2*incr,s0+3*incr,s0+4*incr} subset S then %p A050518 Res:= Res, [s0, incr]; %p A050518 fi %p A050518 od %p A050518 od; %p A050518 od; %p A050518 od: %p A050518 sort([Res],(s,t)->s[1]