

A276457


a(n) is the number of times that a(n1) appears in the concatenation of all numbers from a(0) to a(n2), with a(0) = 0.


4



0, 0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, 0, 8, 0, 9, 0, 10, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 2, 10, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 3, 10, 4, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 4, 10, 5, 5, 6, 5, 7, 5, 8, 5, 9, 5, 10, 6, 6, 7, 6, 8, 6, 9, 6, 10, 7, 7, 8, 7, 9, 7, 10
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OFFSET

0,5


COMMENTS

138, 185 and 199 are three smallest numbers that do not appear among the first 5000 terms of the sequence. They first appear at n = 8776, 5117 and 10580 respectively.
1187 and 1190 are two smallest numbers that do not appear among the first 100000 terms.
Question: will every natural number eventually appear in the sequence?
The sequence can be started with any number a(0). The terms will be different, but for larger n behavior will be similar for all a(0).


LINKS

Yuriy Sibirmovsky, Table of n, a(n) for n = 0..100000
Yuriy Sibirmovsky, Plot for n=0..4999
Yuriy Sibirmovsky, Plot for n=0..59999


FORMULA

a(n) = A142150(n) = A171181(n), if 0<=n<=20.
a(n) = A248034(n19), if 21<=n<=120.  Omar E. Pol, Sep 03 2016


EXAMPLE

From a(0) to a(0), a(1) appears once, thus a(2) = 1.
From a(0) to a(1), a(2) appears 0 times, thus a(3) = 0.
...
From a(0) to a(19), a(20) = 10 appears once, in the form of '1,0'. Thus a(21) = 1.


MATHEMATICA

Nm=100;
A=Table[0, {j, 1, Nm}];
A[[3]]=1;
Do[B=Table[IntegerDigits[A[[l]]], {l, 1, j1}];
A[[j+1]]=SequenceCount[Flatten[B], IntegerDigits[A[[j]]]], {j, 3, Nm1}];
A


CROSSREFS

Cf. A142150, A171181, A248034.
Similar in spirit to van Eck's A181391.
Sequence in context: A257770 A027656 A142150 * A171181 A034948 A135472
Adjacent sequences: A276454 A276455 A276456 * A276458 A276459 A276460


KEYWORD

nonn,base,look,hear


AUTHOR

Yuriy Sibirmovsky, Sep 03 2016


STATUS

approved



