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%I #57 May 12 2015 10:16:09
%S 583200,1166400,1749600,2332800,2916000,3499200,4082400,4665600,
%T 5248800,5832000,6415200,6998400,7581600,8164800,8748000,9331200,
%U 9914400,10497600,11080800,11664000,12247200,12830400,13413600,13996800,14580000,15163200,15746400
%N An arithmetic progression of at least 6 terms having the same value of phi starts at these numbers.
%C From _Mauro Fiorentini_, Apr 12 2015 (Start):
%C The following are all the terms between 13413600 and 10^9 with increment <= 1000:
%C 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 (End)
%C 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 The a.p. of 7 terms starting at 1158419010 with increment 210 have the same value of phi. - _Robert Israel_, Apr 15 2015
%C a(n) = 583200*n for n <= 112, but a(113) = 65621220. - _Robert Israel_, May 10 2015
%H Robert Israel, <a href="/A050518/b050518.txt">Table of n, a(n) for n = 1..114</a> (all the terms <= 6.6*10^7).
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/NonRecursions.html">Non Recursions</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TotientFunction.html">Totient function.</a>
%p N:= 10^7: # to get all terms <= N
%p with(numtheory):
%p Res:= NULL:
%p phis:= {seq(phi(i),i=2..N)}:
%p for m in phis do
%p S:= convert(invphi(m),set);
%p if nops(S) < 6 then next fi;
%p for d from 0 to 4 do
%p Sd[d]:= select(t-> (t mod 5 = d),S, d);
%p nd:= nops(Sd[d]);
%p for i0 from 1 to nd-1 do
%p s0:= Sd[d][i0];
%p if s0 > N then break fi;
%p for i5 from i0+1 to nd do
%p s5:= Sd[d][i5];
%p incr:= (s5 - s0)/5;
%p if {s0+incr,s0+2*incr,s0+3*incr,s0+4*incr} subset S then
%p Res:= Res, [s0, incr];
%p fi
%p od
%p od;
%p od;
%p od:
%p sort([Res],(s,t)->s[1]<t[1]); # gives both A050518 and A050519 entries
%p map2(op,1,%); # _Robert Israel_, Apr 16 2015
%Y Cf. A000010, A050495, A050496, A050497, A050515-A050520.
%Y The increments are in A050519. The values of phi are in A050520.
%K nonn
%O 1,1
%A _Jud McCranie_, Dec 28 1999
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