A248034 -id:A248034 - cp's OEIS frontend

A278288 a(n) is the number of ways to represent a(n-1) as a sum of two distinct terms from a(0) to a(n-2). a(0) = 0. a(1) = a(2) = 1.

0, 1, 1, 1, 2, 3, 3, 4, 6, 2, 4, 8, 3, 8, 4, 12, 6, 10, 10, 11, 12, 15, 9, 12, 19, 7, 15, 17, 10, 18, 22, 17, 12, 22, 21, 22, 25, 25, 26, 22, 26, 26, 27, 32, 25, 30, 27, 35, 21, 23, 31, 31, 32, 37, 37, 38, 37, 39, 37, 40, 40, 41, 45, 28, 37, 42, 38, 50, 33, 43, 58, 34

Offset: 0

Yuriy Sibirmovsky, Nov 16 2016

nonn

a(n) is the number of pairs (j, k) such that a(j) + a(k) = a(n-1), with 0 <= j, k < n-1 and j != k.
For large n, a(n) on average follows linear law a(n) ~ 0.7 n with linear spread (see the plot).
Unlike sequences such as A248034 or A276457, this sequence is base-independent.

			a(2) = a(1) + a(0), so a(3) = 1.
a(3) = a(1) + a(0) = a(2) + a(0), so a(4) = 2.
a(4) = a(3) + a(2) = a(3) + a(1) = a(2) + a(1), so a(5) = 3.

Yuriy Sibirmovsky, Table of n, a(n) for n = 0..2999
Yuriy Sibirmovsky, The plot of a(n) for n=0..2999 (with linear regression)

Cf. A248034, A276457.

Nm=100;
A=Table[1,{n,1,Nm}];
A[[1]]=0;
Do[Nc=0;
Do[If[A[[j]]+A[[k]]==A[[n]] && k!=j,Nc++],{j,1,n-1},{k,1,j}];
A[[n+1]]=Nc,{n,3,Nm-1}];
A

A309872 For each term, take the last digit of the previous term and count all the appearances of that digit up to and including the previous term; the first term is 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, 12, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 2, 10, 2, 11, 16, 3, 3, 4, 3, 5, 3, 6, 4, 4, 5, 4, 6, 5, 5, 6, 6, 7, 3, 7, 4, 7, 5, 7, 6, 8, 3, 8, 4, 8, 5, 8, 6, 9, 3, 9, 4, 9, 5, 9, 6, 10, 3, 10, 4, 10, 5, 10, 6, 11, 23, 11

Offset: 1

Maxim Skorohodov, Aug 21 2019

			To get the eleventh term, you need to get the last digit of the tenth term, which is 1, and then count all the 1's already in the sequence: 1, 1, 2, 1, 3, 1, 4, 1, 5, 1; there are six 1's, so the eleventh term is 6.

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

Cf. A248034 (first term is 0).

Q:= Array(0..9):
A:= Vector(100):
Q[1]:= 1:
A[1]:= 1:
for n from 2 to 100 do
  d:= A[n-1] mod 10;
  A[n]:= Q[d];
  L:= convert(%,base,10);
  for i in L do Q[i]:= Q[i]+1 od
od:
convert(A,list); # Robert Israel, Feb 18 2020

f = vector(base=10); for (n=1, 91, v = if (n==1, 1, f[1+(v%base)]); apply (d -> f[1+d]++, if (v, digits(v, base), [0])); print1 (v ", ")) \\ Rémy Sigrist, Aug 21 2019

s, a, n = "1", [1], 1
while n < 100:
    n = n+1
    d = s[len(s)-1]
    i, aa = 0, 0
    while i < len(s):
        if s[i] == d:
            aa = aa+1
        i = i+1
    s, a = s+str(aa), a+[aa]
for n in range(1, 92): print(a[n-1], end=', ') # A.H.M. Smeets, Aug 22 2019

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])))))

A384309 a(1) = 1. Thereafter a(n) is the cardinality of the set of terms whose leading decimal digit is the same as that of 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, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 2, 8, 2, 9, 2, 10, 21, 11, 22, 12, 23, 13, 24, 14, 25, 15, 26, 16, 27, 17, 28, 18, 29, 19, 30, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 3, 10, 31, 11

Offset: 1

David James Sycamore, May 25 2025

Conjecture: All positive integers appear precisely 9 times in this sequence, except for 1, which appears 10 times. For k >= 1, the last k digit term in the sequence is a(23[k-1]93) = [k]9 where "[m]9" means a run of m nines; see Example.

			a(1) = 1 is given so a(2) = 1, the number of terms having leading decimal digit = 1. Now there are two terms with leading digit = 1, so a(3) = 2. Since a(3) is the only term with leading digit = 2, a(4) = 1.
a(233) = 9 is the last one-digit term, a(2393) = 99 is the last two-digit term, a(23993) = 999 is the last three-digit term, etc.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Log log scatterplot of a(n), n = 1..10^6.

Cf. A000030, A248034.

nn = 120; c[] := 0; d[x] := First@ IntegerDigits[x]; j = 1; {j}~Join~Reap[Do[Sow[k = ++c[d[j]]]; j = k, {n, nn}] ][[-1, 1]] (* Michael De Vlieger, May 25 2025 *)

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    an, c = 1, Counter()
    while True:
        yield an
        leading = str(an)[0]
        c[leading] += 1
        an = c[leading]
print(list(islice(agen(), 80))) # Michael S. Branicky, May 25 2025

cp's OEIS Frontend

A278288 a(n) is the number of ways to represent a(n-1) as a sum of two distinct terms from a(0) to a(n-2). a(0) = 0. a(1) = a(2) = 1.

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A309872 For each term, take the last digit of the previous term and count all the appearances of that digit up to and including the previous term; the first term is 1.

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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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MATLAB

Python

A384309 a(1) = 1. Thereafter a(n) is the cardinality of the set of terms whose leading decimal digit is the same as that of a(n-1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python