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A254409
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a(n) is the least x such that tau(x) divides (x-1)^n but not (x-1)^(n-1), for n >= 2.
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1
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OFFSET
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2,1
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COMMENTS
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So far, all terms appear to be multiples of 15.
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.
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
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
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LINKS
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Table of n, a(n) for n=2..9.
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EXAMPLE
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tau(15) = 4; (15 - 1)^2 = 196 and 196 / 4 = 49.
tau(135) = 8; (135 - 1)^3 = 2406104 and 2406104 / 8 = 300763.
tau(1155) = 16; (1155 - 1)^4 = 1773467504656 and 1773467504656 / 16 = 110841719041.
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MAPLE
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with(numtheory):P:=proc(q) local a, j, k, n; for k from 2 to q do
for n from 1 to q do if not isprime(n) then
if type((n-1)^k/tau(n), integer) then
if not type((n-1)^(k-1)/tau(n), integer) then print(n);
break; fi; fi; fi; od; od; end: P(10^9);
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PROG
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(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
(PARI) isok(k, n) = {my(m = Mod(k-1, numdiv(k))); (m^(n-1) != 0) && (m^n == 0); }
a(n) = {my(k=2); while(!isok(k, n), k++); k}; \\ Michel Marcus, Aug 20 2018
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CROSSREFS
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Cf. A000005 (tau(n)), A234936, A254352.
Sequence in context: A036217 A022643 A125378 * A155648 A126536 A299033
Adjacent sequences: A254406 A254407 A254408 * A254410 A254411 A254412
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KEYWORD
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nonn,more
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AUTHOR
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Paolo P. Lava, Jan 30 2015
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EXTENSIONS
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a(8)-a(9) from Felix Fröhlich, Feb 12 2015
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STATUS
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approved
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