Kester Habermann - cp's OEIS frontend

A377757 Number of possibilities to place "hat" monotiles onto the first n hexagons in a counterclockwise order such that more rings of tiles can be placed around it.

1, 3, 7, 9, 10, 15, 18, 22, 23, 26, 26, 30, 32, 35, 37, 43, 44, 44, 44, 51, 51, 56, 60, 60, 65, 69, 74, 80, 86, 86, 86, 94, 94, 94, 100, 105, 109

Offset: 1

Ruediger Jehn and Kester Habermann, Nov 06 2024

Fig. 1.1 in the paper "An aperiodic monotile" by Smith et al. (2023) shows an example of tiling the plane with "hat" monotiles. A "hat" monotile covers two thirds of a hexagon and one third of two neighboring hexagons, respectively. A hexagon is assigned a tile if the tile covers two thirds of the hexagon. Since the surface of a "hat" tile is 4/3 the surface of a hexagon, every forth hexagon on average is covered by parts of three different tiles. In this case we assign a zero to this hexagon.
We assign the number "1" to a tile that has the orientation of the dark grey tile in Fig. 1.1 of the paper by Smith. Tiles can only be rotated by multiples of 60 deg and for each counterclockwise rotation of the tile by 60 deg, we increase the count by 1 such that unreflected tiles are assigned the numbers from 1 to 6. Reflected tiles are assigned accordingly the numbers from 7 to 12.
Without loss of generality we place a tile "1" in the central hexagon like in the quoted Fig 1.1 (any other tiling can be transformed into this tiling by rotation or reflection). As next hexagon to be filled we choose the right upper neighbor hexagon. The tiling continues through the surrounding hexagons in counterclockwise direction.
A detailed description of the counting of the tiling options is given in the pdf in the links. a(n) is the number of possible assignments of the first n hexagons such that the plane can be filled at least up to heaxagon 919 (end of ring 17). It would be desirable to demand that the tiling can be continued to infinity, but this is much more difficult to prove.
An open question is: What is the asymptotic behavior of a(n) when n tends to infinity?

			a(4)=9:  1-0-1-0, 1-0-1-3, 1-0-1-6, 1-0-8-6, 1-2-2-6, 1-2-3-10, 1-2-7-6, 1-2-12-5 and 1-10-12-0 are the nine tiling options for the first four hexagons such that further tiling is possible.
a(7)=18: there are 18 different ways to tile a ring a round the central hexagon.

Ruediger Jehn, and Kester Habermann, Number of possibilities to tile the plane with hat monotiles
David Smith, Joseph Samuel Myers, Craig S. Kaplan, and Chaim Goodman-Strauss, An aperiodic monotile, arXiv:2303.10798 [math.CO], 2023.

Cf. A342285, A363348, A363445.

Corrected and extended by Ruediger Jehn, Jun 05 2025

A349698 Denominators of the probability that the first player wins the game Super Six if both players have n sticks in their hand and if there are 3 sticks on the lid, assuming optimal play.

Original entry on oeis.org

127838, 364984531847631619, 3212797979972917332633146175485560069226398681488, 21570506042045917755280171226734858792217536499150631950302282702757299436929665640958967552

Offset: 1

Ruediger Jehn, Kester Habermann and Pontus von Brömssen, Nov 25 2021

For the rules of Super Six see A349697.

			a(1) = 127838 because the probability that the first player wins the game Super Six, when both players have 1 stick and there are 3 sticks on the lid, is 78307/127838 (0.612548...).

Ruediger Jehn, Table of n, a(n) for n = 1..11
Rüdiger Jehn, Optimum Strategies for the Game Super Six, arXiv:2109.10700 [math.GM], 2021.
Wikipedia, Super Six (in German)

Cf. A345383, A349697.

A349697 Numerators of the probability that the first player wins the game Super Six if both players have n sticks in their hand and if there are 3 sticks on the lid, assuming optimal play.

Original entry on oeis.org

78307, 186990749618019112, 1614205257536455860879998130775735700828260230275, 10794889897425456513785608689552167481027910004023676512263195628077085109766959836330846217

Offset: 1

Ruediger Jehn, Kester Habermann and Pontus von Brömssen, Nov 25 2021

The rules of Super Six for two players are as follows. The equipment consists of a six-sided die, a number of sticks, and a box whose lid has six holes. The holes numbered 1 through 5 are shallow, and a stick placed in any one of them will stand up in it; hole #6 goes all the way through the lid so that any stick placed in it falls into the box and is out of play. Initially, an even number of sticks are divided evenly between the two players. The goal is to get rid of all one's sticks before the other player does.
The players take turns. On each turn, the active player rolls the die and places a stick in the numbered hole that matches the number on the die (e.g., a player who rolls a 4 then places a stick in hole #4). The player may roll and place a stick for each roll as many times as desired until rolling a number that is already filled by a stick. When this occurs, the player must take that stick in hand, and play passes to the opponent.
The game proceeds with players taking turns and ends when one player has run out of sticks. The only freedom that the players have is the decision whether to continue rolling the die or not after successfully placing a stick.
The optimum strategy and the winning probabilities can be found in "Optimum Strategies for the Game Super Six" (see link below). The terms of this sequence give the numerators and the terms of sequence A349698 give the denominators of the probability that the first player wins if there are 3 sticks on the lid and both players hold n sticks in their hands, assuming optimal play. If n tends to infinity this probability tends to 1/2.

			a(1) = 78307 because the probability that the first player wins the game Super Six, when both players have 1 stick and there are 3 sticks on the lid, is 78307/127838 (0.612548...).

Ruediger Jehn, Table of n, a(n) for n = 1..11
Rüdiger Jehn, Optimum Strategies for the Game Super Six, arXiv:2109.10700 [math.GM], 2021.
Wikipedia, Super Six (in German)

Cf. A345383, A349698.

A342325 a(n) is the position in digits of Pi where k(n) is for the first time leader in the race of digits of Pi. k(n) = 0, 1, ..., 9, 00, 01, ..., 99, 000, 001, ...

Original entry on oeis.org

174999, 4, 187, 1, 274, 11, 990213634, 320741, 108, 59, 16972551346, 98100646316

Offset: 0

Ruediger Jehn and Kester Habermann, Mar 08 2021

48 out of all 100 two-digit combinations are never in the lead when the first 100 billion digits of Pi are analyzed. 02 is the first of these 48, hence a(12) is unknown so far. The results for the 52 two-digit numbers which took the lead at some time are given in the link below.

			a(1) = 4 because the 1 becomes the leader in the race of digits of Pi for the first time after 4 digits: 3141.
a(41) = 2 because k(41) = 31 and 31 is the leader after two digits of Pi.

Ruediger Jehn, Leaderboard considering the first 100 billion digits

Cf. A000796, A195835.

A333666 Smallest k > 0 with gcd(k, rev(k)) = n, where rev(k) is digit reversal of k and with sum of digits of k = n, or 0 if no such k exists.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 209, 48, 4000009, 21182, 5055, 21184, 13328, 288, 12844, 0, 1596, 2398, 13892, 2976, 52675, 45890, 2889, 61768, 178292, 0, 177475, 29984, 42999, 279718, 529865, 29988, 1009009009009, 485678, 1951599, 0, 694499, 655998, 1677688, 658988

Offset: 1

Ruediger Jehn and Kester Habermann, Sep 03 2020

This differs from A069554 in that, additionally, the sum of the digits of a(n) must be equal to n. This is not required in A069554.
As gcd(k, rev(k)) = n, n | k and n | rev(k). - David A. Corneth, Sep 03 2020
Since the sum of the digits of k is n and n | k, all the terms that are not 0 are Niven numbers (A005349). - Amiram Eldar, Sep 03 2020
The first digit of any number of this sequence is less than or equal to the last digit of this number (provided that the last digit is nonzero), because if k satisfies all requirements, also rev(k) does. This means that numbers starting with a "9" are quite rare. So far we have found only 9. But numbers ending with a "1" seem to be even less frequent. Amongst the first 303 terms of this sequence there is none except the trivial solution a(1) = 1. The second term of this sequence ending with a "1", if it exists, is still to be found. - Ruediger Jehn, Sep 20 2020 [Corrected by Pontus von Brömssen, Oct 07 2021]

			a(11) = 209. The sum of the digits is 11 and gcd(209,902) = 11.
a(12) = 48. The sum of the digits is 12 and gcd(48,84) = 12.

Ruediger Jehn, Table of n, a(n) for n = 1..303
Ruediger Jehn, Proofs for difficult terms
Rüdiger Jehn, Porous Numbers, arXiv:2104.02482 [math.GM], 2021.

Cf. A004086, A005349, A007953, A069554.

m = 36; s = Table[0, {m}]; c = 0; n = 1; While[c < m - Quotient[m, 10], g = GCD[n, FromDigits @ Reverse @ (d = IntegerDigits[n])]; If[g <= m && g == Plus @@ d && s[[g]] == 0, c++; s[[g]] = n]; n++]; s (* Amiram Eldar, Sep 03 2020 *)

a(n) = {if ((n % 10) == 0, return(0)); my(k=n); while (! ((sumdigits(k)==n) && (gcd(k, fromdigits(Vecrev(digits(k)))) == n)), k+=n); k;} \\ Michel Marcus, Sep 03 2020

for n in range(11, 20):
    for k in range(n, 1000000000, n):
       s = str(k)
       revk = "" # digit reversal of k
       sum = 0
       for i in range(len(s)):
          revk = revk + s[len(s) - i - 1]
          sum = sum + int(s[i])
       g = gcd(k,int(revk))
       if g == n and sum == n:
          print(n, k, revk, g)
          break

a(10*n) = 0 since all multiples of 10 have a 0 at the end, but their reverse numbers have no 0 at the end and therefore 10*n cannot be their gcd.

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User: Kester Habermann

A377757 Number of possibilities to place "hat" monotiles onto the first n hexagons in a counterclockwise order such that more rings of tiles can be placed around it.

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A349698 Denominators of the probability that the first player wins the game Super Six if both players have n sticks in their hand and if there are 3 sticks on the lid, assuming optimal play.

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A349697 Numerators of the probability that the first player wins the game Super Six if both players have n sticks in their hand and if there are 3 sticks on the lid, assuming optimal play.

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A342325 a(n) is the position in digits of Pi where k(n) is for the first time leader in the race of digits of Pi. k(n) = 0, 1, ..., 9, 00, 01, ..., 99, 000, 001, ...

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A333666 Smallest k > 0 with gcd(k, rev(k)) = n, where rev(k) is digit reversal of k and with sum of digits of k = n, or 0 if no such k exists.

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