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%I #46 Jul 02 2023 02:18:45
%S 15,135,1155,10395,135135,2297295,57432375,1003917915,25097947875
%N a(n) is the least x such that tau(x) divides (x-1)^n but not (x-1)^(n-1), for n >= 2.
%C So far, all terms appear to be multiples of 15.
%C It appears that tau(a(n)) is equal to 2^n. For instance, for n=3, a(3)=135 and 135 has 2^3=8 divisors.
%C The quotients tau(a(n))/(a(n)-1)^n are: 7^2, 67^3, 577^4, 5197^5, 67567^6, 1148647^7, 28716187^8, 501958957^9, ... - _Michel Marcus_, Feb 13 2015
%C Assuming that tau(a(n)) is 2^n (which would appear to follow from the minimality of a(n) and the fact that tau(a(n)) must be divisible by an n-th power), a(n)-1 would have to contain a solitary factor of 2, and so a(n) would be the least number congruent to 3 modulo 4 such that tau(a(n)) = 2^n. The next few terms would appear to be a(9)*25, a(9)*25*29, a(9)*25*29*37, a(9)*25*29*31*43, and a(9)*25*29*31*37*43. - _Charlie Neder_, Aug 19 2018
%e tau(15) = 4; (15 - 1)^2 = 196 and 196 / 4 = 49.
%e tau(135) = 8; (135 - 1)^3 = 2406104 and 2406104 / 8 = 300763.
%e tau(1155) = 16; (1155 - 1)^4 = 1773467504656 and 1773467504656 / 16 = 110841719041.
%p with(numtheory):P:=proc(q) local a,j,k,n; for k from 2 to q do
%p for n from 1 to q do if not isprime(n) then
%p if type((n-1)^k/tau(n),integer) then
%p if not type((n-1)^(k-1)/tau(n),integer) then print(n);
%p break; fi; fi; fi; od; od;end: P(10^9);
%o (PARI) for(n=2, 10, forcomposite(x=1, , if(Mod((x-1)^n, numdiv(x))==0 && Mod((x-1)^(n-1), numdiv(x))!=0, print1(x, ", "); break({1})))) \\ _Felix Fröhlich_, Feb 12 2015
%o (PARI) isok(k, n) = {my(m = Mod(k-1, numdiv(k))); (m^(n-1) != 0) && (m^n == 0);}
%o a(n) = {my(k=2); while(!isok(k, n), k++); k}; \\ _Michel Marcus_, Aug 20 2018
%Y Cf. A000005 (tau(n)), A234936, A254352.
%K nonn,more
%O 2,1
%A _Paolo P. Lava_, Jan 30 2015
%E a(8)-a(9) from _Felix Fröhlich_, Feb 12 2015
%E a(10) from _Amiram Eldar_, Jul 02 2023
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