# [seqfan] an interesting but difficult sequence

Tue Jun 15 11:34:23 CEST 2010

```Dear SeqFans,

I  used a Neil's restricted permission to send interesting sequences in summer hours. Yesterday I have sent the following sequence (now-with some changes)

%S A178785 60,6552,222768,288288,87360,49585536,25486965504,203558400,683289600,
%T A178785 556121548800
%N A178785 a(n) is the smallest n-perfect number of the form 2^(n+1)*l, where l is an odd number with exponents<=n in its prime power factorization, and a(n)=0 if such  n-perfect number does not exist
%C A178785 Let k>=1. In the multiplicative basis Q^(k)={p^(k+1)^j, p runs A000040, j=0,1,...} every positive integer m has unique factorization of the form m=Prod{q is in Q^(k)}q^(m_q), where m_q is in {0,1,...,k}. In particular, in the case of k=1, we have the unique factorization over distinct terms of A050376. Notice that the standard prime basis is the limiting for k tending to infinity, and, by the definition, Q^(infinity)=A000040. Number d is called a k-divisor of m if the exponents d_q in its factorization in basis Q^(k) do not exceed m_q. A number m is called k-perfect if it equals to the sum of its proper positive k-divisors.
Conjecture. a(11)=0.
%D A178785 S.Litsyn and V.Shevelev, On factorization of integers with restrictions on the exponents, INTEGERS: El. J. of Combin. Number Theory, 7(2007),#A33,1-35
%F A178785 m=Prod{q is in Q^(k)}q^(m_q) is k-perfect number iff Prod{q is in Q^(k)}(q^((m_q)+1)-1)/(q-1)=2*m.
%e A178785 In case of n=2, we have the basis ("2-primes"): 2,3,5,7,8,11,13,...By the formula, we construct from the left m and from the right 2*m. By the condition, m begins from "2-prime" 8. From the right we have 8+1=3^2, therefore from the left we have 8*3^2 and from the right 3^2*(3^3-1)/(3-1)=3^2*13. Thus, from the left, it should be 8*3^2*13 and, from the right, 3^2*13*14. Finally,
from the left we obtain m=8*3^2*13*7=6552 and from the right we have 2*m=3^2*13*14*8. By the construction, it is the smallest 2-perfect number of the required form. Thus a(2)=6552.
%Y A178785 A000396 A050376 A007357 A092356
%K A178785 nonn
%O A178785 1,1

If anyone can prove the conjecture and continue the sequence?

Best regards,