A358873 - OEIS

%I #26 Dec 07 2022 09:10:37

%S 1,211,3211211,432112113211211,5432112113211211432112113211211,

%T 654321121132112114321121132112115432112113211211432112113211211

%N a(1) = 1. For n >= 2, to obtain a(n), concatenate the numbers n,...,1,a(1),...,a(n-1).

%C Most of the terms appear to have relatively few and large prime factors.

%C 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.

%C An alternative construction method is using reverse S of A082850.

%C 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).

%e 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.

%e For n = 4, a(4) is concatenation

%e   a(4) =  4 3 2 1   1    211  3211211

%e   parts:  n ... 1  a(1)  a(2)   a(3)

%t 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 *)

%Y Cf. A082850.

%K nonn,base,easy

%O 1,2

%A _Tamas Sandor Nagy_, Dec 04 2022