A356348 -id:A356348 - cp's OEIS frontend

A358851 a(n+1) is the number of occurrences of the largest digit of a(n) among all the digits of [a(0), a(1), ..., a(n)], with a(0)=0.

0, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 11, 13, 2, 2, 3, 3, 4, 2, 4, 3, 5, 2, 5, 3, 6, 2, 6, 3, 7, 2, 7, 3, 8, 2, 8, 3, 9, 2, 9, 3, 10, 15, 4, 4, 5, 5, 6, 4, 6, 5, 7, 4, 7, 5, 8, 4, 8, 5, 9, 4, 9, 5, 10, 17, 6, 6, 7, 7, 8, 6

Offset: 0

Bence Bernáth, Dec 08 2022

Up to a(19)=10, the terms are identical to A248034. The branches (distinct lines of terms indicating the largest digit of the preceding term) can be labeled by the counter digit (shown in the scatter plot). From 9 to 1 the branches gradually get fragmented. Below digit 5 it is harder to disentangle the branches in some places. A repeating pattern also appears (shown in the inset of the scatter plot).

Bence Bernáth, Table of n, a(n) for n = 0..10000
Eric Angelini, Digit-counters updating themselves
Bence Bernáth, Scatter plot for n=0..500000
Michael De Vlieger, Log log scatterplot of a(n) n = 1..10^6.
Michael De Vlieger, Log log scatterplot of a(n), n = 1..10^5, with a color function indicating largest digit of preceding term, where black = 0, red = 1, orange = 2, ..., magenta = 9.

Cf. A248034, A249009, A356348, A336514.

length_seq=150;
sequence(1)=0;
seq_for_digits=(num2str(sequence(1))-'0');
for i1=1:1:length_seq
     sequence(i1+1)=sum(seq_for_digits==max((num2str(sequence(i1))-'0'))');
     seq_for_digits=[seq_for_digits, num2str(sequence(i1+1))-'0'];
end

nn = 79; s[] = 0; a[0] = 0; Do[(Set[d, Max[#]]; Map[s[#1] += #2 & @@ # &, Tally[#] ]) &@ IntegerDigits[a[n - 1]]; a[n] = s[d], {n, nn}]; Array[a, nn + 1, 0] (* _Michael De Vlieger, Dec 28 2022 *)

sequence=[0]
length=150
seq_for_digits=list(map(int, list(str(sequence[0]))))
for ii in range(length):
     sequence.append(seq_for_digits.count(max(list(map(int,list(str(sequence[-1])))))))
     seq_for_digits.extend(list(map(int, list(str(sequence[-1])))))

A357930 a(0) = 0; for n > 0, let S = concatenation of a(0)..a(n-1); a(n) is the number of times the digit at a(n-1) digits back from the end of S appears in S.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 7, 7, 8, 8, 7, 7, 8, 8, 7, 13, 7, 14, 7, 13, 13, 15, 15, 13, 15, 15, 6, 17, 16, 19, 8, 10, 10, 22, 10, 23

Offset: 0

Scott R. Shannon, Oct 21 2022

			a(7) = 3 as a(6) = 3 and the string concatenation of a(0)..a(6) = "0112223", and the digit 3 digits back from the end of the string concatenation is 2, and 2 has appeared three times in the string.

Michael De Vlieger, Table of n, a(n) for n = 0..9999
Michael De Vlieger, Table of n, a(n) for n = 0..10^5
Michael De Vlieger, Scalar scatterplot of a(n), n = 0..2^16, with points assigned a color function as to digit d that is a(n-1) digits back. Black indicates d=0, red d=1, ..., magenta d=9.
Michael De Vlieger, Log-log scatterplot of a(n), n = 0..2^20, with points assigned a color function as to digit d that is a(n-1) digits back. Black indicates d=0, red d=1, ..., magenta d=9.
Scott R. Shannon, Image of the first 5000000 terms.

Cf. A117707, A356348, A000217, A351753.

function a = A357930( max_n )
    a = 0; s = '0'; c = zeros(1,10); c(1) = 1;
    for n = 2:max_n
        k = c(s(end-a(n-1))-47); sk = num2str(k);
        c(sk-47) = c(sk-47)+1; s = [s sk]; a(n) = k;
    end
end % Thomas Scheuerle, Oct 21 2022

A358967 a(n+1) gives the number of occurrences of the smallest digit of a(n) so far, up to and including a(n), with a(0)=0.

Original entry on oeis.org

0, 1, 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, 8, 8, 9, 8, 10, 9, 9, 10, 10, 11, 22, 12, 23, 14

Offset: 0

Bence Bernáth, Dec 08 2022

Up to a(103)=12, the terms are identical to A248034.

Bence Bernáth, Table of n, a(n) for n = 0..10000
Eric Angelini, Digit-counters updating themselves

Cf. A248034, A249009, A356348, A336514, A358851.

length_seq=150;
sequence(1)=0;
seq_for_digits=(num2str(sequence(1))-'0');
for i1=1:1:length_seq
     sequence(i1+1)=sum(seq_for_digits==min((num2str(sequence(i1))-'0'))');
     seq_for_digits=[seq_for_digits, num2str(sequence(i1+1))-'0'];
end

sequence=[0]
length=150
seq_for_digits=list(map(int, list(str(sequence[0]))))
for ii in range(length):
   sequence.append(seq_for_digits.count(min(list(map(int,list(str(sequence[-1])))))))
   seq_for_digits.extend(list(map(int, list(str(sequence[-1])))))

A359031 a(n+1) gives the number of occurrences of the mode of the digits of a(n) among all the digits of [a(0), a(1), ..., a(n)], with a(0)=0.

Original entry on oeis.org

0, 1, 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

Bence Bernáth, Dec 12 2022

The mode is the most frequently occurring value among the digits of a(n). When there are multiple values occurring equally frequently, the mode is the smallest of those values.

Up to a(464)=110, the terms are identical to A358967.

Bence Bernáth, Table of n, a(n) for n = 0..20000
Eric Angelini, Digit-counters updating themselves

Cf. A248034, A249009, A356348, A336514, A358967, A358851, A322182.

length_seq=470;
sequence(1)=0;
seq_for_digits=(num2str(sequence(1))-'0');
for i1=1:1:length_seq
     sequence(i1+1)=sum(seq_for_digits==mode((num2str(sequence(i1))-'0'))');
     seq_for_digits=[seq_for_digits, num2str(sequence(i1+1))-'0'];
end

import statistics as stat
sequence=[0]
length=470
seq_for_digits=list(map(int, list(str(sequence[0]))))
for ii in range(length):
    sequence.append(seq_for_digits.count(stat.mode(list(map(int, list(str(sequence[-1])))))))
    seq_for_digits.extend(list(map(int, list(str(sequence[-1])))))

A360257 a(1) = 1; for n > 1, a(n) is the number of preceding terms having the same sum of divisors as a(n-1).

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 2, 2, 3, 2, 4, 2, 5, 2, 6, 3, 3, 4, 3, 5, 3, 6, 4, 4, 5, 4, 6, 5, 5, 6, 6, 7, 2, 7, 3, 7, 4, 7, 5, 7, 6, 8, 2, 8, 3, 8, 4, 8, 5, 8, 6, 9, 2, 9, 3, 9, 4, 9, 5, 9, 6, 10, 2, 10, 3, 10, 4, 10, 5, 10, 6, 11, 12, 1, 12, 2, 11, 13, 1

Offset: 1

Scott R. Shannon, Jan 31 2023

nonn

The most common sum of divisor count of all previous terms changes as n increases; these values, up to 2 million terms, are 1, 12, 24, 12, 24, 72, 720, 72, 720, 72, 720, 72, 720. The value 72 holds the record from a(6998) = 71 to a(1271035) = 563. After a(1285242) = 264 the divisor sum 720 becomes the most common sum to well beyond 10 million terms. It is likely the record becomes arbitrarily large as n increases.

			a(22) = 2 as a(21) = 11 and 11 has a divisor sum of A000203(11) = 12. However, A000203(6) also equals 12, and as a(11) = 6 there are two previous terms with a divisor sum of 12.

Scott R. Shannon, Image of the first 2000000 terms.

Cf. A000203, A276457, A356348.

lista(nn) = my(va = vector(nn), vs=vector(nn)); va[1] = 1; vs[1] = 1; for (n=2, nn, va[n] = #select(x->(x==vs[n-1]), vs); vs[n] = sigma(va[n]);); va; \\ Michel Marcus, Jan 31 2023

A378852 a(1) = 1. For n > 1 a(n) is the number of terms a(i); 1 <= i <= n-1 such that d(a(i)) >= d(a(n-1)), where d is the decimal digital sum function A007953.

Original entry on oeis.org

1, 1, 2, 1, 4, 1, 6, 1, 8, 1, 10, 11, 5, 3, 5, 4, 6, 3, 9, 1, 20, 13, 9, 2, 16, 4, 12, 15, 7, 5, 11, 23, 12, 20, 26, 4, 18, 3, 24, 11, 32, 16, 8, 6, 14, 20, 38, 1, 48, 1, 50, 23, 24, 17, 9, 6, 20, 47, 3, 40, 35, 12, 43, 16, 17, 13, 40, 41, 34, 19, 4, 45, 9, 10, 74, 4, 49, 1, 78, 1, 80

Offset: 1

David James Sycamore, Dec 09 2024

d(a(n-1)) >= d(a(i)); 1 <= i <= n-1 implies a(n) = 1. a(n) <= n-1 for all n > 1, with equality iff d(a(n-1)) = 1.

Compare with A356348 and A378782.

			a(1) = 1 so a(2) also = 1 since there is only one term up to and including a(1) = 1 which has digit sum >= 1. Then a(3) = 2 because now there are two terms having digit sum >= 1. a(11) = 10 so a(12) = 11 since all terms up to and including a(11) have digit sum >= 1. a(19) = 9, whose digit sum (9) sets a record, thus a(20) = 1, which means a(21) = 20.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A007953, A378293, A378782, A356348.

R:= 1: dR:= 1:
for n from 2 to 100 do
  v:= nops(select(i -> dR[i] >= dR[n-1], [$1..n-1]));
  R:= R,v; dR:= dR, convert(convert(v,base,10),`+`);
od:
R; # Robert Israel, Feb 09 2025

first(n) = {
	my(res = vector(n), digs = vector(n));
	res[1] = 1; digs[1] = 1;
	for(i = 2, n,
		s = 1 + sum(j = 1, i-2, digs[j] >= digs[i-1]);
		res[i] = s;
		digs[i] = sumdigits(s)
	);
	res
} \\ David A. Corneth, Dec 24 2024

More terms from David A. Corneth, Dec 24 2024

cp's OEIS Frontend

A358851 a(n+1) is the number of occurrences of the largest digit of a(n) among all the digits of [a(0), a(1), ..., a(n)], with a(0)=0.

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A357930 a(0) = 0; for n > 0, let S = concatenation of a(0)..a(n-1); a(n) is the number of times the digit at a(n-1) digits back from the end of S appears in S.

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A358967 a(n+1) gives the number of occurrences of the smallest digit of a(n) so far, up to and including a(n), with a(0)=0.

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A359031 a(n+1) gives the number of occurrences of the mode of the digits of a(n) among all the digits of [a(0), a(1), ..., a(n)], with a(0)=0.

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A360257 a(1) = 1; for n > 1, a(n) is the number of preceding terms having the same sum of divisors as a(n-1).

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A378852 a(1) = 1. For n > 1 a(n) is the number of terms a(i); 1 <= i <= n-1 such that d(a(i)) >= d(a(n-1)), where d is the decimal digital sum function A007953.

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