A060310 -id:A060310 - cp's OEIS frontend

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

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

A093095 Start with a(1) = 2, a(2) = 3, then apply the rule of A093094.

Original entry on oeis.org

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

Offset: 1

Bodo Zinser, Mar 20 2004

			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)=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, A093096, A093097, A096381.

a093095 n = a093095_list !! (n-1)
a093095_list = f [2,3] 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 08 2013

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

from itertools import islice
from collections import deque
def agen(): # generator of terms
    a = deque([2, 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

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

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

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A093095 Start with a(1) = 2, a(2) = 3, 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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