OEIS/A322469 - Revision history

imported>Gfis: Gfis moved page OEIS/Div3Mult2 Sequence to OEIS/A322469 without leaving a redirect

2019-08-03T21:07:14Z

Gfis moved page OEIS/Div3Mult2 Sequence to OEIS/A322469 without leaving a redirect

← Older revision	Revision as of 21:07, 3 August 2019
(No difference)

imported>Gfis at 11:43, 10 December 2018

2018-12-10T11:43:33Z

← Older revision		Revision as of 11:43, 10 December 2018
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	Example:		Example:
	i \| j = 1 2 3 4 5 6 7		The table T(i, j) begins:
	~~----+-------------------------~~		i\j 1 2 3 4 5 6 7
	1 \| 3 1 2		1 \| 3 1 2
	2 \| 7		2 \| 7

imported>Gfis: coverage

2018-12-09T12:09:12Z

coverage

← Older revision		Revision as of 12:09, 9 December 2018
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	The residues m in each column and therefore the T(i, j) are all disjoint.		The residues m in each column and therefore the T(i, j) are all disjoint.
	For numbers which contain a sufficiently high power of 3, the length of		For numbers which contain a sufficiently high power of 3, the length of
	the rows grows beyond any limit, and the numbers containing any power of 2		the rows in T grows beyond any limit, and the numbers containing any power of 2
	will finally be covered		will finally be covered.
	(End)		(End)
	~~The lengths grow slowly; for~~ example, the ~~first number not~~ in a(~~1..10^7~~) ~~is 65536~~.		All numbers > 0 up to and including 2^(2*j + 1) appear in the rows in T up to and including
			A066443(j). For example, 4096 and 8192 are the trailing elements in row 398581 = A066443(6).

	Example:		Example:
	i \| j = 1 2 3 4 5 6		i \| j = 1 2 3 4 5 6 7
	----+----------------------		----+-------------------------
	1 \| 3 1 2		1 \| 3 1 2
	2 \| 7		2 \| 7
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	5 \| 19		5 \| 19
	6 \| 23		6 \| 23
	7 \| 27 9 18 6 12 4		7 \| 27 9 18 6 12 4 8

	Prog:		Prog:

imported>Gfis: formatting

2018-12-09T10:13:50Z

formatting

← Older revision		Revision as of 10:13, 9 December 2018
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	(c.f. the example below) which has rows í >= 1 consisting of subsequences		(c.f. the example below) which has rows í >= 1 consisting of subsequences
	of varying length as defined by the following algorithm:		of varying length as defined by the following algorithm:
	T(i, 1) := 4i - 1; j := 1;		T(i, 1) := 4 * i - 1; j := 1;
	while T(i, j) divisible by 3 do		while T(i, j) divisible by 3 do
	T(i, j + 1) := T(i, j) / 3;		T(i, j + 1) := T(i, j) / 3;
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	the operations in the algorithm. They are the following:		the operations in the algorithm. They are the following:
	e = 3^floor(j / 2)		e = 3^floor(j / 2)
	f = A066443(floor(j / 4)) with A066443(0) = 0, A066443(n) = 9 * A066443(n-1) - 2		f = A066443(floor(j / 4)) with A066443(0) = 0, A066443(n) = 9 * A066443(n - 1) - 2
	g = 2^floor((j + 3) / 2)		g = 2^floor((j + 3) / 2)
	m = 2^floor((j - 1) / 4) * A084101(j + 1 mod 4) with A084101(0..3) = (1, 3, 3, 1)		m = 2^floor((j - 1) / 4) * A084101(j + 1 mod 4) with A084101(0..3) = (1, 3, 3, 1)
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	Example:		Example:
	i \| j = 1 2 3 4 5 6		i \| j = 1 2 3 4 5 6
	----+--------------------		----+----------------------
	1 \| 3 1 2		1 \| 3 1 2
	2 \| 7		2 \| 7
Line 65:		Line 65:

	Prog:		Prog:
	(PARI) n=1; for(i=1,~~100~~, a=4*i-1; print1(a); while(a%3==0, a=a/3;\		(PARI) n=1; for(i=1,1000, a=4*i-1; print1(a); while(a%3==0, a=a/3;\
	print1(" ", a); a=a*2; print1(" ", a)); print;)		print1(" ", a); a=a*2; print1(" ", a)); print;)
	(Perl) use integer; my $n = 1; my $i = 1;		(Perl) use integer; my $n = 1; my $i = 1; while ($i <= 1000) {
	while ($i < 1000) {		my $an = 4 * $i - 1; print "$n $an\n"; $n ++;
	my $an = 4 * $i - 1; print "$n $an\n"; $n ++;		while ($an % 3 == 0) {
	while ($an % 3 == 0) {		$an /= 3; print "$n $an\n"; $n ++;
	$an /= 3; print "$n $an\n"; $n ++;		$an *= 2; print "$n $an\n"; $n ++;
	$an *= 2; print "$n $an\n"; $n ++;		} $i ++;
	} $i ++;		}
	}

	Crossrefs: Cf. A066443, A084101		Crossrefs: Cf. A066443, A084101
	Keywords: tabf,easy		Keywords: tabf,easy
	Author: Georg Fischer		Author: Georg Fischer

imported>Gfis: Algorithm in the name

2018-12-09T09:44:09Z

Algorithm in the name

← Older revision		Revision as of 09:44, 9 December 2018
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	Name: Permutation of the natural numbers: ~~Step~~ by 4 and ~~replace factors 3~~ by 2.		Name: Permutation of the natural numbers: Start with 3, divide by 3 and multiply by 2 as long as possible, then increase the start value by 4.

	3, 1, 2, 7, 11, 15, 5, 10, 19, 23, 27, 9, 18, 6, 12, 4, 8, 31, 35, 39, 13,		3, 1, 2, 7, 11, 15, 5, 10, 19, 23, 27, 9, 18, 6, 12, 4, 8, 31, 35, 39, 13,
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	j := j + 2;		j := j + 2;
	end while		end while
	The algorithm successively ~~replaces any factor~~ 3 by ~~a factor~~ 2,		The algorithm successively tries to divide by 3 and multiply by 2,
	~~and therefore~~ it always stops. The first rows which are longer		as long as the number contains a factor 3. Therefore it always stops.
	than any previous row are 1, 7, 61, 547, 4921 ... (A066443).		The first rows which are longer than any previous row are 1, 7, 61, 547, 4921 ... (A066443).

	Property: The sequence is a permutation of the natural numbers > 0.		Property: The sequence is a permutation of the natural numbers > 0.
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	Crossrefs: Cf. A066443, A084101		Crossrefs: Cf. A066443, A084101
	Keywords: ~~nonn,~~tabf,easy		Keywords: tabf,easy
	Author: Georg Fischer		Author: Georg Fischer

imported>Gfis: blocks

2018-12-09T00:05:36Z

blocks

← Older revision		Revision as of 00:05, 9 December 2018
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	Name: Permutation of the natural numbers: Step by 4 and replace factors 3 by 2.		Name: Permutation of the natural numbers: Step by 4 and replace factors 3 by 2.

	3, 1, 2, 7, 11, 15, 5, 10, 19, 23, 27, 9, 18, 6, 12, 4, 8, 31, 35, 39, 13,		3, 1, 2, 7, 11, 15, 5, 10, 19, 23, 27, 9, 18, 6, 12, 4, 8, 31, 35, 39, 13,
	26, 43, 47, 51, 17, 34, 55, 59, 63, 21, 42, 14, 28, 67, 71, 75, 25, 50, 79,		26, 43, 47, 51, 17, 34, 55, 59, 63, 21, 42, 14, 28, 67, 71, 75, 25, 50, 79,
	83, 87, 29, 58, 91, 95, 99, 33, 66, 22, 44, 103, 107, 111, 37, 74		83, 87, 29, 58, 91, 95, 99, 33, 66, 22, 44, 103, 107, 111, 37, 74

	Offset: 1,1		Offset: 1,1

	Comments: The sequence is the flattened form of an irregular table T(i, j)		Comments: The sequence is the flattened form of an irregular table T(i, j)
	(c.f. the example below) which has rows í >= 1 consisting of subsequences		(c.f. the example below) which has rows í >= 1 consisting of subsequences
Line 50:		Line 52:
	(End)		(End)
	The lengths grow slowly; for example, the first number not in a(1..10^7) is 65536.		The lengths grow slowly; for example, the first number not in a(1..10^7) is 65536.

	Example:		Example:
	i \| j = 1 2 3 4 5 6		i \| j = 1 2 3 4 5 6
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	6 \| 23		6 \| 23
	7 \| 27 9 18 6 12 4		7 \| 27 9 18 6 12 4

	Prog:		Prog:
	(PARI) n=1; for(i=1,100, a=4*i-1; print1(a); while(a%3==0, a=a/3;\		(PARI) n=1; for(i=1,100, a=4*i-1; print1(a); while(a%3==0, a=a/3;\

imported>Gfis: finished?

2018-12-09T00:03:51Z

finished?

← Older revision		Revision as of 00:03, 9 December 2018
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	Offset: 1,1		Offset: 1,1
	Comments: The sequence is the flattened form of a table T(i, j)		Comments: The sequence is the flattened form of an irregular table T(i, j)
	(c.f. the example below) which has rows í >= 1 consisting of		(c.f. the example below) which has rows í >= 1 consisting of subsequences
	~~subsequences~~ of varying length as defined by the		of varying length as defined by the following algorithm:
	following algorithm:
	T(i, 1) := 4i - 1; j := 1;		T(i, 1) := 4i - 1; j := 1;
	while T(i, j) divisible by 3 do		while T(i, j) divisible by 3 do
Line 15:		Line 14:
	j := j + 2;		j := j + 2;
	end while		end while
	The algorithm ~~always stops, and in one row,~~		The algorithm successively replaces any factor 3 by a factor 2,
	it successively replaces any factor 3 by a factor 2.		and therefore it always stops. The first rows which are longer
			than any previous row are 1, 7, 61, 547, 4921 ... (A066443).

	Property: The sequence is a permutation of the natural numbers > 0.		Property: The sequence is a permutation of the natural numbers > 0.
	Proof: (Start)		Proof: (Start)
	The values in columns j ~~in the table's rows~~ i of the form i = f * k + g,		The values in the columns j of T for row indexes i of the form
	k >= 0, if such columns are present, have the following residues modulo		i = e * k + f, k >= 0, if such columns are present, have the following
	~~powers~~ of 2:		residues modulo some power of 2:

	j \| Op. \| Form of i \| T(i, j) \| Residues \| Residues not yet covered		j \| Op. \| Form of i \| T(i, j) \| Residues \| Residues not yet covered
	---\|-----\| ----------\|-----------\|------------\|-------------------------		---+------+ -------------+--------------+------------+-------------------------
	1 \| \| 1k + 1 \| 4k + 3 \| 3 mod 4 \| 0, 1, 2 mod 4		1 \| \| 1 * k + 1 \| 4 * k + 3 \| 3 mod 4 \| 0, 1, 2 mod 4
	2 \| /3 \| 3k + 1 \| 4k + 1 \| 1 mod 4 \| 0, 2, 4, 6 mod 8		2 \| / 3 \| 3 * k + 1 \| 4 * k + 1 \| 1 mod 4 \| 0, 2, 4, 6 mod 8
	3 \| *2 \| 3k + 1 \| 8k + 2 \| 2 mod 8 \| 0, 4, 6 mod 8		3 \| * 2 \| 3 * k + 1 \| 8 * k + 2 \| 2 mod 8 \| 0, 4, 6 mod 8
	4 \| /3 \| 9k + 7 \| 8k + 6 \| 6 mod 8 \| 0, 4, 8, 12 mod 16		4 \| / 3 \| 9 * k + 7 \| 8 * k + 6 \| 6 mod 8 \| 0, 4, 8, 12 mod 16
	5 \| *2 \| 9k + 7 \| ~~16k~~ + 12 \| 12 mod 16 \| 0, 4, 8 mod 16		5 \| * 2 \| 9 * k + 7 \| 16 * k + 12 \| 12 mod 16 \| 0, 4, 8 mod 16
	6 \| /3 \| ~~27k~~ + 7 \| ~~16k~~ + 4 \| 4 mod 16 \| 0, 8, 16, 24 mod 32		6 \| / 3 \| 27 * k + 7 \| 16 * k + 4 \| 4 mod 16 \| 0, 8, 16, 24 mod 32
	7 \| *2 \| ~~27k~~ + 7 \| ~~32k~~ + 8 \| 8 mod 32 \| 0, 16, 24 mod 32		7 \| * 2 \| 27 * k + 7 \| 32 * k + 8 \| 8 mod 32 \| 0, 16, 24 mod 32
	8 \| /3 \| ~~81k~~ + 61 \| ~~32k~~ + 24 \| 24 mod 32 \| 0, 16, 32, 48 mod 64		8 \| / 3 \| 81 * k + 61 \| 32 * k + 24 \| 24 mod 32 \| 0, 16, 32, 48 mod 64
	9 \| *2 \| ~~81k~~ + 61 \| ~~64k~~ + 48 \| 48 mod 64 \| 0, 16, 32 mod 64		9 \| * 2 \| 81 * k + 61 \| 64 * k + 48 \| 48 mod 64 \| 0, 16, 32 mod 64
	...\| \| f*k + g \| ... ~~+ r \| \|~~		...\| ... \| e * k + f \| g * k + m \| m mod g \| 0, ...

	~~In general, f is a power of 3 increasing~~ in ~~every second column~~,		The variables in the last, general line can be computed from the
	~~and g is taken from~~ A066443 (a(0) = ~~1; a~~(n) = 9 * a(n-1) - 2		the operations in the algorithm. They are the following:
	~~for n >~~ 0). ~~g increases in every fourth column~~. ~~The residues r~~		e = 3^floor(j / 2)
	~~follow the pattern~~ (3, 1, 1~~, 3~~) * 2^h, where h increases in		f = A066443(floor(j / 4)) with A066443(0) = 0, A066443(n) = 9 * A066443(n-1) - 2
	~~every second column.~~		g = 2^floor((j + 3) / 2)
			m = 2^floor((j - 1) / 4) * A084101(j + 1 mod 4) with A084101(0..3) = (1, 3, 3, 1)

	The residues in each column and therefore the ~~values~~ are all disjoint,		The residues m in each column and therefore the T(i, j) are all disjoint.
	and ~~with i (and j, k) going to infinity, they cover all natural~~ numbers ~~> 0.~~		For numbers which contain a sufficiently high power of 3, the length of
			the rows grows beyond any limit, and the numbers containing any power of 2
			will finally be covered
	(End)		(End)
			The lengths grow slowly; for example, the first number not in a(1..10^7) is 65536.

	Example:		Example:
	i\| j=1 2 3 4 5 6		i \| j = 1 2 3 4 5 6
	-+--------------------		----+--------------------
	1: 3 1 2		1 \| 3 1 2
	2: 7		2 \| 7
	3: 11		3 \| 11
	4: 15 5 10		4 \| 15 5 10
	5: 19		5 \| 19
	6: 23		6 \| 23
	7: 27 9 18 6 12 4		7 \| 27 9 18 6 12 4

	Prog: (~~Perl~~)		Prog:
	use integer; my $n = 1; my $i = 1;		(PARI) n=1; for(i=1,100, a=4*i-1; print1(a); while(a%3==0, a=a/3;\
			print1(" ", a); a=a*2; print1(" ", a)); print;)
			(Perl) use integer; my $n = 1; my $i = 1;
	while ($i < 1000) {		while ($i < 1000) {
	my $an = 4 * $i - 1; print "$n $an\n"; $n ++;		my $an = 4 * $i - 1; print "$n $an\n"; $n ++;
Line 66:		Line 72:
	$an *= 2; print "$n $an\n"; $n ++;		$an *= 2; print "$n $an\n"; $n ++;
	} $i ++;		} $i ++;
	} ~~# A323232.pl, Georg Fischer Dec. 9 2018~~		}

	Crossrefs: ~~C.f~~. A066443, A084101		Crossrefs: Cf. A066443, A084101
	Keywords: nonn,tabf,easy		Keywords: nonn,tabf,easy
	Author: Georg Fischer		Author: Georg Fischer

imported>Gfis: preparation for OEIS

2018-12-08T09:56:19Z

preparation for OEIS

New page

Name: Permutation of the natural numbers: Step by 4 and replace factors 3 by 2.
3, 1, 2, 7, 11, 15, 5, 10, 19, 23, 27, 9, 18, 6, 12, 4, 8, 31, 35, 39, 13,
26, 43, 47, 51, 17, 34, 55, 59, 63, 21, 42, 14, 28, 67, 71, 75, 25, 50, 79,
83, 87, 29, 58, 91, 95, 99, 33, 66, 22, 44, 103, 107, 111, 37, 74

Offset: 1,1
Comments: The sequence is the flattened form of a table T(i, j)
(c.f. the example below) which has rows í >= 1 consisting of
subsequences of varying length as defined by the
following algorithm:
T(i, 1) := 4i - 1; j := 1;
while T(i, j) divisible by 3 do
T(i, j + 1) := T(i, j) / 3;
T(i, j + 2) := T(i, j + 1) * 2;
j := j + 2;
end while
The algorithm always stops, and in one row,
it successively replaces any factor 3 by a factor 2.

Property: The sequence is a permutation of the natural numbers > 0.
Proof: (Start)
The values in columns j in the table's rows i of the form i = f * k + g,
k >= 0, if such columns are present, have the following residues modulo
powers of 2:

j | Op. | Form of i | T(i, j) | Residues | Residues not yet covered
---|-----| ----------|-----------|------------|-------------------------
1 | | 1k + 1 | 4k + 3 | 3 mod 4 | 0, 1, 2 mod 4
2 | /3 | 3k + 1 | 4k + 1 | 1 mod 4 | 0, 2, 4, 6 mod 8
3 | *2 | 3k + 1 | 8k + 2 | 2 mod 8 | 0, 4, 6 mod 8
4 | /3 | 9k + 7 | 8k + 6 | 6 mod 8 | 0, 4, 8, 12 mod 16
5 | *2 | 9k + 7 | 16k + 12 | 12 mod 16 | 0, 4, 8 mod 16
6 | /3 | 27k + 7 | 16k + 4 | 4 mod 16 | 0, 8, 16, 24 mod 32
7 | *2 | 27k + 7 | 32k + 8 | 8 mod 32 | 0, 16, 24 mod 32
8 | /3 | 81k + 61 | 32k + 24 | 24 mod 32 | 0, 16, 32, 48 mod 64
9 | *2 | 81k + 61 | 64k + 48 | 48 mod 64 | 0, 16, 32 mod 64
...| | f*k + g | ... + r | |

In general, f is a power of 3 increasing in every second column,
and g is taken from A066443 (a(0) = 1; a(n) = 9 * a(n-1) - 2
for n > 0). g increases in every fourth column. The residues r
follow the pattern (3, 1, 1, 3) * 2^h, where h increases in
every second column.

The residues in each column and therefore the values are all disjoint,
and with i (and j, k) going to infinity, they cover all natural numbers > 0.
(End)

Example:
i| j=1 2 3 4 5 6
-+--------------------
1: 3 1 2
2: 7
3: 11
4: 15 5 10
5: 19
6: 23
7: 27 9 18 6 12 4

Prog: (Perl)
use integer; my $n = 1; my $i = 1;
while ($i < 1000) {
my $an = 4 * $i - 1; print "$n $an\n"; $n ++;
while ($an % 3 == 0) {
$an /= 3; print "$n $an\n"; $n ++;
$an *= 2; print "$n $an\n"; $n ++;
} $i ++;
} # A323232.pl, Georg Fischer Dec. 9 2018

Crossrefs: C.f. A066443, A084101
Keywords: nonn,tabf,easy
Author: Georg Fischer