A364697 Lexicographically earliest permutation of the positive integers such that the successive cumulative products reproduce the sequence itself, digit by digit.

1, 11, 2, 25, 50, 27, 500, 7, 4, 2500, 3, 71, 250000, 259, 8, 750000, 10, 39, 5000000, 2598, 7500000000, 77, 9, 6, 2500000000, 5, 53, 533, 75000000001, 38, 383, 43, 75000000000000, 35, 84, 13, 103, 12, 5000000000000, 28, 67, 30, 48, 25000000000000000, 21, 504, 78, 61, 87

Offset

1,2

Comments

If we want the sequence to be the lexicographically earliest permutation of the integers > 0, we must start with a(1) = 1 and a(2) = 11. With a(2) < 11, the sequence stops immediately.

Links

Table of n, a(n) for n=1..49.

Eric Angelini, Cumulative Sums, Personal blog.

Example

a(1) = 1
a(1) * a(2) = 11
a(1) * a(2) * a(3) = 22
a(1) * a(2) * a(3) * a(4) = 550
a(1) * a(2) * a(3) * a(4) * a(5) = 27500
a(1) * a(2) * a(3) * a(4) * a(5) * a(6) = 742500; etc.
The succession of the above results is:
  1, 11, 22, 550, 27500, 742500, ...
The first terms of the sequence are:
  1, 11, 2, 25, 50, 27, 500, 7, 4, 2500,, ...
We see that the successive digits are the same in the two sequences.

Mathematica

Nest[(a=#; AppendTo[a, (new=Flatten[IntegerDigits/@Table[Times@@a[[;; i]], {i, Length@a}]][[Length@Flatten[IntegerDigits/@a]+1;; ]];
k=1; While[MemberQ[a, FromDigits@new[[;; k]]]||new[[k+1]]==0, k++]; FromDigits@new[[;; k]])])&, {1, 11, 2, 25}, 45] (* Giorgos Kalogeropoulos, Aug 05 2023 *)

Crossrefs

Cf. A309151, A364664.

Sequence in context: A336904 A051309 A077344 * A298439 A095157 A110767

Adjacent sequences: A364694 A364695 A364696 * A364698 A364699 A364700

Keyword

base,nonn

Author

Eric Angelini, Aug 03 2023

Status

approved