A276457 a(n) is the number of times that a(n-1) appears in the concatenation of all numbers from a(0) to a(n-2), with a(0) = 0.

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

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(n-19), 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, j-1}];
A[[j+1]]=SequenceCount[Flatten[B], IntegerDigits[A[[j]]]], {j, 3, Nm-1}];
A

Crossrefs

Cf. A142150, A171181, A248034.

Similar in spirit to van Eck's A181391.

Sequence in context: A257770 A027656 A142150 * A171181 A364788 A309261

Adjacent sequences: A276454 A276455 A276456 * A276458 A276459 A276460

Keyword

nonn,base,look,hear

Author

Yuriy Sibirmovsky, Sep 03 2016

Status

approved