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Start of n consecutive indices k such that phi(k) contains distinct number of divisors.
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%I #18 May 13 2013 01:49:55

%S 1,2,5,29,56,56,59,424,424,1351,1353,1353,4004,4004,4004,15212,40725,

%T 64098,76662,76662,192998,251887,489989,489991,1013057,4143368,

%U 4431511,4431511,4431511,8309350,30951255,35867405,55131136,102123612,144869833,148753758

%N Start of n consecutive indices k such that phi(k) contains distinct number of divisors.

%e a(6) = 56 because:

%e phi(56) = 24 = 2 ^ 3 * 3;

%e phi(57)= 36 = 2 ^ 2 * 3 ^ 2;

%e phi(58) = 28 = 2 ^ 2 * 7;

%e phi(59) = 58 = 2 * 29;

%e phi(60) = 16 = 2 ^ 4;

%e phi(61) = 60 = 2 ^ 2 * 3 * 5.

%e All have distinct number of divisors: 8, 9, 6, 4, 5 and 12, respectively.

%p with(numtheory): for n from 1 to 22 do: i:=0:for k from 1 to 500000 while(i=0) do: lst:={}:for p from 0 to n-1 do :x:= phi(k+p):y:=divisors(x):n1:=nops(y):lst:= lst union {n1}:od:if nops(lst)=n then printf(`%d, `,k): i:=1:else fi:od:od:

%o (PARI) v=vectorsmall(10^7,n,numdiv(eulerphi(n)));

%o a(n,startAt=1)=n--;for(k=startAt,#v-n,for(i=k,k+n-1,for(j=i+1,k+n,if(v[i]==v[j],next(3))));return(k))

%o k=1;for(n=1,30,k=a(n,k);print1(k", ")) \\ _Charles R Greathouse IV_, Aug 09 2011

%Y Cf. A000010.

%K nonn,hard

%O 1,2

%A _Michel Lagneau_, Aug 08 2011

%E a(25)-a(32) from _Charles R Greathouse IV_, Aug 09 2011

%E a(33)-a(36) from _Donovan Johnson_, Sep 26 2011