A214365 -id:A214365 - 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

A339359 Irregular triangle read by rows; the first row simply contains the value 1; given the succession of digits of the n-th row, say [d_0, ..., d_k], the (n+1)-th row is [d_0, d_0+d_1, d_1+d_2, ..., d_{k-1}+d_k, d_k].

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 6, 1, 1, 1, 5, 6, 1, 1, 7, 12, 7, 2, 2, 6, 11, 7, 1, 1, 8, 8, 3, 9, 9, 4, 8, 7, 2, 8, 8, 1, 1, 9, 16, 11, 12, 18, 13, 12, 15, 9, 10, 16, 9, 1, 1, 10, 10, 7, 7, 2, 2, 3, 3, 9, 9, 4, 4, 3, 3, 6, 14, 10, 1, 1, 7, 15, 10, 1

Offset: 0

Rémy Sigrist, Dec 02 2020

This sequence combines features of Pascal's triangle (A007318) and of A214365.
Rows 0 to 5 match that of Pascal's triangle, thereafter the values differ.
Every column is eventually periodic.

			The first rows are:
    1
    1, 1
    1, 2, 1
    1, 3, 3, 1
    1, 4, 6, 4, 1
    1, 5, 10, 10, 5, 1
    1, 6, 6, 1, 1, 1, 5, 6, 1
    1, 7, 12, 7, 2, 2, 6, 11, 7, 1
    1, 8, 8, 3, 9, 9, 4, 8, 7, 2, 8, 8, 1
    1, 9, 16, 11, 12, 18, 13, 12, 15, 9, 10, 16, 9, 1

Paolo Xausa, Table of n, a(n) for n = 0..15582 (rows 0..30 of triangle, flattend).

See A339379 for a similar sequence.

Cf. A007318, A214365.

NestList[Map[Total, Partition[Flatten[IntegerDigits[#]], 2, 1, {-1, 1}, 0]] &, {1}, 10] (* Paolo Xausa, Aug 19 2025 *)

{ r=[1]; for (n=0, 10, apply (v -> print1(v", "), r); d=concat(apply(digits, r)); r=vector(#d+1, k, if (k==1, d[1], k==#d+1, d[#d], d[k-1]+d[k]))) }

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.

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