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%I #57 Mar 26 2022 14:30:39
%S 1,5,28,513,3126,2176875649,823544,50331649,774840979,
%T 100000000020000000001,285311670612,158993694406808436568227841,
%U 302875106592254,123476695691247958050243432972289,191751059232884087544279144287109376,73786976294838206465
%N Number of divisors of n^(n^n).
%C An infinite number of squares are terms of this sequence.
%C Proof: for any n of the form (p*q)^k (with p and q distinct primes), a(n) = (k * n^n + 1)^2.
%C 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.
%H Jon E. Schoenfield, <a href="/A249784/b249784.txt">Table of n, a(n) for n = 1..155</a>
%F a(n) = A000005(A002488(n)).
%F a(n) = Product_{j=1..m} (e_j * n^n + 1)
%F where m = number of distinct prime factors of n
%F and e_j = multiplicity of the j-th prime factor.
%F If n is a prime p, then m=1 and e_1=1, so
%F a(p) = p^p + 1 = A000312(p) + 1 = A014566(p).
%F If n=10^L, then m=2 and e_1=e_2=L, so
%F a(10^L) = (L * 10^(L * 10^L) + 1)^2.
%e 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.
%o (Magma) // program to generate b-file
%o for n in [1..155] do
%o k:=n^n;
%o F:=Factorization(n);
%o prod:=1;
%o for j in [1..#F] do
%o prod*:=F[j,2]*k + 1;
%o end for;
%o n, prod;
%o end for;
%o (Sage)
%o def A249784(n):
%o n_exp_n = n^n
%o return mul(exp[1]*n_exp_n + 1 for exp in factor(n))
%o [A249784(n) for n in (1..16)] # _Peter Luschny_, Nov 08 2014
%o (PARI) a(n)=my(v=factor(n)[,2]*n^n); prod(i=1,#v,v[i]+1) \\ _Charles R Greathouse IV_, Jul 21 2015
%Y Cf. A000005, A000312, A002488, A014566, A033934, A225739.
%K nonn
%O 1,2
%A _Jon E. Schoenfield_, Nov 05 2014
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