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Finding a pattern, I'm stuck
Finding pattern (2,7,8,3,5)Finding pattern- $6$ $11$ $25$What is a permutation pattern?Pattern finding algorithmsNumber patterns - finding the patternPalindromic Numbers - Pattern “inside” Prime Numbers?Collatz conjecture pattern (3n + 1 problem).Completing sequence patternWhat is the general term of this series (obtained from an iterated mean)?Find the number pattern
$begingroup$
I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?
binomial-coefficients pattern-recognition pattern-matching
asked 8 hours ago
KaræthonKaræthon
183
New contributor
Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?
binomial-coefficients pattern-recognition pattern-matching
asked 8 hours ago
KaræthonKaræthon
183
New contributor
Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
$endgroup$
– Ak19
8 hours ago
$begingroup$
Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
$endgroup$
– Karæthon
8 hours ago
add a comment |
$begingroup$
I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?
binomial-coefficients pattern-recognition pattern-matching
asked 8 hours ago
KaræthonKaræthon
183
New contributor
Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a friend who knows how much I love math, (I imagine new problems just to do the math behind them and to see if I can expand my understanding) and so he brings me the stuff that stumps him. Usually I end up finding the answer for him and then explaining how to him, but he gave me one 2 weeks ago and I cant figure it out. It's part of some math for calculating the area (or maybe perimeter?) of an elipse, the part he wanted me to figure out is: there is a part that goes π * (a + b) * (1 + 1/4 * h + 1/64 * h^2 + 1/256 * h^3 + 1/16384 * h^4 + 1/ 65536 * h^5 + 1/1048576 * h^6 ...) we only have those six, and he wanted me to find the pattern to calculate the rest of the denominators to extend it. But everything I have tried adds extra values within the first six or skips values within the first six. Does anyone know what the pattern is and what formula can be used to calculate the rest?
binomial-coefficients pattern-recognition pattern-matching
binomial-coefficients pattern-recognition pattern-matching
asked 8 hours ago
KaræthonKaræthon
183
New contributor
Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
KaræthonKaræthon
183
New contributor
Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
KaræthonKaræthon
183
New contributor
Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
KaræthonKaræthon
183
asked 8 hours ago
KaræthonKaræthon
183
183
New contributor
Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Karæthon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
$endgroup$
– Ak19
8 hours ago
$begingroup$
Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
$endgroup$
– Karæthon
8 hours ago
add a comment |
$begingroup$
I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
$endgroup$
– Ak19
8 hours ago
$begingroup$
Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
$endgroup$
– Karæthon
8 hours ago
$begingroup$
I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
$endgroup$
– Ak19
8 hours ago
$begingroup$
I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
$endgroup$
– Ak19
8 hours ago
$begingroup$
Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
$endgroup$
– Karæthon
8 hours ago
$begingroup$
Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
$endgroup$
– Karæthon
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.
Considering it has to do with ellipses, the sequence of denominators should be A056982.
answered 8 hours ago
FimpellizieriFimpellizieri
17.3k11836
$endgroup$
$begingroup$
Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
$endgroup$
– Karæthon
8 hours ago
$begingroup$
The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
$endgroup$
– Fimpellizieri
8 hours ago
$begingroup$
Ok, I think I understand, it works on my calculator so...
$endgroup$
– Karæthon
8 hours ago
add a comment |
$begingroup$
The $n^textth$ denominator appears to be
$$4^large2n-text(the number of 1s in the base-2 representation of n)$$
starting with $n=0.$
answered 8 hours ago
Mitchell SpectorMitchell Spector
7,1572824
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.
Considering it has to do with ellipses, the sequence of denominators should be A056982.
answered 8 hours ago
FimpellizieriFimpellizieri
17.3k11836
$endgroup$
$begingroup$
Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
$endgroup$
– Karæthon
8 hours ago
$begingroup$
The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
$endgroup$
– Fimpellizieri
8 hours ago
$begingroup$
Ok, I think I understand, it works on my calculator so...
$endgroup$
– Karæthon
8 hours ago
add a comment |
$begingroup$
If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.
Considering it has to do with ellipses, the sequence of denominators should be A056982.
answered 8 hours ago
FimpellizieriFimpellizieri
17.3k11836
$endgroup$
$begingroup$
Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
$endgroup$
– Karæthon
8 hours ago
$begingroup$
The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
$endgroup$
– Fimpellizieri
8 hours ago
$begingroup$
Ok, I think I understand, it works on my calculator so...
$endgroup$
– Karæthon
8 hours ago
add a comment |
$begingroup$
If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.
Considering it has to do with ellipses, the sequence of denominators should be A056982.
answered 8 hours ago
FimpellizieriFimpellizieri
17.3k11836
$endgroup$
If I could give some advice, these kinds of questions generally benefit from a look at the Online Encyclopedia of Integer Sequences.
Considering it has to do with ellipses, the sequence of denominators should be A056982.
answered 8 hours ago
FimpellizieriFimpellizieri
17.3k11836
answered 8 hours ago
FimpellizieriFimpellizieri
17.3k11836
answered 8 hours ago
FimpellizieriFimpellizieri
17.3k11836
answered 8 hours ago
FimpellizieriFimpellizieri
17.3k11836
17.3k11836
$begingroup$
Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
$endgroup$
– Karæthon
8 hours ago
$begingroup$
The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
$endgroup$
– Fimpellizieri
8 hours ago
$begingroup$
Ok, I think I understand, it works on my calculator so...
$endgroup$
– Karæthon
8 hours ago
add a comment |
$begingroup$
Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
$endgroup$
– Karæthon
8 hours ago
$begingroup$
The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
$endgroup$
– Fimpellizieri
8 hours ago
$begingroup$
Ok, I think I understand, it works on my calculator so...
$endgroup$
– Karæthon
8 hours ago
$begingroup$
Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
$endgroup$
– Karæthon
8 hours ago
$begingroup$
Oooo, never seen that site before.... Me likey, and yes that does appear to be the sequence, but I'm not understanding what it is telling me to do to calculate those numbers...
$endgroup$
– Karæthon
8 hours ago
$begingroup$
The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
$endgroup$
– Fimpellizieri
8 hours ago
$begingroup$
The sequence is given as $4$ to the power of A005187, which in turn is given as a recurrence relation (or other non-simple descriptions). It does not appear to have a simple formula involving only sums, products and exponentiation, which is often the case in maths. There are other ways to describe it though (with words, via generating functions, or with formulas using other more complex operations).
$endgroup$
– Fimpellizieri
8 hours ago
$begingroup$
Ok, I think I understand, it works on my calculator so...
$endgroup$
– Karæthon
8 hours ago
$begingroup$
Ok, I think I understand, it works on my calculator so...
$endgroup$
– Karæthon
8 hours ago
add a comment |
$begingroup$
The $n^textth$ denominator appears to be
$$4^large2n-text(the number of 1s in the base-2 representation of n)$$
starting with $n=0.$
answered 8 hours ago
Mitchell SpectorMitchell Spector
7,1572824
$endgroup$
add a comment |
$begingroup$
The $n^textth$ denominator appears to be
$$4^large2n-text(the number of 1s in the base-2 representation of n)$$
starting with $n=0.$
answered 8 hours ago
Mitchell SpectorMitchell Spector
7,1572824
$endgroup$
add a comment |
$begingroup$
The $n^textth$ denominator appears to be
$$4^large2n-text(the number of 1s in the base-2 representation of n)$$
starting with $n=0.$
answered 8 hours ago
Mitchell SpectorMitchell Spector
7,1572824
$endgroup$
The $n^textth$ denominator appears to be
$$4^large2n-text(the number of 1s in the base-2 representation of n)$$
starting with $n=0.$
answered 8 hours ago
Mitchell SpectorMitchell Spector
7,1572824
answered 8 hours ago
Mitchell SpectorMitchell Spector
7,1572824
answered 8 hours ago
Mitchell SpectorMitchell Spector
7,1572824
answered 8 hours ago
Mitchell SpectorMitchell Spector
7,1572824
7,1572824
add a comment |
add a comment |
Karæthon is a new contributor. Be nice, and check out our Code of Conduct.
Karæthon is a new contributor. Be nice, and check out our Code of Conduct.
Karæthon is a new contributor. Be nice, and check out our Code of Conduct.
Karæthon is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I think it follows the increment pattern of 1,2,1,3,1,2,1,3... in the powers of 4 (starting from 0- 0,1,3,4,7,8,10,11...)
$endgroup$
– Ak19
8 hours ago
$begingroup$
Yeah, i noticed a pattern like that, but can't figure out the logic behind it.
$endgroup$
– Karæthon
8 hours ago