OFFSET
1,3
COMMENTS
This sequence can be calculated by a recursive algorithm:
Let B1 be an array of finite length, the "1" denotes that it is the first generation. Let B1' be the reversed version of B1. Let C be the element-wise product C = B1 * B1'. Then B2 is a concatenation of taking each element of B1 and add all divisors of the corresponding element in C. If we start with B1 = {1} then we get this sequence of arrays: B2 = {2}, B3 = {3, 4, 6}, ... . a(n) is the length of the array Bn. In short the length of Bn+1 and so a(n+1) is the sum over A000005(Bn * Bn').
The transform used in the definition of this sequence is its own inverse, so if c = S(b) then b = S(c). The eigensequence is 2^n = S(2^n).
There exist some transformation pairs of infinite sequences in the database:
These transformation pairs are conjectured:
A349079 <--> A------.
("A------" means not yet in the database.)
Some sequences in the lists above may need offset adjustment to force a beginning with 1,2,... in the transformation.
EXAMPLE
a(4) = 17. The 17 transformation pairs of length 4 are:
{1, 2, 3, 4} = S({1, 2, 6, 24}).
{1, 2, 3, 5} = S({1, 2, 6, 15}).
{1, 2, 3, 6} = S({1, 2, 6, 12}).
{1, 2, 3, 9} = S({1, 2, 6, 9}).
{1, 2, 3, 12} = S({1, 2, 6, 8}).
{1, 2, 3, 21} = S({1, 2, 6, 7}).
{1, 2, 4, 5} = S({1, 2, 4, 20}).
{1, 2, 4, 6} = S({1, 2, 4, 12}).
{1, 2, 4, 8} = S({1, 2, 4, 8}).
{1, 2, 4, 12} = S({1, 2, 4, 6}).
{1, 2, 4, 20} = S({1, 2, 4, 5}).
{1, 2, 6, 7} = S({1, 2, 3, 21}).
{1, 2, 6, 8} = S({1, 2, 3, 12}).
{1, 2, 6, 9} = S({1, 2, 3, 9}).
{1, 2, 6, 12} = S({1, 2, 3, 6}).
{1, 2, 6, 15} = S({1, 2, 3, 5}).
{1, 2, 6, 24} = S({1, 2, 3, 4}).
b(1) = 1 by definition, b(2) = 1+1 as 1 has only 1 as divisor.
a(3) = A000005(b(2)*b(2)) = 3.
The divisors of b(2) are 1,2,4. So b(3) can be b(2)+1, b(2)+2 and b(2)+4.
CROSSREFS
Cf. A000005.
KEYWORD
more,nonn
AUTHOR
Thomas Scheuerle, Aug 19 2022
STATUS
approved