A329447 -id:A329447 - cp's OEIS frontend

A377905 Irregular triangle read by rows: T(n,k) (n >= 0, k >= 0, T[n,k]>0) is the total number of occurrences of the digit k in terms 0 to n of A329447.

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

Offset: 0

N. J. A. Sloane, Nov 19 2024

A329447(n) is constructed by finding the smallest number in row n-1 of this triangle, and in case of a tie, choosing the leftmost. If this is an entry c in column d, then A329447(n) = 10*c+d.

Row n has sum 2*n+1.

			[See links for initial rows of the triangle]

N. J. A. Sloane, Rows 0 to 15
N. J. A. Sloane, Rows 0 to 100
N. J. A. Sloane, Naive Maple code for A329447 and the present sequence

Cf. A329447.

Maple
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[See link]
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A376779 The terms of A329447 sorted into increasing order.

Original entry on oeis.org

0, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 33, 34, 35, 36, 37, 38, 39, 40, 44, 45, 46, 47, 48, 49, 50, 55, 56, 57, 58, 59, 60, 66, 67, 68, 69, 70, 77, 78, 79, 80, 88, 89, 90, 99, 100, 112, 113, 114, 115, 116, 117, 118, 119, 120, 123, 124, 125, 126, 127, 128, 129, 130, 134, 135, 136, 137, 138, 139, 140, 145, 146, 147, 148

Offset: 1

N. J. A. Sloane, Nov 11 2024

Is there an independent characterization of these numbers? The following seems like a promising attack. Let the sequence {b(n)} be obtained by deleting the right-most digit of each a(n). Then for k from 1 though 10, k appears 11-k times in {b(n)}, and the numbers 10*k+i missing from {a(n)} correspond to i = 1,2,...,k-1.

Cf. A329447.

A329448 Numbers n which appear twice in the sequence starting with n and subsequent terms given by 10c + d where c is the number of times (> 0) the least frequent digit d has appeared so far, smallest d in case of tie.

Original entry on oeis.org

10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 34, 35, 36, 37, 38, 39, 40, 44, 45, 46, 47, 48, 49, 50, 55, 56, 57, 58, 59, 60, 66, 67, 68, 69, 70, 77, 78, 79, 80, 88, 89, 90, 99, 100, 102, 103, 104, 105, 106, 107, 108, 109, 115, 122, 123, 124, 125, 126, 127, 128

Offset: 1

Eric Angelini and M. F. Hasler, Nov 14 2019

Sequence A329447 is defined as above with n = 0, but one can take any other initial term n. This sequence lists those n which reappear later in the resulting sequence. To know whether n reappears, it is sufficient to compute the sequence up to the point where the last digit of n has appeared at least n/10 times. (The sequence may be not strictly increasing, but each subsequence of terms ending in the same digit is, ignoring the initial term.)
There are 369 integers < 1000 and 2696 integers < 10^4 in this sequence.
All powers of 10 appear to be in the sequence, but this is only verified up to 10^7, not proved in general.

			Single-digit numbers cannot reappear: all terms are of the form 10c + d, c > 0.
10 does reappear, because after this initial value, the next term is 10 for "one '1' [appeared so far]".
11 does not reappear, because after this initial value, the digit 1 has appeared already twice, and a term 11 would mean that only "one 1" would have appeared so far.
The numbers n = 12, ..., 19 reappear because they are, as initial term, followed by 11 for "one '1' [has appeared so far]"; thereafter the second digit of n will be the least frequent one, having appeared once, so the next term will be n for "one (second digit of n) [has appeared so far]".
The number 20 reappears because it is followed by 10 (one '0'), then 11 (one '1'), then 12 (one '2'), and then 20 (two '0's).
The number 21 does not reappear because it is followed by 11 (one '1'); at this point three '1's have already appeared and we can never again have 21 meaning that only "two '1's" have appeared so far.

Eric Angelini, Look left and say, Nov 14 2019.

Cf. A329447.

select( is(n)={my(c=Vec(0,10),t=-n,i, n0=n%10+1, n1=n\10); while(t!=n&&c[n0]<=n1, for(j=1,#i=vecsort(c,,1), if(c[i[j]], t=c[i[j]]*10+i[j]-1; break)); for(j=1, #i=digits(t), c[i[j]+1]++));t==n&&n}, [0..130])

A329818 Look left and tell the least frequent digit string; a(0) = 0, a(n) = concat(c,s), where c > 0 is the number of times the string s, composed of one or more digits, has appeared in a(0) to a(n-1) such that a(n) is minimal. Digit strings with leading zeros are ignored.

Original entry on oeis.org

0, 10, 11, 20, 12, 22, 30, 13, 23, 33, 40, 14, 24, 34, 44, 50, 15, 25, 35, 45, 55, 60, 16, 26, 36, 46, 56, 66, 70, 17, 27, 37, 47, 57, 67, 77, 80, 18, 28, 38, 48, 58, 68, 78, 88, 90, 19, 29, 39, 49, 59, 69, 79, 89, 99, 100, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 122

Offset: 0

Scott R. Shannon, Nov 21 2019

This is a variation on A329447. Instead of just considering the number of times the digits 0 to 9 have appeared so far in the sequence, we consider all numbers formed by the substrings of each previous entry; from single digits to the entire entry. Each value of a(n) is determined by considering the count of all the substrings occurring in a(0) to a(n-1) and then choosing the one which forms the lowest number when the count of that substring is concatenated with the substring itself.
The only restriction in the substring counting is that any substrings which have leading zeros are ignored. For example if a(n) was 2001, this would lead to incrementing the count of '0' by two, the count of '1' by one, the count of '2' by one, the count of '20' by 1, the count of '200' by 1, and the count of '2001' by 1. The substrings '00','01', and '001' are ignored and do not increment the count of '0' or '1'.
Unlike A329447, which is fairly constant in its growth, this sequence can have large decreases in its values from one entry to the next. These tend to be bunched and are followed by long series of entries with steady growth. For n up to 200000 the largest entry is 434397, the sequence decreases 4690 times, and the largest drop from one value to the next is 419882 which occurs at n = 199902.
This sequence is the same as A329447 up to a(55) = 100. After that, as a(2) = 11 and '11' has only appeared once, the next smallest value is that of 'one 11', that is 111.

			a(56) = 111 as a(2) = 11, and as 11 has only appeared once, the next smallest value is that of 'one 11', that is 111. After this entry the string '11' has now appeared three times, and the string '111' has appeared once.
a(199903) = 14342 as a(199902) = 434224 and as that contains the first appearance of the digit string '4342' in any entry, the next smallest value is that of 'one 4342', that is 14342. This leads to the largest drop in value for the first 200000 terms of the sequence. As a comparison after 199902 terms the digit '1' has occurred 153333 times.

Scott R. Shannon, Table of n, a(n) for n = 0..20000
Eric Angelini, Look left and say, Nov 14 2019.
Scott R. Shannon, Scatter plot for the first 200000 terms.

Cf. A005150, A045918, A329447.

A330005 a(n) counts the square-words immediately before a(n), with a(1) = 0.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 1, 2, 0, 0, 1, 0, 0, 1, 1, 2, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 1, 2, 0, 0, 1, 0, 0, 1, 1, 3, 0, 0, 1, 0, 0, 1, 1, 2, 0, 0, 1, 0, 0, 1, 1, 2, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 1, 2, 0, 0, 1, 0, 0, 1, 1, 3, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 1, 2, 0, 0, 1, 0, 0, 1, 1, 3, 1, 1, 1, 0, 0, 1, 0

Offset: 1

Eric Angelini and Jean-Marc Falcoz, Nov 26 2019

Conjecture: Every integer will appear in the sequence. (Although very slowly: after 10000 terms, the biggest term is 5.)
Only the square-words are counted, not the cubes or the higher powers. The segment 1,1,1 will thus be followed by 0, and the same 0 will follow the segment 0,1,0,1,0,1,0,1 (because the pattern 0,1 is present four times instead of the required two, and because we don't accept splitting 0,1,0,1,0,1,0,1 in order to form two 0,1,0,1 substrings).
Overlaps are admitted: a typical case is the start with a(1) = 1. This produces 1,0,0,1,0,0,2,... with a(7) = 2 because there are indeed two squares immediately to the left of a(7): the square {0,0} and the square {100,100}.

			S = 0, ... We start with a(1) = 0 as this 0 means: "I don't see any square on my immediate left";
S = 0,0, ... We extend S with a(2) = 0 as this 0 says the same as above: "I don't see any square on my immediate left". Note that no other integer would fit without contradiction [a(2) = 1, for instance, would say that there is one square on the immediate left of "1", which is not the case - a lonely symbol cannot form a square].
S = 0,0,1, ... As we now see the square {00}, S is extended with a(3) = 1
S = 0,0,1,0, ... No square in sight, we thus extend S with a(4) = 0 [the word "immediate" is important: "There is no square on the immediate left of a(4) = 0"];
S = 0,0,1,0,0, ... No visible square again, so a(5) = 0;
S = 0,0,1,0,0,1, ... We see here that a(6) must be 1, as this "1" is immediately after the square {00}
S = 0,0,1,0,0,1,1, ... Yes: a(7) = 1 because a(7) sees 1 square on its immediate left, which is formed by {001,001};
S = 0,0,1,0,0,1,1,2, ... We see a third symbol entering S, the integer "2"; the reason is that 2 sees two squares on its immediate left, the first one being {11} and the second one {00} (again, the remote square {00} that opens S is not "immediately" to the left of 2, so we don't count it);
etc.

Jean-Marc Falcoz, Table of n, a(n) for n = 1..10001

Cf. A090822, A091970, A329447. The sequence A330023 does the same with cubes.

A330005_upto(N,A=0)={for(n=2,#A=Vec(A,N), A[n]=sum(L=1,(n-1)\2, if(A[n-L..n-1]==A[n-L*2..n-L-1] &&(L*3>=n || A[n-L..n-1]!=A[n-L*3..n-L*2-1]), A[n-L*2+1]+1)));A} \\ Optional arg A specifies starting value. - M. F. Hasler, Nov 28 2019

A330023 a(n) counts the cube-words immediately before a(n), with a(1) = 0.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1

Offset: 1

Eric Angelini and Jean-Marc Falcoz, Nov 28 2019

Only the cube-words are counted, not the higher powers. The segment 1,1,1,1 will thus be followed by 0, and the same 0 will follow the segment 0,1,0,1,0,1,0,1,0,1,0,1 (because the pattern 0,1 is present six times instead of the required three, and because we don't accept to split 0,1,0,1,0,1,0,1,0,1,0,1 in order to form three 0,1,0,1 substrings).
Overlaps are admitted: the term "2" in the segment 1,0,0,0,1,0,0,0,1,0,0,0,2 counts the cube 000 and the cube 100010001000.
Needs a large b-file! - N. J. A. Sloane, Dec 18 2019

			S = 0, ... We start with a(1) = 0 as this 0 means: "I don't see any cube on my immediate left";
S = 0,0, ... We extend S with a(2) = 0 as this 0 says the same as above: "I don't see any cube on my immediate left".
S = 0,0,0, ... Same as above - note that no other integer would fit without contradiction [a(3) = 1, for instance, would say that there is one cube on the immediate left of "1", which would not the case because 00 is a square, not a cube].
S = 0,0,0,1, ... As we now see the cube 000, S is extended with a(4) = 1.
S = 0,0,0,1,0, ... No cube in sight, we thus extend S with a(5) = 0 [the word "immediate" is important: "There is no cube on the immediate left of a(5) = 0" - indeed the first cube 000 is separated from a(5) by a(4) = 1];
S = 0,0,0,1,0,0, ... No visible cube again, so a(6) = 0;
S = 0,0,0,1,0,0,0, ... No visible cube again, so a(7) = 0;
S = 0,0,0,1,0,0,0,1 ... a(8) = 1 as this "1" is immediately after the cube {000}; etc.

Cf. A090822, A329447. The sequence A330005 does the same with squares.

cp's OEIS Frontend

A377905 Irregular triangle read by rows: T(n,k) (n >= 0, k >= 0, T[n,k]>0) is the total number of occurrences of the digit k in terms 0 to n of A329447.

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A376779 The terms of A329447 sorted into increasing order.

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A329448 Numbers n which appear twice in the sequence starting with n and subsequent terms given by 10c + d where c is the number of times (> 0) the least frequent digit d has appeared so far, smallest d in case of tie.

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A329818 Look left and tell the least frequent digit string; a(0) = 0, a(n) = concat(c,s), where c > 0 is the number of times the string s, composed of one or more digits, has appeared in a(0) to a(n-1) such that a(n) is minimal. Digit strings with leading zeros are ignored.

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A330005 a(n) counts the square-words immediately before a(n), with a(1) = 0.

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PARI

A330023 a(n) counts the cube-words immediately before a(n), with a(1) = 0.

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