A093095 -id:A093095 - cp's OEIS frontend

A093094 "Products into digits": start with a(1)=2, a(2)=2; adjoin digits of product of a(k) and a(k+1) for k from 1 to infinity.

2, 2, 4, 8, 3, 2, 2, 4, 6, 4, 8, 2, 4, 2, 4, 3, 2, 1, 6, 8, 8, 8, 1, 2, 6, 2, 6, 4, 8, 6, 4, 6, 4, 8, 2, 1, 2, 1, 2, 1, 2, 2, 4, 3, 2, 4, 8, 2, 4, 2, 4, 2, 4, 3, 2, 1, 6, 2, 2, 2, 2, 2, 2, 4, 8, 1, 2, 6, 8, 3, 2, 1, 6, 8, 8, 8, 8, 8, 1, 2, 6, 2, 6, 1, 2, 4, 4, 4, 4, 4, 8, 3, 2, 8, 2, 1, 2, 4, 8, 2, 4

Offset: 1

Bodo Zinser, Mar 20 2004

Only the digits 1,2,3,4,6,8 occur, infinitely often. The sequence is not periodic. Around a(800) there are many 8's.
From Giovanni Resta, Mar 16 2006: (Start)
Proof that sequence is not periodic:
Let us assume that somewhere in the sequence there is a subsequence of 3 adjacent 8': ...,8,8,8,....(which is true).
Then we know that in the following there will be the subsequence ...,6,4,6,4.. (i.e. 8x8, 8x8) again, there will be somewhere ...,2,4,2,4,2,4,... (i.e. 6x4, 4x6, 6x4) and finally ...,8,8,8,8,8,...
Analogously, starting from 8,8,8,8 we obtain 6,4,6,4,6,4 then 2,4,2,4,2,4,2,4,2,4 and finally 8,8,8,8,8,8,8,8,8.
Generalizing, if somewhere appears a run of k>2 8's, then in some future position will appear a run of at least 4*k-7 8's (where since k>2, 4*k-7>k).
So the sequence will contain arbitrary long runs of 8's, without being constantly equal to 8, thus it cannot be periodic. (End)
Essentially the same as A045777. [R. J. Mathar, Sep 08 2008]

			a(3)=a(1)*a(2), a(4)=a(2)*a(3), a(5)=first digit of (a(3)*a(4)), a(6)=2nd digit of (a(3)*a(4)), a(9)=a(6)*a(7)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A045777, A093086, A093087, A093088, A093089, A093090, A093091, A093095, A093096, A093097, A096381.

a093094 n = a093094_list !! (n-1)
a093094_list = f [2,2] where
   f (u : vs@(v : _)) = u : f (vs ++
     if w < 10 then [w] else uncurry ((. return) . (:)) $ divMod w 10)
        where w = u * v
-- Reinhard Zumkeller, Aug 08 2013

Fold[Join[#, IntegerDigits[Times @@ #[[#2;; #2+1]]]] &, {2, 2}, Range[100]] (* Paolo Xausa, Aug 18 2025 *)

from itertools import islice
from collections import deque
def agen(): # generator of terms
    a = deque([2, 2])
    while True:
        a.extend(list(map(int, str(a[0]*a[1]))))
        yield a.popleft()
print(list(islice(agen(), 101))) # Michael S. Branicky, Feb 15 2024

Definition revised by Franklin T. Adams-Watters, Mar 16 2006

A093097 Start with a(1) = 3, a(2) = 7; then apply the rule of A093094.

Original entry on oeis.org

3, 7, 2, 1, 1, 4, 2, 1, 4, 8, 2, 4, 3, 2, 1, 6, 8, 1, 2, 6, 2, 6, 4, 8, 8, 2, 1, 2, 1, 2, 1, 2, 2, 4, 3, 2, 6, 4, 1, 6, 2, 2, 2, 2, 2, 2, 4, 8, 1, 2, 6, 1, 2, 2, 4, 4, 6, 1, 2, 4, 4, 4, 4, 4, 8, 3, 2, 8, 2, 1, 2, 6, 2, 4, 8, 1, 6, 2, 4, 6, 2, 8, 1, 6, 1, 6, 1, 6, 1, 6, 3, 2, 2, 4, 6, 1, 6, 1, 6, 2, 2, 1, 2, 1, 2

Offset: 1

Bodo Zinser, Mar 20 2004

			a(3)=first digit of (a(1)*a(2)), a(4)=2nd digit of (a(1)*a(2)), a(5)=first digit of (a(2)*a(3)), a(6)=2nd digit of (a(2)*a(3))

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A045777, A060310, A093094, A093095, A093096, A096381.

Fold[Join[#, IntegerDigits[Times @@ #[[#2;; #2+1]]]] &, {3, 7}, Range[100]] (* Paolo Xausa, Aug 18 2025 *)

from itertools import islice
from collections import deque
def agen(): # generator of terms
    a = deque([3, 7])
    while True:
        a.extend(list(map(int, str(a[0]*a[1]))))
        yield a.popleft()
print(list(islice(agen(), 105))) # Michael S. Branicky, Aug 18 2025

A093096 Start with a(1) = a(2) = 3; then apply the rule of A093094.

Original entry on oeis.org

3, 3, 9, 2, 7, 1, 8, 1, 4, 7, 8, 8, 4, 2, 8, 5, 6, 6, 4, 3, 2, 8, 1, 6, 4, 0, 3, 0, 3, 6, 2, 4, 1, 2, 6, 1, 6, 8, 6, 2, 4, 0, 0, 0, 0, 1, 8, 1, 2, 8, 4, 2, 1, 2, 6, 6, 4, 8, 4, 8, 1, 2, 8, 0, 0, 0, 0, 0, 8, 8, 2, 1, 6, 3, 2, 8, 2, 2, 1, 2, 3, 6, 2, 4, 3, 2, 3, 2, 3, 2, 8, 2, 1, 6, 0, 0, 0, 0, 0, 0, 6, 4, 1, 6, 2

Offset: 1

Bodo Zinser, Mar 20 2004

Essentially the same as A060310. [From R. J. Mathar, Sep 08 2008]

			a(3)=a(1)*a(2), a(4)=first digit of (a(2)*a(3)), a(5)=2nd digit of (a(2)*a(3)), a(6)=first digit of (a(3)*a(4)), a(7)=2nd digit of (a(3)*a(4))

Hugo Steinhaus, Studentenfutter, Urania, Leipzig 1991, #1.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A045777, A060310, A093094, A093095, A093097, A096381.

a093096 n = a093096_list !! (n-1)
a093096_list = f [3,3] where
   f (u : vs@(v : _)) = u : f (vs ++
     if w < 10 then [w] else uncurry ((. return) . (:)) $ divMod w 10)
        where w = u * v
-- Reinhard Zumkeller, Aug 13 2013

Fold[Join[#, IntegerDigits[Times @@ #[[#2;; #2+1]]]] &, {3, 3}, Range[100]] (* Paolo Xausa, Aug 18 2025 *)

from itertools import islice
from collections import deque
def agen(): # generator of terms
    a = deque([3, 3])
    while True:
        a.extend(list(map(int, str(a[0]*a[1]))))
        yield a.popleft()
print(list(islice(agen(), 105))) # Michael S. Branicky, Aug 18 2025

A096381 Beginning with 2, 7, multiply successive pairs of members and adjoin the result as the next one or two members of the sequence, depending on whether the product is a one- or two-digit number.

Original entry on oeis.org

2, 7, 1, 4, 7, 4, 2, 8, 2, 8, 8, 1, 6, 1, 6, 1, 6, 6, 4, 8, 6, 6, 6, 6, 6, 3, 6, 2, 4, 3, 2, 4, 8, 3, 6, 3, 6, 3, 6, 3, 6, 1, 8, 1, 8, 1, 2, 8, 1, 2, 6, 8, 3, 2, 2, 4, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 6, 8, 8, 8, 8, 2, 1, 6, 8, 2, 1, 2, 4, 8, 2, 4, 6, 4, 8, 4, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 4, 8

Offset: 1

Gerry Myerson, Aug 04 2004

Berzsenyi sets the puzzle of showing that 6 occurs infinitely often in the sequence. It is easy to compose variations on the sequence, e.g., vary a(1) and a(2), or use a base other than 10, or use the product of three successive members instead of 2.

			a(1)a(2) = 14, so a(3) = 1 and a(4) = 4.

George Berzsenyi, Competition Corner problem 468, The Mathematics Student (published by NCTM), Vol. 26, No. 2, November 1978.
Loren C. Larson, Problem-Solving Through Problems, Springer, 1983, page 8, Problem 1.1.6.

Robert Israel, Table of n, a(n) for n = 1..10000
George Berzsenyi, István Laukó, and Gabriella Pintér, The Competition Corner in the Mathematics Student, 2021. See p. 2, problem 1/2/1.

Cf. A045777, A060310, A093094, A093095, A093096, A093097.

a=2:7:concat[(if x*y>9then[x*y`div`10]else[])++[x*y`mod`10]|(x,y)<-a`zip`tail a] -- Paul Stoeber (pstoeber(AT)uni-potsdam.de), Oct 08 2005

R:= 2,7: count:= 2:
for i from 1 while count < 200 do
  t:= R[i]*R[i+1];
  if t >= 10 then R:= R, floor(t/10),t mod 10; count:= count+2 else R:= R, t;
count:= count+1 fi;
od:
R; # Robert Israel, Jan 16 2018

Fold[Join[#, IntegerDigits[Times @@ #[[#2;;#2+1]]]] &, {2, 7}, Range[100]] (* Paolo Xausa, Aug 17 2025 *)

from itertools import islice
from collections import deque
def agen(): # generator of terms
    a = deque([2, 7])
    while True:
        a.extend(list(map(int, str(a[0]*a[1]))))
        yield a.popleft()
print(list(islice(agen(), 105))) # Michael S. Branicky, Aug 18 2025

Corrected by Robert Israel, Jan 16 2018

A060310 a(1)=1, a(2)=3; then append digits of a(n-1)*a(n).

Original entry on oeis.org

1, 3, 3, 9, 2, 7, 1, 8, 1, 4, 7, 8, 8, 4, 2, 8, 5, 6, 6, 4, 3, 2, 8, 1, 6, 4, 0, 3, 0, 3, 6, 2, 4, 1, 2, 6, 1, 6, 8, 6, 2, 4, 0, 0, 0, 0, 1, 8, 1, 2, 8, 4, 2, 1, 2, 6, 6, 4, 8, 4, 8, 1, 2, 8, 0, 0, 0, 0, 0, 8, 8, 2, 1, 6, 3, 2, 8, 2, 2, 1, 2, 3, 6, 2, 4, 3, 2, 3, 2, 3, 2, 8, 2, 1, 6, 0, 0, 0, 0, 0, 0, 6, 4, 1, 6

Offset: 1

Jason Earls, Mar 27 2001

			1*3=3, 3*3=9, 3*9=27, 9*2=18, ...

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A045777, A093094, A093095, A093096, A093097, A096381.

Fold[Join[#, IntegerDigits[Times @@ #[[#2;; #2+1]]]] &, {1, 3}, Range[100]] (* Paolo Xausa, Aug 18 2025 *)

from itertools import islice
from collections import deque
def agen(): # generator of terms
    a = deque([1, 3])
    while True:
        a.extend(list(map(int, str(a[0]*a[1]))))
        yield a.popleft()
print(list(islice(agen(), 105))) # Michael S. Branicky, Aug 18 2025

cp's OEIS Frontend

A093094 "Products into digits": start with a(1)=2, a(2)=2; adjoin digits of product of a(k) and a(k+1) for k from 1 to infinity.

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A093097 Start with a(1) = 3, a(2) = 7; then apply the rule of A093094.

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A093096 Start with a(1) = a(2) = 3; then apply the rule of A093094.

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A096381 Beginning with 2, 7, multiply successive pairs of members and adjoin the result as the next one or two members of the sequence, depending on whether the product is a one- or two-digit number.

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A060310 a(1)=1, a(2)=3; then append digits of a(n-1)*a(n).

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Programs

Mathematica

Python