Leo Crabbe - cp's OEIS frontend

User: Leo Crabbe

Leo Crabbe has authored 3 sequences.

A381135 Numbers of the form d_1 d_2 d_3 ... where the sum of their digits multiplied by their digit positions is equal to their number of digits.

1, 20, 110, 300, 1010, 2100, 4000, 10010, 12000, 20100, 31000, 50000, 100010, 111000, 200100, 220000, 301000, 410000, 600000, 1000010, 1020000, 1101000, 1300000, 2000100, 2110000, 3001000, 3200000, 4010000, 5100000, 7000000, 10000010, 10110000, 11001000, 12100000

Offset: 1

Leo Crabbe, Feb 14 2025

The digits for these numbers are 1-indexed.

Numbers k such that A156207(k) = A055642(k).

			301000 is a member of this sequence because it has 3*1 + 1*3 = 6 digits.

Michael S. Branicky, Table of n, a(n) for n = 1..10000
Michael S. Branicky, Python program for OEIS A381135

Cf. A055642, A156207.

isok(k) = my(d=digits(k)); sum(i=1, #d, i*d[i]) == #d; \\ Michel Marcus, Feb 17 2025

def ok(n): return sum(int(d)*(i+1) for i, d in enumerate(str(n))) == len(str(n))

# see linked program for a different algorithm
from itertools import count, islice
from sympy.utilities.iterables import partitions
def agen(): # generator of terms
    for d in count(1): yield from sorted(int("".join(str(p[k]) if k in p else "0" for k in range(1, d+1))) for p in partitions(d, m=d, k=d) if max(p.values()) < 10 if 1 in p and p[1] != 0)
print(list(islice(agen(), 34))) # Michael S. Branicky, Feb 19 2025

More terms from Michel Marcus, Feb 17 2025

A380985 Numbers whose k-th digit indicates the number of digits which occur k times.

Original entry on oeis.org

1, 20, 110, 2100, 20100, 200100, 2000100, 20000100, 200000100, 2000000100, 20000000100, 200000000100, 2000000000100, 20000000000100, 200000000000100, 2000000000000100, 20000000000000100, 200000000000000100, 2000000000000000100, 20000000000000000100

Offset: 1

Leo Crabbe, Feb 11 2025

The most significant digit is digit position k=1, meaning that it counts how many numbers appear 1 time.

			110 is a term because 1 number appears once (0), 1 number appears twice (1) and 0 numbers appear 3 times.

Cf. A046043.

from collections import Counter
def ok(n):
    d = list(map(int, str(n)))
    c = Counter(Counter(d).values())
    return all(dk == c[k] for k, dk in enumerate(d, 1)) # Michael S. Branicky, Feb 18 2025

a(n) = 2*10^(n-1) + 100 for n >= 4. - Andrew Howroyd, Feb 22 2025

A349595 Number of self-counting sequences of length n (sequences indexed from 0 to (n-1) where a(i) is the number of times i appears in the sequence).

Original entry on oeis.org

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

Offset: 1

Leo Crabbe, Nov 22 2021

There is exactly one self-counting sequence for any n > 6 and it will always be of the form [n-4, 2, 1, ..., 1, ...], where the second "1" is at the (n-4)th index, and the "..."s are filled with the appropriate number of zeros. For example, the only self-counting sequence of length 7 is [3, 2, 1, 1, 0, 0, 0], and the only self-counting sequence of length 15 is [11, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0].
Proof:
Let a be a self-counting sequence of finite length. Firstly, observe that for all such sequences, the sum of a(i) - 1 over all i must equal 0. The contributions to this sum from those i for which a(i) = 0 will be some number of -1s. How many such -1s? a(0). The contribution from the i=0 case will be a(0)-1 (note that a(0) cannot be equal to 0, so these two contribution cases are disjoint). Thus, if we sum a(i) - 1 over just those nonzero i for which a(i) is nonzero, we will get a(0) - (a(0) - 1) = 1. These a(i) - 1 values are all whole numbers >= 0, so for them to sum to 1 is precisely for one of them to be 1 and the rest to be 0. And thus, we have precisely one nonzero i at which a(i) is 2, and at the other nonzero i, we have a(i) is either 1 or 0, and the only configuration of these ones and zeros that works is to have a(1) = 2, a(2) = 1, a(a(0)) = 1 and a(i) = 0 for all other nonzero i.

			n = 4 is the smallest length that a self-counting sequence can have (without considering the empty sequence of length 0, which could be self-counting or not, depending on the definition). There are two self-counting sequences of length 4, namely [1, 2, 1, 0] and [2, 0, 2, 0]. We can verify the first one by counting the times each number appears: 0 appears once, 1 appears twice, 2 appears once and 3 appears zero times.

Index entries for linear recurrences with constant coefficients, signature (1).

G.f.: x^4*(2 - x - x^2 + x^3)/(1 - x). - Stefano Spezia, Dec 08 2021

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User: Leo Crabbe

A381135 Numbers of the form d_1 d_2 d_3 ... where the sum of their digits multiplied by their digit positions is equal to their number of digits.

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A380985 Numbers whose k-th digit indicates the number of digits which occur k times.

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A349595 Number of self-counting sequences of length n (sequences indexed from 0 to (n-1) where a(i) is the number of times i appears in the sequence).

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