Julia Zimmerman - cp's OEIS frontend

User: Julia Zimmerman

Julia Zimmerman has authored 4 sequences.

A367733 Numbers k such that the sum of digits of k is equal to the sum of digits of k+9.

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 84, 85, 86, 87, 88, 89, 101

Offset: 1

Julia Zimmerman, Nov 28 2023

Numbers k which end in the digits ...xy with x!=9 and y!=0.
Differs from A052382, as there are terms with 0 here, the first being a(82)=101. First differs from A067251 at a(82)=101, A067251(82)=91. Similarly to A067251, A209931 includes 91-99 as terms whereas they are not in this sequence. A043095(1)=0 and A023804(1)=0 whereas 0 is not a term in this sequence (there are additional differences, such as the term that comes after 89 in A023804 and A043095 being 99).
This sequence is defined as follows: |digsum(k + seed) - digsum(k)| = r where digsum is the digital sum (A007953), with seed = 9 and r = 0.
The way this sequence looks has to do with using base 10: if you choose 8 as a seed and 1 as the sought difference (r), or 7 as a seed and 2 as the sought difference, you will get similar long, full sequences. However if you choose 8 as a seed and 0 as the sought difference, you'll get no terms.

			For k=3, 3 + 9 = 12. Sum of 1 + 2 = 3. Since the sum of the digits in 3 and the sum of the digits in 12 are the same, 3 is a term of the sequence.

Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1).

Cf. A007953.

Select[Range[100], Equal @@ Plus @@@ IntegerDigits[{#, # + 9}] &] (* Amiram Eldar, Nov 28 2023 *)

is(n) = sumdigits(n) == sumdigits(n+9) \\ David A. Corneth, Nov 28 2023

def A367733(n): return n + (n-1)//9 + ((n-1)//81)*10

a(n) = n + floor((n-1)/9) + floor((n-1)/81)*10.

A363054 Look and say sequence: describe the previous term (method A, starting with 20).

Original entry on oeis.org

20, 1210, 11121110, 31123110, 132112132110, 11131221121113122110, 311311222112311311222110, 1321132132211213211321322110, 11131221131211132221121113122113121113222110, 3113112221131112311332211231131122211311123113322110

Offset: 1

Julia Zimmerman, May 15 2023

			The term after 1210 is given by saying "I see one 1, one 2, one 1, and one 0", and then writing down the digits as 11-12-11-10, yielding 11121110.

Michael De Vlieger, Table of n, a(n) for n = 1..21
Julia Witte Zimmerman, Denis Hudon, Kathryn Cramer, Alejandro J. Ruiz, Calla Beauregard, Ashley Fehr, Mikaela Irene Fudolig, Bradford Demarest, Yoshi Meke Bird, Milo Z. Trujillo, Christopher M. Danforth, and Peter Sheridan Dodds, Tokens, the oft-overlooked appetizer: Large language models, the distributional hypothesis, and meaning, arXiv:2412.10924 [cs.CL], 2024. See pp. 21, 28.

Other look-and-say sequences: A005150, A001155, A006751, A006715, A001140, A001141, A001143, A001145, A001151, A007651, A001387, A098595, A045918.

NestList[FromDigits@ Flatten@ Map[Reverse@ Tally[#][[1]] &, Split@ IntegerDigits[#] ] &, 20, 12] (* Michael De Vlieger, Jul 05 2023 *)

from itertools import count, groupby, islice
def LS(n): return int(''.join(str(len(list(g)))+k for k, g in groupby(str(n))))
def agen(an=20): yield an; yield from (an:=LS(an) for n in count(1))
print(list(islice(agen(), 10))) # Michael S. Branicky, May 15 2023

A363377 Largest positive integer having n holes that can be made using the fewest possible digits.

Original entry on oeis.org

7, 9, 8, 98, 88, 988, 888, 9888, 8888, 98888, 88888, 988888, 888888, 9888888, 8888888, 98888888, 88888888, 988888888, 888888888, 9888888888, 8888888888, 98888888888, 88888888888, 988888888888, 888888888888, 9888888888888, 8888888888888, 98888888888888, 88888888888888, 988888888888888

Offset: 0

Julia Zimmerman, May 29 2023

Each decimal digit has 0, 1 or 2 holes so that n holes requires A065033(n) digits.

			For n=0, the largest integer with no holes in it that is as short as possible is 7 (9 is larger, but has 1 hole; 11 is larger and has no holes, but is longer at length 2 > length 1).
For n=1, the largest integer with 1 hole that is as short as possible is 9 (following the same kind of reasoning as with n=0).

Index entries for sequences related to holes in digits.
Index entries for linear recurrences with constant coefficients, signature (1,10,-10).

Cf. A002281 and A002282 (number of holes), A065033 (digits required).
Cf. A249572 and A250256 (smallest number).
Cf. A337099 (largest 7-segment).

CoefficientList[Series[(7 + 2 x - 71 x^2 + 70 x^3)/((1 - x) (1 - 10 x^2)), {x, 0, 30}], x] (* Michael De Vlieger, Jul 05 2023 *)

A363377=lambda n: (8+n%2*81)*10**(n>>1)//9 if n else 7
print([A363377(n) for n in range(30)]) # Natalia L. Skirrow, Jun 26 2023

From Natalia L. Skirrow, Jun 26 2023: (Start)
a(n) = (89*(10^((n-1)/2))-8)/9 for odd n; a(n) = 8*(10^(n/2)-1)/9 for even n >= 2.
a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3), for n >= 4.
G.f.: (7+2*x-71*x^2+70*x^3)/((1-x)*(1-10*x^2)).
E.g.f.: (80*cosh(sqrt(10)*x) + 89*sqrt(10)*sinh(sqrt(10)*x) - 80*e^x)/90 + 7. (End)

A338819 The entries in the rows of the n X n identity matrix, multiplied by the size of the matrix (n).

Original entry on oeis.org

1, 2, 0, 0, 2, 3, 0, 0, 0, 3, 0, 0, 0, 3, 4, 0, 0, 0, 0, 4, 0, 0, 0, 0, 4, 0, 0, 0, 0, 4, 5, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 5, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 7, 0

Offset: 1

Julia Zimmerman, Nov 10 2020

			For every identity matrix of size n starting with n=1, append n*(each entry of each row of the matrix), e.g., n=1 -> 1, n=2 -> 2,0,0,2, so the first 5 terms of the sequence are 1,2,0,0,2.

Cf. A191747 (with 1's), A191748 (locations of nonzero terms), A056520 (start location of each matrix).

import numpy as np
def n_id_sequence(iterations):
    sequence = []
    for i in range(1,iterations+1):
        matrix = i*(np.identity(i))
        for row in matrix:
            for entry in row:
                sequence.append(int(entry))
    return sequence

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User: Julia Zimmerman

A367733 Numbers k such that the sum of digits of k is equal to the sum of digits of k+9.

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A363054 Look and say sequence: describe the previous term (method A, starting with 20).

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A363377 Largest positive integer having n holes that can be made using the fewest possible digits.

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A338819 The entries in the rows of the n X n identity matrix, multiplied by the size of the matrix (n).

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Author

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Python