%I #32 May 14 2015 23:25:27
%S 155520,311040,466560,622080,777600,933120,933120,1244160,1399680,
%T 1555200,1555200,1866240,1866240,1866240,2332800,2488320,2488320,
%U 2799360,2799360,3110400,2799360,3110400,3421440,3732480,3888000
%N Values of phi in arithmetic progression of at least 6 terms having the same value of phi in A050518.
%C The values of phi for terms between 13413600 and 10^9 (see comment on A050518) are 3732480, 3888000, 3732480, 4199040, 3732480, 4354560, 4665600, 4665600, 4976640, 4665600, 14999040, 19595520, 29998080, 44130240, 39191040, 44997120, 58786560, 59996160, 88260480, 78382080, 132390720, 134648640, 145313280, 176520960, 220651200, 237948480, 264781440. - _Mauro Fiorentini_, Apr 17 2015
%H Robert Israel, <a href="/A050520/b050520.txt">Table of n, a(n) for n = 1..114</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/NonRecursions.html">Non Recursions</a>
%p N:= 10^7: # to get all terms <= N in A050518
%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, m];
%p fi
%p od
%p od;
%p od;
%p od:
%p map2(op,2,sort([Res], (s, t)->s[1]<t[1])); # _Robert Israel_, May 10 2015
%Y Cf. A000010, A050495, A050496, A050497, A050515-A050520.
%K nonn
%O 1,1
%A _Jud McCranie_, Dec 28 1999
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