A392141 a(0) = 1 and a(1) = 2, then each subsequent term is obtained by multiplying the two previous terms and then deleting repeated digits, keeping only the first occurrence of each digit.

1, 2, 2, 4, 8, 32, 256, 8192, 209715, 1798520, 3716280, 683095, 2538760, 1734260, 40286917, 6987420, 28150694, 19670248, 53712, 10562837, 56731094, 592418736, 3608529714, 2137609854, 713628509, 152493768, 1082396, 1650842, 1786432, 2941675, 521036, 1532780, 7986350, 1243750, 9302815

Offset

0,2

Comments

The first repeated number is 481653 = a(2085) = a(3165).

From Michael S. Branicky, Jan 01 2026: (Start)

Maximum term is 9876523140 = a(230773).

Periodic with period 167227 beginning at (a(455403), a(455404)) = (92407138, 2904378651) = (a(622630), a(622631)). (End)

Links

Michael S. Branicky, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, order 167227.

Formula

For n >= 1: a(n+1) = A137564(a(n-1)*a(n)). - Michael S. Branicky, Jan 01 2026

Example

For n = 8 we have that 256 * 8192 = 2097152, then since 2 is repeated a(8) = 209715.
For n = 9 we have that 8192 * 209715 = 1717985280, then without repeated digits a(9) = 1798520.

Mathematica

a[n_] := a[n] = FromDigits[DeleteDuplicates[IntegerDigits[a[n-1]*a[n-2]]]]; a[0] = 1; a[1] = 2; Array[a, 35, 0] (* Amiram Eldar, Jan 01 2026 *)

Prog

(Python)
from itertools import islice
def f(n): # A137564
    seen, out, s = set(), "", str(n)
    for d in s:
        if d not in seen: out += d; seen.add(d)
    return int(out)
def A392141(): # generator of terms
    anm1, an = 1, 2
    yield anm1
    while True:
        yield an
        anm1, an = an, f(anm1*an)
print(list(islice(A392141(), 35))) # Michael S. Branicky, Jan 01 2026

Crossrefs

Cf. A000301, A137564.

Sequence in context: A070323 A109213 A109214 * A000301 A360808 A124439

Adjacent sequences: A392138 A392139 A392140 * A392142 A392143 A392144

Keyword

nonn,base,easy

Author

Rodolfo Kurchan, Jan 01 2026

Status

approved