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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Jonathan F. Waldmann

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A349482 Bases where the n-th Goodstein sequence starting in base 3 (instead of base 2) reaches 0.

Original entry on oeis.org

4, 5, 7, 9, 11, 15, 19, 23, 63, 159, 2047, 10239, 49151, 1048575, 20971519, 402653183, 1180591620717411303423, 233840261972944466912589573234605283144949206876159
Offset: 1

Author

Jonathan F. Waldmann, Nov 19 2021

Keywords

Comments

a(A056004(n)) lists the bases where the n-th Goodstein sequence starting in base 2 reaches 0. That sequence goes 3, 5, 7, 3*2^402653211 - 1, ...
The Goodstein function is sometimes given as the base where the sequence last has a nonzero value. Following this definition decreases each term in the above sequence by 1.
Like the Goodstein function (which starts in base 2), this sequence appears to grow faster than f_alpha if and only if alpha is smaller than epsilon_0.
As given by the formula below, the sequence continues with a(20,...,26) = 3*2^391 - 1, 4*2^2057 - 1, 5*2^10251 - 1, 3*2^49166 - 1, 4*2^1048594 - 1, 5*20971540 - 1, 3*402653211 - 1.

Examples

			For n=6, we write 6 = 2*3.
Then in each step, we increase the base by 1, then subtract 1. This sequence goes
2*3 -> 1*4 + 3 -> 1*5 + 2 -> 1*6 + 1 -> 1*7 + 0 -> 0*8 + 7 ->...-> 0*14 + 1 -> 0*15, so 0 is reached at base 15.
		

Crossrefs

Cf. A056004.

Formula

For n<27, i.e. n = a*3^2 + b*3 + c with a, b, c < 3, a(n) = f_2^a (f_1^b (f_0^c (4) ) ) - 1.

A344845 a(n)/2 is the smallest possible area of a non-obtuse triangle with integer coordinates in the plane and with shortest side of length sqrt(A001481(n+1)).

Original entry on oeis.org

1, 2, 4, 5, 8, 9, 10, 12, 16, 15, 18, 20, 23, 24, 28, 32, 30, 36, 34, 38, 39, 42, 49, 45, 48, 52, 55, 58, 56
Offset: 1

Author

Jonathan F. Waldmann, May 29 2021

Keywords

Comments

Since sqrt(m) is a distance between two points in Z^2 iff m is the sum of two squares, the "shortest side length" requirement can only be met for m in A001481\{0}. Thus this sequence cannot be extended to more arguments.
It has been shown that A344710(n) can be described and computed as min{a(k) | n <= A001481(k+1) < 5n/4}.
For the following comments, let m := A001481(n+1).
The validity of a triangle can be checked via four Diophantine inequalities: three Euclidean norms for the distances, and one derived from the law of cosines 2*max{|AB|,|BC|,|CA|}<=|AB|+|BC|+|CA| for the angles.
It was proved that a valid triangle with a sidelength of sqrt(2m) or larger has at least area m/2, and that a valid triangle with area m/2 always exists.
All valid triangles with an area smaller than m/2 can be found by checking for triangles with no sidelength of sqrt(2m) or longer, and at most one sidelength of sqrt(5m/4) or longer.
These criteria can be used to set A to (0,0), and look for B in the set of points X with |AX| = sqrt(m), and C in the set of points X with sqrt(m) <= |AX| < sqrt(5m/4). Furthermore, B can be assumed to be in the first octant, and C in a different octant but at most 2 octants away. It has been shown that this suffices to find congruent versions of all valid triangles with an area below m/2.
The sequence c(n) := min{a(k) | k >= n} is an upper bound on the sequence b(n) := ceiling(1/x(n)), where x(n) is the supremum on the density of marked points ("dots") in the discrete plane Z^2 with increasing pairwise minimum distance. Up to n=10, the two sequences have been shown to contain the same terms.
It has been shown that an alternative interpretation of the problem described in b(n) is the packing of circles with increasing diameter with centers in the discrete plane Z^2.
For [b(n)], it is conjectured that 1/x(n) is always an integer, making the ceiling function omittable for the definition.

Examples

			[a(n)]: For n = 3, i.e. A001481(n+1) = 4, a triangle with a shortest sidelength of sqrt(4) and the minimal area of 4/2 = 2 can be placed at A=(0,0), B=(2,0), and C=(0,2). Alternatively, C can be placed at (1,2) or (2,2). -> a(3) = 4.
[b(n)]: For the smallest 3 minimum distances sqrt(1), sqrt(2), and sqrt(4), the following repeating patterns (X for dots, O for empty spaces) achieve the highest possible densities of 1, 1/2, and 1/4 respectively:
   XXXXXX   OXOXOX   OXOXOX
   XXXXXX   XOXOXO   OOOOOO
   XXXXXX   OXOXOX   OXOXOX
   XXXXXX   XOXOXO   OOOOOO
		

Crossrefs

Formula

Let m := A001481(n+1).
a(n) = min{m, 2*min{area(ABC) | A, B, C in Z^2;
A = (0,0);
B = (a,b) with a >= b >= 0;
|AB| = sqrt(m);
C = (c,d) with c >= 0 or d >= c;
sqrt(m) <= |AC| < sqrt(5m/4);
sqrt(m) <= |BC| < sqrt(2m) } } (proved in "A more nuanced upright triangle sequence").
a(n) <= m.
a(n) > sqrt(3/4)*m (conjectured).

A344710 a(n)/2 is the smallest possible area of a non-obtuse triangle with coordinates in Z^2 and no side shorter than sqrt(n).

Original entry on oeis.org

1, 2, 4, 4, 5, 8, 8, 8, 9, 10, 12, 12, 12, 15, 15, 15, 15, 18, 20, 20, 23, 23, 23, 23, 23, 24, 28, 28, 28, 30, 30, 30, 30, 30, 34, 34, 34, 38, 38, 38, 39, 42, 42, 42, 42, 45, 45, 45, 45
Offset: 1

Author

Jonathan F. Waldmann, May 26 2021

Keywords

Comments

The validity of a triangle can be checked via four Diophantine inequalities: three Euclidean norms for the distances, and one derived from the law of cosines 2*max{|AB|,|BC|,|CA|}<=|AB|+|BC|+|CA| for the angles.
If n is not in A001481, a(n) = a(n+1) because a distance of sqrt(n) can never occur in Z^2.
The smallest area of a non-obtuse triangle with an exact rather than a minimum shortest sidelength is documented in A344845. For A344845, only the terms corresponding to A001481/{0} can exist, as only those are norms in Z^2.
For n in A001481, it was proved that a valid triangle with a sidelength of sqrt(2n) or larger has at least area n/2, and that for n in A001481 a valid triangle with area n/2 always exists.
For n in A001481, all valid triangles with an area smaller than n/2 can be found by checking for triangles with no sidelength of sqrt(2n) or longer, and at most one sidelength of sqrt(5n/4) or longer.
These criteria can be used to set A to (0,0), and look for B and C in the set of points X with sqrt(n) <= |AX| < sqrt(5n/4). Furthermore, B can be assumed to be in the first octant, and C in a different octant but at most 2 octants away. Lastly, |AB| <= |AC| can be assumed. It has been shown that for n in A001481, this suffices to find congruent versions of all valid triangles with an area below n/2.
Up to at least n=17, the following sequence [b(n)] has been proved to have the same terms: b(n) = ceiling(1/x(n)), where x(n) is the supremum on the density of marked points ("dots") in the discrete plane Z^2 with pairwise minimum distance sqrt(n).
If it is not identical for larger n, [a(n)] has been shown to at least be an upper bound on [b(n)].
It has been shown that an alternative interpretation of the problem described in b(n) is the packing of circles with diameter sqrt(n) with centers in the discrete plane Z^2.
For [b(n)], it is conjectured that 1/x(n) is always an integer, so the ceiling function can be omitted.
Conjectured next terms (from the conjectured inequality a(n) > sqrt(3/4)*n) are a(50,...,64) ?= 45, 48, 48, 52, 55, 55, 55, 55, 55, 56, 56, 56, 56, 56, 56.

Examples

			[a(n)]: For n=4, a triangle with the minimal area of 4/2 = 2 can be placed at A=(0,0), B=(2,0), and C=(0,2). Alternatively, C can be placed at (1,2) or (2,2).
[b(n)]: For n=1, n=2, and n=3, the following repeating patterns (X for dots, O for empty spaces) achieve the highest possible densities of 1, 1/2, and 1/4 respectively:
   XXXXXX   OXOXOX   OXOXOX
   XXXXXX   XOXOXO   OOOOOO
   XXXXXX   OXOXOX   OXOXOX
   XXXXXX   XOXOXO   OOOOOO
		

Crossrefs

Formula

a(n) = min{A344845(k) | n <= A001481(k+1) < 5n/4}.
a(n+1) >= a(n).
a(n) <= n if n is in A001481.
a(n) > sqrt(3/4)*n (conjectured).