

A249784


Number of divisors of n^(n^n).


1



1, 5, 28, 513, 3126, 2176875649, 823544, 50331649, 774840979, 100000000020000000001, 285311670612, 158993694406808436568227841, 302875106592254, 123476695691247958050243432972289, 191751059232884087544279144287109376, 73786976294838206465
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OFFSET

1,2


COMMENTS

An infinite number of squares are terms of this sequence.
Proof: for any n of the form (p*q)^k (with p and q distinct primes), a(n) = (k * n^n + 1)^2.
It seems likely that the only nontrivial palindromes in this sequence comprise a subset of these squares and occur at n = 10^(10^M) for M>=0; at such values of n, a(n) = (10^(10^(10^M + M) + M) + 1)^2 = A033934(10^(10^M + M) + M). The actual decimal expansion of each of these numbers is of the form 1000...0002000...0001, where the total number of zero digits on each side of the 2 is 10^(10^M + M) + M  1.
Simple implementations that compute n^(n^n) first and then evaluate its number of divisors are likely to become impractical well before yielding all the terms listed in the Data section; e.g., to obtain a(15) using such an approach would require computing 15^(15^15), which exceeds 3*10^515003176870815367.


LINKS

Jon E. Schoenfield, Table of n, a(n) for n = 1..155


FORMULA

a(n) = A000005(A002488(n)).
a(n) = prod_{j=1..m} (e_j * n^n + 1)
where m = number of distinct prime factors of n
and e_j = multiplicity of the jth prime factor.
If n is a prime p, then m=1 and e_1=1, so
a(p) = p^p + 1 = A000312(p) + 1 = A014566(p).
If n=10^L, then m=2 and e_1=e_2=L, so
a(10^L) = (L * 10^(L * 10^L) + 1)^2.


EXAMPLE

12 = 2^2 * 3^1 (two distinct prime factors, with multiplicities e_1=2 and e_2=1), so a(12) = (2*k+1)*(1*k+1) = 2*k^2 + 3*k + 1 where k = 12^12, so a(12) = 158993694406808436568227841.


PROG

(MAGMA) // program to generate bfile
for n in [1..155] do
k:=n^n;
F:=Factorization(n);
prod:=1;
for j in [1..#F] do
prod*:=F[j, 2]*k + 1;
end for;
n, prod;
end for;
(Sage)
def A249784(n):
n_exp_n = n^n
return mul(exp[1]*n_exp_n + 1 for exp in list(factor(n)))
[A249784(n) for n in (1..16)] # Peter Luschny, Nov 08 2014
(PARI) a(n)=my(v=factor(n)[, 2]*n^n); prod(i=1, #v, v[i]+1) \\ Charles R Greathouse IV, Jul 21 2015


CROSSREFS

Cf. A000005, A000312, A002488, A014566, A033934, A225739.
Sequence in context: A174464 A024068 A308593 * A057642 A125137 A132550
Adjacent sequences: A249781 A249782 A249783 * A249785 A249786 A249787


KEYWORD

nonn


AUTHOR

Jon E. Schoenfield, Nov 05 2014


STATUS

approved



