A225224 -id:A225224 - cp's OEIS frontend

A327036 a(0)=0; for n > 0, a(n) is the number of distinct digit strings that occur at least twice (not counting overlapping occurrences) in the concatenation of all previous terms.

0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 12, 13, 13, 14, 15, 15, 16, 17, 17, 18, 19, 19, 21, 22, 24, 24, 25, 26, 26, 27, 28, 28, 29, 30, 30, 31, 32, 33, 36, 36, 37, 38, 38, 39, 40, 40, 41, 43, 44, 47, 47, 48, 49, 49, 51, 51, 54, 55, 58, 58, 59, 60, 60

Offset: 0

Scott R. Shannon, Nov 28 2019

To calculate a(n) take all the terms from a(0) to a(n-1), concatenate them, and then count the distinct digit strings that have at least two nonoverlapping occurrences. For example, if the concatenated terms formed the string '2210102240404' then the next term would be 8 as the strings '0','1','2','4','04','10','22','40' have all occurred at least twice. Note that the string '404' is not counted as its two occurrences overlap.

In the first 20000 terms, the largest increase in consecutive terms is from a(16496) = 45375 to a(16497) = 45410, an increase of 35; the concatenation of a(16496) to the previous terms results in the longest repeated string, '245352453624537'.

			a(1) = 0 as there have been no repeated strings prior to a(1).
a(2) = 1 as there has been one repeat of '0', which occurs in a(0) and a(1).
a(3) = 1 as only '0' has repeated up to a(2).
a(4) = 2 as now both '0' and '1' have repeated up to a(3).
a(22) = 12 as the strings '0' to '10' have repeated, but '10,10' contains the string '01' which also appears from a(1) to a(2), thus '01' has also repeated.
a(23) = 13 as a(22) = '12' created a repeat of the string '12', with a(3) and a(4). Even though the terms '10,10,12' contain two occurrences of the string '101' they overlap so are not counted. The string '10,12' also contains another '01' but that has already repeated and been counted previously so is ignored as the count is of distinct strings.

Cf. A225224, A329818, A329448.

Cf. A330015 (same sequence, but overlapping instances of the same digit string are counted among the repeats).

A330015 a(0) = 0; for n > 0, a(n) is the number of distinct digit strings that occur at least twice (including any overlapping occurrences) in the concatenation of all previous terms.

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 12, 14, 14, 15, 17, 17, 18, 20, 20, 21, 24, 24, 25, 27, 27, 28, 30, 30, 31, 33, 34, 36, 36, 37, 39, 39, 40, 40, 41, 43, 45, 46, 46, 47, 49, 49, 51, 52, 54, 56, 57, 57, 58, 60, 60, 61, 63, 63, 65, 67, 68, 68, 70, 70, 71, 73, 75

Offset: 0

Scott R. Shannon, Nov 27 2019

To calculate a(n) take all the terms from a(0) to a(n-1), concatenate them, and then count the distinct digit strings that have at least two occurrences (including any that overlap). For example, if the concatenated terms formed the string '2210102240404' then the next term would be 9 as the strings '0','1','2','4','04','10','22','40','404' have all occurred at least twice. (The two occurrences of '404' overlap.)

In the first 20000 terms, the largest increase in consecutive terms is from a(13058) = 47849 to a(13059) = 47885, an increase of 36, and the concatenation of a(13006) = 47623 to the previous terms results in the longest repeated string, '47611476114761147'.

			a(1) = 0 as there have been no repeated strings prior to a(1).
a(2) = 1 as there has been one repeat of '0', which occurs in a(0) and a(1).
a(3) = 1 as only '0' has repeated up to a(2).
a(4) = 2 as now both '0' and '1' have repeated up to a(3).
a(22) = 12 as the strings '0' to '10' have repeated, but '10,10' contains the string '01' which also appears from a(1) to a(2), thus '01' has also repeated.
a(23) = 14 as a(22) = '12' created a repeat of the string '12', with a(3) and a(4), and also created an overlapping repeat of the string '101' which is in both '10,10' and '10,12'. Note this later string also contains '01' but that has already repeated and been counted previously so is ignored as the count is of distinct strings.

Cf. A225224, A329818, A329448.

Cf. A327036 (same sequence, but overlapping instances of the same digit string are not counted among the repeats).

A225334 Look-and-repeat: similar to look-and-say except frequency is repeated (seed 3).

Original entry on oeis.org

3, 113, 221113, 222331113, 332223331113, 223332333331113, 222333112553331113, 332333221112225333331113, 223112333222331332115553331113, 222113221112333332223111223112221335333331113

Offset: 1

Jean-Christophe Hervé, May 12 2013

Look-and-repeat sequence A225329 with seed 3.
Contains 1, 2, 3 and 5, but not 4.
All terms end with 3 (the seed) and, starting at the fourth, with 3331113, which makes the 5 appear.
All terms except the second begin with 2 or 3; it is a direct consequence of the look-and-repeat rule.
Sequence A225333, the look-and-repeat sequence with seed 2 is almost the same, differing only in the last digit of each term (2 instead of 3 here). Indeed, any one-digit seed except 1 leads to essentially the same sequence with all terms identical except the last one = the seed.

			The term after 113 is obtained by saying (repeating) two two 1, one one 3, which gives 221113.

Cf. A225329 (seed one), A225333 (seed two), A225332 (continuous look-and-repeat, seed 2), A225330, A225331 (continuous look-and-repeat, seed 1).

Cf. A005150 (original look-and-say), A225224, A221646, A225212 (continuous look-and-say versions).

a(n) = A225333(n) + 1.

A330027 a(0) = 0; for n > 0, a(n) is the total number of repeated strings in the concatenation of terms a(0) to a(n-1). Repeated strings can overlap.

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 12, 16, 18, 20, 22, 25, 28, 30, 32, 34, 38, 40, 42, 44, 47, 50, 52, 55, 59, 62, 64, 66, 69, 72, 74, 76, 79, 82, 85, 88, 91, 96, 99, 103, 112, 118, 125, 130, 134, 140, 144, 150, 155, 161, 168, 173, 177, 183, 188, 194, 199, 205, 210

Offset: 0

Scott R. Shannon, Nov 27 2019

This sequence uses the same rule as A330015 but, instead of the count of unique repeated strings, here the total number of repeated strings seen in the concatenation of a(0) to a(n-1) forms the next term. After a(19) all single digits have been seen, so from that entry all new terms increase the previous term by at least the number of digits in the previous term. In this sequence repeated strings can be overlapping, thus a new entry '444' would add at least two to the total repeated count if '44' was already in the sequence, or at least one otherwise as '444' contains a repeat of '44' itself.

			a(4) = 2 as both '0' and '1' have repeated in the terms a(0) to a(3).
a(21) = 12, as a(11) = 10 which added another appearance of '0' and '1', so the total repeated string count incremented by two.
a(22) = 16, as a(21) = 12 which added another appearance of '1','2','12','01', so the total repeated string count incremented by four.

Cf. A225224, A329818, A329448, A330015, A327036.

A225335 A continuous "look-and-repeat" sequence (seed 3).

Original entry on oeis.org

3, 1, 1, 3, 2, 2, 1, 1, 1, 3, 2, 2, 2, 3, 3, 1, 1, 1, 3, 3, 3, 2, 2, 2, 3, 3, 3, 1, 3, 3, 3, 3, 3, 2, 3, 3, 3, 1, 1, 1, 5, 5, 3, 1, 1, 2, 3, 3, 3, 3, 3, 1, 2, 2, 5, 1, 1, 3, 2, 2, 1, 1, 1, 2, 5, 5, 3, 1, 1, 1, 2, 2

Offset: 1

Jean-Christophe Hervé, May 12 2013

The 'look-and-repeat' sequence A225330, with seed 3. The variant A225331 with the same seed 3 gives this same sequence.
It describes at each step the preceding digits by repeating the frequency number.
The sequence is determined by triples of digits. The first two terms of a triple are the repeated frequency and the last term is the digit.
a(n) is always equal to 1, 2, 3, or 5. No series of four identical digits happens in the sequence, nor any of five 5's.
Applying the look-and-repeat principle to the sequence itself, it is simply shift one rank to the left.

			a(1) = 3, you then see "one 3" and repeating "one", a(2) = a(3) = 1 (one) and a(4) = 3; you have then two 1's, so 2,2,1 ; then one 3, so 1,1,3, etc.

Cf. A225330, A225331 (two variants with seed 1), A225331 (seed 2), A225329, A225333, A225334 (look-and-repeat by block with resp. seed 1, seed 2, seed 3).

Cf. A005150 (original look-and-say), A225224, A221646, A225212 (continuous look-and-say versions).

cp's OEIS Frontend

A327036 a(0)=0; for n > 0, a(n) is the number of distinct digit strings that occur at least twice (not counting overlapping occurrences) in the concatenation of all previous terms.

Views

Author

Keywords

Comments

Examples

Crossrefs

A330015 a(0) = 0; for n > 0, a(n) is the number of distinct digit strings that occur at least twice (including any overlapping occurrences) in the concatenation of all previous terms.

Views

Author

Keywords

Comments

Examples

Crossrefs

A225334 Look-and-repeat: similar to look-and-say except frequency is repeated (seed 3).

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A330027 a(0) = 0; for n > 0, a(n) is the total number of repeated strings in the concatenation of terms a(0) to a(n-1). Repeated strings can overlap.

Views

Author

Keywords

Comments

Examples

Crossrefs

A225335 A continuous "look-and-repeat" sequence (seed 3).

Views

Author

Keywords

Comments

Examples

Crossrefs