Revision History for A286347
(Underlined text is an addition;
strikethrough text is a deletion.)
Showing entries 1-10
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#13 by Alois P. Heinz at Mon Nov 19 11:01:58 EST 2018
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#12 by Alois P. Heinz at Tue Nov 13 16:47:47 EST 2018
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#11 by Alois P. Heinz at Mon May 08 11:44:45 EDT 2017
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#10 by Alois P. Heinz at Mon May 08 11:43:47 EDT 2017
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Alois P. Heinz, <a href="/A286347/b286347.txt">Table of n, a(n) for n = 0..852</a>
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#9 by Alois P. Heinz at Mon May 08 11:42:33 EDT 2017
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allocated for Alois P. Heinz
Number of tilings of a 6 X n rectangle using pentominoes of shapes X, Y, T and monominoes.
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| DATA
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1, 1, 15, 188, 3763, 54368, 790504, 11771179, 176285795, 2627027061, 39172906410, 584774465361, 8725442013820, 130117410190768, 1940656499251689, 28950613608637003, 431889355175650435, 6442566792491842258, 96102610815876310611, 1433558139553752297236
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| OFFSET
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0,3
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Wikipedia, <a href="https://en.wikipedia.org/wiki/Pentomino">Pentomino</a>
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Cf. A247125, A278330, A278456, A278657, A278877, A278964, A286391.
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| KEYWORD
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allocated
nonn
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| AUTHOR
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Alois P. Heinz, May 08 2017
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approved
editing
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#8 by Alois P. Heinz at Mon May 08 11:42:33 EDT 2017
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allocated for Alois P. Heinz
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recycled
allocated
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#7 by N. J. A. Sloane at Mon May 08 11:03:09 EDT 2017
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#6 by N. J. A. Sloane at Mon May 08 11:03:06 EDT 2017
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Sequence of sizes of repeated strings of zeroes of (10^d)Digit(2^L).
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| DATA
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2, 4, 6, 9, 11, 13, 16, 18, 20, 23, 25, 27, 30, 32, 34, 37, 39, 41, 44, 46, 48, 51, 53, 55, 58, 60, 62, 65, 67, 69, 72, 74, 76, 79, 81, 83, 86, 88, 90, 93, 95, 97, 100, 102, 104, 107, 109, 111, 114, 116, 118, 121, 123, 125
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| OFFSET
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1,1
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| COMMENTS
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Lets define a new function named DthDigit as
(10^d)Digit(any function)=([any function mod 10^d]-[any function mod 10^(d-1)])/10^(d-1), for d>0 and,
for d=0 (10^0)Digit(any function) is equal to [any function mod 10].
Now, if our any function is f(L)=2^L and L varies from 1 to infinite, f(L) will construct a infinite table with numbers. Lets align all of them by unit digit. One number below the other. Example of the beginning of the table where 11>L>0:
d = 3 2 1 0
L
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1 o o o 2
2 o o 0 4
3 o 0 0 8
4 0 0 1 6
5 0 0 3 2
6 0 0 6 4
7 0 1 2 8
8 0 2 5 6
9 0 5 1 2
10 1 0 2 4
Then, when we varies d from 1 to infinite, (10^d)Digit(2^L)=([2^L mod 10^d]-[2^L mod 10^(d-1)])/10^(d-1) will generate infinite strings of digits. The strings will be vertical in our table. All strings will have 0 as first element. For example, when d=2, the string is 0000001250…
Now, lets also define [(10^d)Digit(function)]S#(criteria) as string size of the digits generated by (10^d)Digit(function) under the criteria defined.
And also, lets define as the repeated strings of zeroes of (10^d)Digit(2^L) as the string in the column d that starts from the zero at line L=d+1 (the zero where L=d+1) until the zero just before any digit not 0. In our example, when d=2 the first string of zeroes of (10^2)Digit(2^L) is 000000, but the first repeated strings of zeroes is 0000 with the first zero is in the line L=d+1=2+1=3.
So, the formula that count the numbers of zeroes witch repeat infinitely in the string of (10^d)Digit(2^L), for each d>0, is:
[(10^d)Digit(2^L)]S#(repeated strings of zeroes)=a(d-3)+10-d where a(-2)=-7; a(-1)=-4; a(0)=-1; d>0; and L>0.
The repeating sequence of [(10^d)Digit(2^n)]S#(repeated strings of zeroes) mod 9 will be 246024702570357135813681468 size of 27 elements.
It is important to note that even in this new sequence of sizes of repeated strings of zeroes of (10^d)Digit(2^n) can be done the same concept:
[(10^d)Digit{[(10^d)Digit(2^n)]S#(repeated strings of zeroes)}]S#(repeated strings of zeroes)
and we will find a new sequence of repeated strings of zeroes. And so on, and so on. And, due to we are working on mod, and as mod is cyclic, I conjecture that in some point will return to initial combination of the digits again between mod 10 with mod 9.
Also, these concepts can be done to any function.
The size of repetitive strings for each d will be given by SRS(d)=10^d/2^(d-2) that it is A005054 for d>0. And A005054 mod 9 is 421578 size 6 (also do not have triade (3;6;9) as do not have 2^n, and also we can do the same procedure to it).
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| LINKS
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Charles Kusniec, <a href="/A286347/a286347.pdf">DthDigit(2^n)</a>
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| FORMULA
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a(d)=a(d-3)+10-d where a(-2)=-7; a(-1)=-4; a(0)=-1; d>0;
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| CROSSREFS
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Cf. A005054
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| KEYWORD
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nonn,changed
recycled
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| AUTHOR
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Charles Kusniec, May 07 2017
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#5 by Charles Kusniec at Mon May 08 09:01:21 EDT 2017
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| COMMENTS
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And also, lets define as the repeated strings of zeroes of (10^d)Digit(2^L) as the string in the column d that starts from the zero at line L=d+1 (the zero where L=d+1) until the zero just before any digit not 0. In our example, when d=2 the first string of zeroes of (10^2)Digit(2^L) is 000000, andbut the first repeated strings of zeroes is 0000 with the first zero is in the line L=d+1=2+1=3.
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#4 by Charles Kusniec at Mon May 08 08:53:21 EDT 2017
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| COMMENTS
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d = 3 2 1 0
1 o o o 2
2 o o 0 4
3 o 0 0 8
4 0 0 1 6
5 0 0 3 2
6 0 0 6 4
7 0 1 2 8
8 0 2 5 6
9 0 5 1 2
10 1 0 2 4
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