A093087 -id:A093087 - cp's OEIS frontend

A093086 "Fibonacci in digits": start with a(0)=0, a(1)=1; repeatedly adjoin the digits of the sum of the next two terms.

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

Offset: 0

Bodo Zinser, Mar 20 2004

Formally, define strings of digits S_i as follows. S_0={0}, S_1={0,1}. For n >= 1, let S_n={t_0, t_1, ..., t_z}. Then S_{n+1} is obtained by adjoining the digits of t_{n-1}+t_n to S_n. The sequence gives the limiting string S_oo.

All digits appear infinitely often, although the sequence is not periodic.

			After S_6 = {0,1,1,2,3,5,8} we have 5+8 = 13, so we get
S_7 = {0,1,1,2,3,5,8,1,3}. Then 8+1 = 9, so we get
S_8 = {0,1,1,2,3,5,8,1,3,9}. Then 1+3 = 4, so we get
S_9 = {0,1,1,2,3,5,8,1,3,9,4}, and so on.

Hakan Icoz, Table of n, a(n) for n = 0..20000

Cf. A093087-A093098, A105967, A102085, A214365.

with(linalg): A:=matrix(1,2,[0,1]): for n from 1 to 100 do if A[1,n]+A[1,n+1]<10 then A:=concat(A,matrix(1,1,A[1,n]+A[1,n+1])) else A:=concat(A,matrix(1,2,[1,A[1,n]+A[1,n+1]-10])) fi od: matrix(A); # Emeric Deutsch, May 31 2005

Fold[Join[#, IntegerDigits[Total[#[[#2;; #2+1]]]]] &, {0, 1}, Range[100]] (* Paolo Xausa, Aug 18 2025 *)

Edited by N. J. A. Sloane, Mar 20 2010

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.

Original entry on oeis.org

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

A093088 "Fibonacci-like in digits": start with a(1)=1, a(2)=4; repeatedly adjoin digits of sum of previous two terms.

Original entry on oeis.org

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

Offset: 1

Bodo Zinser, Mar 20 2004

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

Cf. A093086, A093087.

Fold[Join[#, IntegerDigits[Total[#[[#2;; #2+1]]]]] &, {1, 4}, Range[100]] (* Paolo Xausa, Aug 18 2025 *)

A093089 "Fibonacci in pairs": start with a(1)=1, a(2)=1; repeatedly adjoin sum of previous two terms but chopped from the right into pairs of 2 digits.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 1, 44, 90, 45, 1, 34, 1, 35, 46, 35, 35, 36, 81, 81, 70, 71, 1, 17, 1, 62, 1, 51, 1, 41, 72, 18, 18, 63, 63, 52, 52, 42, 1, 13, 90, 36, 81, 1, 26, 1, 15, 1, 4, 94, 43, 14, 1, 3, 1, 26, 1, 17, 82, 27, 27, 16, 16, 5, 98, 1, 37, 57, 15, 4, 4, 27, 27, 18

Offset: 1

Bodo Zinser, Mar 20 2004

Do all pairs of digits appear infinitely often? The sequence is not periodic.

			... a(11)=a(9)+a(10), a(12)=left pair of (a(10)+a(11)=55+89=1 44), a(13)=right pair of (a(10)+a(11)=55+89=1 44), a(14)=a(11)+a(12) ...

Cf. A093086, A093087, A093088.

A093090 Start with a(1)=1, a(2)=3, apply rule of A093089.

Original entry on oeis.org

1, 3, 4, 7, 11, 18, 29, 47, 76, 1, 23, 77, 24, 1, 0, 1, 1, 25, 1, 1, 2, 26, 26, 2, 3, 28, 52, 28, 5, 31, 80, 80, 33, 36, 1, 11, 1, 60, 1, 13, 69, 37, 12, 12, 61, 61, 14, 82, 1, 6, 49, 24, 73, 1, 22, 75, 96, 83, 7, 55, 73, 97, 74, 23, 97, 1, 71, 1, 79, 90, 62, 1, 28, 1, 70, 1, 71, 97, 1, 20

Offset: 1

Bodo Zinser, Mar 20 2004

Cf. A093086, A093087, A093088, A093089.

More terms from Bodo Zinser, Mar 21 2004

A093091 "Fibonacci in pairs from left": start with a(1)=1, a(2)=1; repeatedly adjoin sum of previous two terms but chopped from the left into pairs of 2 digits.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 14, 4, 10, 3, 18, 14, 13, 21, 32, 27, 34, 53, 59, 61, 87, 11, 2, 12, 0, 14, 8, 98, 13, 14, 12, 14, 22, 10, 6, 11, 1, 27, 26, 26, 36, 32, 16, 17, 12, 28, 53, 52, 62, 68, 48, 33, 29, 40, 81, 10, 5, 11, 4, 13, 0, 11, 6, 81, 62, 69, 12, 1, 91, 15, 16

Offset: 1

Bodo Zinser, Mar 20 2004

Do all pairs of digits appear infinitely often? The sequence is not periodic.

			... a(11)=a(9)+a(10), a(12)=left pair of (a(10)+a(11)=55+89=14 4),
a(13)=right pair of (a(10)+a(11)=55+89=14 4),
a(14)=left pair of (a(11)+a(12)=89+14=10 3),
a(15)=right pair of (a(11)+a(12)=89+14=10 3), a(16)=a(12)+a(13) ...

Cf. A093086, A093087, A093088, A093089, A093090.

A093092 "Fibonacci in digits - up and down": start with a(1)=1, a(2)=1; repeatedly adjoin either the sum of the two previous terms (if that sum happens to be even) or else adjoin digits of the sum of previous two terms (if that sum happens to be odd).

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 1, 3, 9, 4, 12, 1, 3, 16, 1, 3, 4, 1, 9, 1, 7, 4, 7, 5, 10, 10, 8, 1, 1, 1, 1, 12, 1, 5, 20, 18, 9, 2, 2, 2, 1, 3, 1, 3, 6, 2, 5, 38, 2, 7, 1, 1, 4, 4, 3, 4, 4, 4, 9, 8, 7, 4, 3, 40, 9, 8, 2, 5, 8, 7, 7, 8, 8, 1, 3, 1, 7, 1, 5, 1, 1, 7, 4, 3, 4, 9, 1, 7, 10, 7, 1, 3, 1, 5, 14, 1, 5, 16, 9, 4

Offset: 1

Bodo Zinser, Mar 20 2004

			... a(6)=a(4)+a(5), a(7)=left digit of (a(5)+a(6)=5+8=1 3) as 13 is odd, a(8)=right digit of (a(5)+a(6)=5+8=1 3) as 13 is odd, a(11)=a(8)+a(9) as even ...

Cf. A093086, A093087, A093088, A093089, A093090, A093091.

a = {0, 1}; f[n_] := Block[{k = a[[n - 1]] + a[[n - 2]]}, If[ EvenQ[k], AppendTo[a, k], a = Join[a, IntegerDigits[k]] ]]; Do[ f[n], {n, 3, 100}]; a (* Robert G. Wilson v, Mar 27 2004 *)

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A093086 "Fibonacci in digits": start with a(0)=0, a(1)=1; repeatedly adjoin the digits of the sum of the next two terms.

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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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A093088 "Fibonacci-like in digits": start with a(1)=1, a(2)=4; repeatedly adjoin digits of sum of previous two terms.

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A093089 "Fibonacci in pairs": start with a(1)=1, a(2)=1; repeatedly adjoin sum of previous two terms but chopped from the right into pairs of 2 digits.

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A093090 Start with a(1)=1, a(2)=3, apply rule of A093089.

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A093091 "Fibonacci in pairs from left": start with a(1)=1, a(2)=1; repeatedly adjoin sum of previous two terms but chopped from the left into pairs of 2 digits.

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A093092 "Fibonacci in digits - up and down": start with a(1)=1, a(2)=1; repeatedly adjoin either the sum of the two previous terms (if that sum happens to be even) or else adjoin digits of the sum of previous two terms (if that sum happens to be odd).

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