OFFSET
1,2
COMMENTS
Most of the terms appear to have relatively few and large prime factors.
For single-digit n, the digit length of a(n) is 2^n-1, with each digit n, n-1, ..., 1 occurring 2^0, 2^1, ..., 2^(n-1) times. However, this simple regularity will not hold for n >= 10.
An alternative construction method is using reverse S of A082850.
Yet another way is doubling a(n-1) and prepending n to get a(n). This formula is valid only for n <= 10: a(n) = n*10^(2^n-2) + a(n-1)*10^(2^(n-1)-1) + a(n-1).
EXAMPLE
For n = 3, to get a(3) we first concatenate the n's in reverse order: 3, 2 and 1, then we continue concatenating to this chain the a(n)'s in forward order up to a(2): 1, 211. Thus the resulting chain a(3) = 3211211.
For n = 4, a(4) is concatenation
a(4) = 4 3 2 1 1 211 3211211
parts: n ... 1 a(1) a(2) a(3)
MATHEMATICA
a[1] = 1; a[n_] := a[n] = FromDigits[Join @@ IntegerDigits[Join[Range[n, 1, -1], Array[a, n - 1]]]]; Array[a, 6] (* Amiram Eldar, Dec 05 2022 *)
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Tamas Sandor Nagy, Dec 04 2022
STATUS
approved