finding a recursive formula for a sequence, if possible [closed]
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Hi everyone: I was looking over an NYS Algebra 2 Regents, and two training programs for a long-distance race were compared. (The answer they wanted was the recursive formula for a simple arithmetic sequence) It is the sequence they did NOT need the answer too which fascinated me. The pattern is quite easy to see, even without math training, but I was wondering if a recursive (or any other) formula could be written for it.
The pattern unfolds over 14 weeks (though it could easily continue), and week one, or a1 = 21 miles. a2 = 22 miles. a3 goes back to 21 miles...
Here's the outputs, sequentially, beginning with week 1: (21,22,21,22,23,22,23,24,25, 24 25 26 27 28...)
Thank you, VJ
recursive-algorithms
asked Aug 13 at 3:20
Victor Jaroslaw
212
closed as unclear what you're asking by Lord Shark the Unknown, Brahadeesh, Claude Leibovici, Adrian Keister, Shailesh Aug 14 at 0:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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Hi everyone: I was looking over an NYS Algebra 2 Regents, and two training programs for a long-distance race were compared. (The answer they wanted was the recursive formula for a simple arithmetic sequence) It is the sequence they did NOT need the answer too which fascinated me. The pattern is quite easy to see, even without math training, but I was wondering if a recursive (or any other) formula could be written for it.
The pattern unfolds over 14 weeks (though it could easily continue), and week one, or a1 = 21 miles. a2 = 22 miles. a3 goes back to 21 miles...
Here's the outputs, sequentially, beginning with week 1: (21,22,21,22,23,22,23,24,25, 24 25 26 27 28...)
Thank you, VJ
recursive-algorithms
asked Aug 13 at 3:20
Victor Jaroslaw
212
closed as unclear what you're asking by Lord Shark the Unknown, Brahadeesh, Claude Leibovici, Adrian Keister, Shailesh Aug 14 at 0:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
0
down vote
favorite
up vote
0
down vote
favorite
Hi everyone: I was looking over an NYS Algebra 2 Regents, and two training programs for a long-distance race were compared. (The answer they wanted was the recursive formula for a simple arithmetic sequence) It is the sequence they did NOT need the answer too which fascinated me. The pattern is quite easy to see, even without math training, but I was wondering if a recursive (or any other) formula could be written for it.
The pattern unfolds over 14 weeks (though it could easily continue), and week one, or a1 = 21 miles. a2 = 22 miles. a3 goes back to 21 miles...
Here's the outputs, sequentially, beginning with week 1: (21,22,21,22,23,22,23,24,25, 24 25 26 27 28...)
Thank you, VJ
recursive-algorithms
asked Aug 13 at 3:20
Victor Jaroslaw
212
Hi everyone: I was looking over an NYS Algebra 2 Regents, and two training programs for a long-distance race were compared. (The answer they wanted was the recursive formula for a simple arithmetic sequence) It is the sequence they did NOT need the answer too which fascinated me. The pattern is quite easy to see, even without math training, but I was wondering if a recursive (or any other) formula could be written for it.
The pattern unfolds over 14 weeks (though it could easily continue), and week one, or a1 = 21 miles. a2 = 22 miles. a3 goes back to 21 miles...
Here's the outputs, sequentially, beginning with week 1: (21,22,21,22,23,22,23,24,25, 24 25 26 27 28...)
Thank you, VJ
recursive-algorithms
asked Aug 13 at 3:20
Victor Jaroslaw
212
asked Aug 13 at 3:20
Victor Jaroslaw
212
asked Aug 13 at 3:20
Victor Jaroslaw
212
asked Aug 13 at 3:20
Victor Jaroslaw
212
212
closed as unclear what you're asking by Lord Shark the Unknown, Brahadeesh, Claude Leibovici, Adrian Keister, Shailesh Aug 14 at 0:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Lord Shark the Unknown, Brahadeesh, Claude Leibovici, Adrian Keister, Shailesh Aug 14 at 0:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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I don't think the pattern is easy to see. If you subtract $20$ from each entry you get
$$1,2\1,2,3\2,3,4,5\4,5,6,7,8$$
and I don't really know how to continue. Clearly the next row has $6$ numbers in a row in it, but does it start with $8$ because you double the previous starting number (ignoring the two $1$s together at the start), with $7$ because you subtract $1$ from the previous ending number, with $8$ because the last number should be the next Fibonacci number, or what? All of these can be encoded into formulas, some easier than others.
Added: If I believe dxiv's claim that the starting number in each row is one less than the ending number in the previous row we can generate the table as long as we want. We can then find an algorithm to compute the $n^th$ term of the sequence. We have
$$1,2\1,2,3\2,3,4,5\4,5,6,7,8\7,8,9,10,11,12\
11,12,13,14,15,16,17\16,17,18,19,20,21,22,23$$
Note that the starting numbers in each row are one more than the triangular numbers. The positions of the first numbers of each row in the sequence are $1,3,6,10,15ldots$, the triangular numbers as well.
Let $n$ be the position in the sequence that we want to find the value of. The row number is $r=lfloor frac 12(sqrt 1+8n-1)rfloor$. The position in the row, counting from $0$ is $p=n-frac 12r(r+1)$. The first number in row $r$ is $frac 12(r-2)(r-1)+1$ and the number in position $n$ is $frac 12(r-2)(r-1)+1+p$
Now add back the$20$ I subtracted
answered Aug 13 at 3:58
Ross Millikan
277k21187352
I believe the pattern is "$n$ steps forward, $1$ step back", that said I don't see much hope for a "nice" recurrence to it (or closed form, for that matter).
– dxiv
Aug 13 at 4:42
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I don't think the pattern is easy to see. If you subtract $20$ from each entry you get
$$1,2\1,2,3\2,3,4,5\4,5,6,7,8$$
and I don't really know how to continue. Clearly the next row has $6$ numbers in a row in it, but does it start with $8$ because you double the previous starting number (ignoring the two $1$s together at the start), with $7$ because you subtract $1$ from the previous ending number, with $8$ because the last number should be the next Fibonacci number, or what? All of these can be encoded into formulas, some easier than others.
Added: If I believe dxiv's claim that the starting number in each row is one less than the ending number in the previous row we can generate the table as long as we want. We can then find an algorithm to compute the $n^th$ term of the sequence. We have
$$1,2\1,2,3\2,3,4,5\4,5,6,7,8\7,8,9,10,11,12\
11,12,13,14,15,16,17\16,17,18,19,20,21,22,23$$
Note that the starting numbers in each row are one more than the triangular numbers. The positions of the first numbers of each row in the sequence are $1,3,6,10,15ldots$, the triangular numbers as well.
Let $n$ be the position in the sequence that we want to find the value of. The row number is $r=lfloor frac 12(sqrt 1+8n-1)rfloor$. The position in the row, counting from $0$ is $p=n-frac 12r(r+1)$. The first number in row $r$ is $frac 12(r-2)(r-1)+1$ and the number in position $n$ is $frac 12(r-2)(r-1)+1+p$
Now add back the$20$ I subtracted
answered Aug 13 at 3:58
Ross Millikan
277k21187352
I believe the pattern is "$n$ steps forward, $1$ step back", that said I don't see much hope for a "nice" recurrence to it (or closed form, for that matter).
– dxiv
Aug 13 at 4:42
add a comment |Â
up vote
1
down vote
I don't think the pattern is easy to see. If you subtract $20$ from each entry you get
$$1,2\1,2,3\2,3,4,5\4,5,6,7,8$$
and I don't really know how to continue. Clearly the next row has $6$ numbers in a row in it, but does it start with $8$ because you double the previous starting number (ignoring the two $1$s together at the start), with $7$ because you subtract $1$ from the previous ending number, with $8$ because the last number should be the next Fibonacci number, or what? All of these can be encoded into formulas, some easier than others.
Added: If I believe dxiv's claim that the starting number in each row is one less than the ending number in the previous row we can generate the table as long as we want. We can then find an algorithm to compute the $n^th$ term of the sequence. We have
$$1,2\1,2,3\2,3,4,5\4,5,6,7,8\7,8,9,10,11,12\
11,12,13,14,15,16,17\16,17,18,19,20,21,22,23$$
Note that the starting numbers in each row are one more than the triangular numbers. The positions of the first numbers of each row in the sequence are $1,3,6,10,15ldots$, the triangular numbers as well.
Let $n$ be the position in the sequence that we want to find the value of. The row number is $r=lfloor frac 12(sqrt 1+8n-1)rfloor$. The position in the row, counting from $0$ is $p=n-frac 12r(r+1)$. The first number in row $r$ is $frac 12(r-2)(r-1)+1$ and the number in position $n$ is $frac 12(r-2)(r-1)+1+p$
Now add back the$20$ I subtracted
answered Aug 13 at 3:58
Ross Millikan
277k21187352
I believe the pattern is "$n$ steps forward, $1$ step back", that said I don't see much hope for a "nice" recurrence to it (or closed form, for that matter).
– dxiv
Aug 13 at 4:42
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I don't think the pattern is easy to see. If you subtract $20$ from each entry you get
$$1,2\1,2,3\2,3,4,5\4,5,6,7,8$$
and I don't really know how to continue. Clearly the next row has $6$ numbers in a row in it, but does it start with $8$ because you double the previous starting number (ignoring the two $1$s together at the start), with $7$ because you subtract $1$ from the previous ending number, with $8$ because the last number should be the next Fibonacci number, or what? All of these can be encoded into formulas, some easier than others.
Added: If I believe dxiv's claim that the starting number in each row is one less than the ending number in the previous row we can generate the table as long as we want. We can then find an algorithm to compute the $n^th$ term of the sequence. We have
$$1,2\1,2,3\2,3,4,5\4,5,6,7,8\7,8,9,10,11,12\
11,12,13,14,15,16,17\16,17,18,19,20,21,22,23$$
Note that the starting numbers in each row are one more than the triangular numbers. The positions of the first numbers of each row in the sequence are $1,3,6,10,15ldots$, the triangular numbers as well.
Let $n$ be the position in the sequence that we want to find the value of. The row number is $r=lfloor frac 12(sqrt 1+8n-1)rfloor$. The position in the row, counting from $0$ is $p=n-frac 12r(r+1)$. The first number in row $r$ is $frac 12(r-2)(r-1)+1$ and the number in position $n$ is $frac 12(r-2)(r-1)+1+p$
Now add back the$20$ I subtracted
answered Aug 13 at 3:58
Ross Millikan
277k21187352
I don't think the pattern is easy to see. If you subtract $20$ from each entry you get
$$1,2\1,2,3\2,3,4,5\4,5,6,7,8$$
and I don't really know how to continue. Clearly the next row has $6$ numbers in a row in it, but does it start with $8$ because you double the previous starting number (ignoring the two $1$s together at the start), with $7$ because you subtract $1$ from the previous ending number, with $8$ because the last number should be the next Fibonacci number, or what? All of these can be encoded into formulas, some easier than others.
Added: If I believe dxiv's claim that the starting number in each row is one less than the ending number in the previous row we can generate the table as long as we want. We can then find an algorithm to compute the $n^th$ term of the sequence. We have
$$1,2\1,2,3\2,3,4,5\4,5,6,7,8\7,8,9,10,11,12\
11,12,13,14,15,16,17\16,17,18,19,20,21,22,23$$
Note that the starting numbers in each row are one more than the triangular numbers. The positions of the first numbers of each row in the sequence are $1,3,6,10,15ldots$, the triangular numbers as well.
Let $n$ be the position in the sequence that we want to find the value of. The row number is $r=lfloor frac 12(sqrt 1+8n-1)rfloor$. The position in the row, counting from $0$ is $p=n-frac 12r(r+1)$. The first number in row $r$ is $frac 12(r-2)(r-1)+1$ and the number in position $n$ is $frac 12(r-2)(r-1)+1+p$
Now add back the$20$ I subtracted
answered Aug 13 at 3:58
Ross Millikan
277k21187352
edited Aug 13 at 5:47
answered Aug 13 at 3:58
Ross Millikan
277k21187352
answered Aug 13 at 3:58
Ross Millikan
277k21187352
answered Aug 13 at 3:58
Ross Millikan
277k21187352
277k21187352
I believe the pattern is "$n$ steps forward, $1$ step back", that said I don't see much hope for a "nice" recurrence to it (or closed form, for that matter).
– dxiv
Aug 13 at 4:42
add a comment |Â
I believe the pattern is "$n$ steps forward, $1$ step back", that said I don't see much hope for a "nice" recurrence to it (or closed form, for that matter).
– dxiv
Aug 13 at 4:42
I believe the pattern is "$n$ steps forward, $1$ step back", that said I don't see much hope for a "nice" recurrence to it (or closed form, for that matter).
– dxiv
Aug 13 at 4:42
I believe the pattern is "$n$ steps forward, $1$ step back", that said I don't see much hope for a "nice" recurrence to it (or closed form, for that matter).
– dxiv
Aug 13 at 4:42
add a comment |Â
