[math-fun] Sequence relating to the Erdős–Straus conjecture
Let A(n) be the number of ways of expressing 4/n as the sum of three integer reciprocals, where the mere permutation of a sum is regarded as not making a difference. Plainly 4/1 = 4 cannot be expressed as the sum of three reciprocals, so A(1) = 0. 4/2 = 2 = 1/1 + 1/2 + 1/2, and there are no other solutions, so A(2) = 1. 4/3 = 1 + 1/4 + 1/12 = 1+ 1/6 + 1/6 = 1/2 + 1/2 + 1/3; I am pretty sure that A(3) = 3. The Erdős–Straus conjecture is that A(n) > 0 for all n > 1. Of course I wanted to know if A was in OEIS. I calculated a few more terms, and what I had was 0, 1, 3, 3, 2, 8 ... I was pretty confident in my enumeration, so I calculated enough entries, and discovered to my surprise that the sequence was missing. Then I searched for "Straus", and quickly found A192787, which claims to be my A. The trouble is, A192787(4) = 4, and I say A(4) = 3. Bear with me while I list my solutions, and then somebody tell me what I missed. 4/4 = 1, so the problem is to partition 1 into three reciprocals. I have the following solutions: 1/2 + 1/3 + 1/6 1/2 + 1/4 + 1/4 1/3 + 1/3 + 1/3 A192787 seems to be claiming that I missed one. Charles R. Greathouse IV was the sequence author, and I think he's a funster, so, Charles, if you're listening, can you tell me the missing dissection?
1/1 ?? R. On Tue, 19 Feb 2013, Allan Wechsler wrote:
Let A(n) be the number of ways of expressing 4/n as the sum of three integer reciprocals, where the mere permutation of a sum is regarded as not making a difference.
Plainly 4/1 = 4 cannot be expressed as the sum of three reciprocals, so A(1) = 0.
4/2 = 2 = 1/1 + 1/2 + 1/2, and there are no other solutions, so A(2) = 1.
4/3 = 1 + 1/4 + 1/12 = 1+ 1/6 + 1/6 = 1/2 + 1/2 + 1/3; I am pretty sure that A(3) = 3.
The Erdős–Straus conjecture is that A(n) > 0 for all n > 1.
Of course I wanted to know if A was in OEIS. I calculated a few more terms, and what I had was 0, 1, 3, 3, 2, 8 ... I was pretty confident in my enumeration, so I calculated enough entries, and discovered to my surprise that the sequence was missing.
Then I searched for "Straus", and quickly found A192787, which claims to be my A. The trouble is, A192787(4) = 4, and I say A(4) = 3.
Bear with me while I list my solutions, and then somebody tell me what I missed.
4/4 = 1, so the problem is to partition 1 into three reciprocals. I have the following solutions:
1/2 + 1/3 + 1/6 1/2 + 1/4 + 1/4 1/3 + 1/3 + 1/3
A192787 seems to be claiming that I missed one. Charles R. Greathouse IV was the sequence author, and I think he's a funster, so, Charles, if you're listening, can you tell me the missing dissection? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I see your point, but if 1/1 is permitted, then 1/2 + 1/2 should be permitted too, and that would make A(4) = 5, not 4. On Tue, Feb 19, 2013 at 4:45 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
1/1 ?? R.
On Tue, 19 Feb 2013, Allan Wechsler wrote:
Let A(n) be the number of ways of expressing 4/n as the sum of three
integer reciprocals, where the mere permutation of a sum is regarded as not making a difference.
Plainly 4/1 = 4 cannot be expressed as the sum of three reciprocals, so A(1) = 0.
4/2 = 2 = 1/1 + 1/2 + 1/2, and there are no other solutions, so A(2) = 1.
4/3 = 1 + 1/4 + 1/12 = 1+ 1/6 + 1/6 = 1/2 + 1/2 + 1/3; I am pretty sure that A(3) = 3.
The Erdős–Straus conjecture is that A(n) > 0 for all n > 1.
Of course I wanted to know if A was in OEIS. I calculated a few more terms, and what I had was 0, 1, 3, 3, 2, 8 ... I was pretty confident in my enumeration, so I calculated enough entries, and discovered to my surprise that the sequence was missing.
Then I searched for "Straus", and quickly found A192787, which claims to be my A. The trouble is, A192787(4) = 4, and I say A(4) = 3.
Bear with me while I list my solutions, and then somebody tell me what I missed.
4/4 = 1, so the problem is to partition 1 into three reciprocals. I have the following solutions:
1/2 + 1/3 + 1/6 1/2 + 1/4 + 1/4 1/3 + 1/3 + 1/3
A192787 seems to be claiming that I missed one. Charles R. Greathouse IV was the sequence author, and I think he's a funster, so, Charles, if you're listening, can you tell me the missing dissection? ______________________________**_________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/**cgi-bin/mailman/listinfo/math-**fun<http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>
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Oh, and I wanted to make sure I said this: if it turns out that A192787 is wrong, then before we correct it, we should figure out what the existing sequence really is, because it's probably interesting and shouldn't be discarded. On Tue, Feb 19, 2013 at 4:49 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I see your point, but if 1/1 is permitted, then 1/2 + 1/2 should be permitted too, and that would make A(4) = 5, not 4.
On Tue, Feb 19, 2013 at 4:45 PM, Richard Guy <rkg@cpsc.ucalgary.ca> wrote:
1/1 ?? R.
On Tue, 19 Feb 2013, Allan Wechsler wrote:
Let A(n) be the number of ways of expressing 4/n as the sum of three
integer reciprocals, where the mere permutation of a sum is regarded as not making a difference.
Plainly 4/1 = 4 cannot be expressed as the sum of three reciprocals, so A(1) = 0.
4/2 = 2 = 1/1 + 1/2 + 1/2, and there are no other solutions, so A(2) = 1.
4/3 = 1 + 1/4 + 1/12 = 1+ 1/6 + 1/6 = 1/2 + 1/2 + 1/3; I am pretty sure that A(3) = 3.
The Erdős–Straus conjecture is that A(n) > 0 for all n > 1.
Of course I wanted to know if A was in OEIS. I calculated a few more terms, and what I had was 0, 1, 3, 3, 2, 8 ... I was pretty confident in my enumeration, so I calculated enough entries, and discovered to my surprise that the sequence was missing.
Then I searched for "Straus", and quickly found A192787, which claims to be my A. The trouble is, A192787(4) = 4, and I say A(4) = 3.
Bear with me while I list my solutions, and then somebody tell me what I missed.
4/4 = 1, so the problem is to partition 1 into three reciprocals. I have the following solutions:
1/2 + 1/3 + 1/6 1/2 + 1/4 + 1/4 1/3 + 1/3 + 1/3
A192787 seems to be claiming that I missed one. Charles R. Greathouse IV was the sequence author, and I think he's a funster, so, Charles, if you're listening, can you tell me the missing dissection? ______________________________**_________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/**cgi-bin/mailman/listinfo/math-**fun<http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>
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Allan, I think you're right -- my script was double-counting. I'll correct the sequence shortly. Charles Greathouse Analyst/Programmer Case Western Reserve University On Tue, Feb 19, 2013 at 4:36 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Let A(n) be the number of ways of expressing 4/n as the sum of three integer reciprocals, where the mere permutation of a sum is regarded as not making a difference.
Plainly 4/1 = 4 cannot be expressed as the sum of three reciprocals, so A(1) = 0.
4/2 = 2 = 1/1 + 1/2 + 1/2, and there are no other solutions, so A(2) = 1.
4/3 = 1 + 1/4 + 1/12 = 1+ 1/6 + 1/6 = 1/2 + 1/2 + 1/3; I am pretty sure that A(3) = 3.
The Erdős–Straus conjecture is that A(n) > 0 for all n > 1.
Of course I wanted to know if A was in OEIS. I calculated a few more terms, and what I had was 0, 1, 3, 3, 2, 8 ... I was pretty confident in my enumeration, so I calculated enough entries, and discovered to my surprise that the sequence was missing.
Then I searched for "Straus", and quickly found A192787, which claims to be my A. The trouble is, A192787(4) = 4, and I say A(4) = 3.
Bear with me while I list my solutions, and then somebody tell me what I missed.
4/4 = 1, so the problem is to partition 1 into three reciprocals. I have the following solutions:
1/2 + 1/3 + 1/6 1/2 + 1/4 + 1/4 1/3 + 1/3 + 1/3
A192787 seems to be claiming that I missed one. Charles R. Greathouse IV was the sequence author, and I think he's a funster, so, Charles, if you're listening, can you tell me the missing dissection? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 2/19/2013 1:36 PM, Allan Wechsler wrote:
Let A(n) be the number of ways of expressing 4/n as the sum of three integer reciprocals, where the mere permutation of a sum is regarded as not making a difference.
Plainly 4/1 = 4 cannot be expressed as the sum of three reciprocals, so A(1) = 0.
4/2 = 2 = 1/1 + 1/2 + 1/2, and there are no other solutions, so A(2) = 1.
4/3 = 1 + 1/4 + 1/12 = 1+ 1/6 + 1/6 = 1/2 + 1/2 + 1/3; I am pretty sure that A(3) = 3.
The Erdős–Straus conjecture is that A(n) > 0 for all n > 1.
Of course I wanted to know if A was in OEIS. I calculated a few more terms, and what I had was 0, 1, 3, 3, 2, 8 ... I was pretty confident in my enumeration, so I calculated enough entries, and discovered to my surprise that the sequence was missing.
Then I searched for "Straus", and quickly found A192787, which claims to be my A. The trouble is, A192787(4) = 4, and I say A(4) = 3.
Bear with me while I list my solutions, and then somebody tell me what I missed.
4/4 = 1, so the problem is to partition 1 into three reciprocals. I have the following solutions:
1/2 + 1/3 + 1/6 1/2 + 1/4 + 1/4 1/3 + 1/3 + 1/3
1/1 + 1/2 + 1/2 Brent Meeker
A192787 seems to be claiming that I missed one. Charles R. Greathouse IV was the sequence author, and I think he's a funster, so, Charles, if you're listening, can you tell me the missing dissection? _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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1/1 + 1/-1 + 1/1, perhaps? They say integer, not natural number. On Tue, Feb 19, 2013 at 3:12 PM, meekerdb <meekerdb@verizon.net> wrote:
On 2/19/2013 1:36 PM, Allan Wechsler wrote:
Let A(n) be the number of ways of expressing 4/n as the sum of three integer reciprocals, where the mere permutation of a sum is regarded as not making a difference.
Plainly 4/1 = 4 cannot be expressed as the sum of three reciprocals, so A(1) = 0.
4/2 = 2 = 1/1 + 1/2 + 1/2, and there are no other solutions, so A(2) = 1.
4/3 = 1 + 1/4 + 1/12 = 1+ 1/6 + 1/6 = 1/2 + 1/2 + 1/3; I am pretty sure that A(3) = 3.
The Erdős–Straus conjecture is that A(n) > 0 for all n > 1.
Of course I wanted to know if A was in OEIS. I calculated a few more terms, and what I had was 0, 1, 3, 3, 2, 8 ... I was pretty confident in my enumeration, so I calculated enough entries, and discovered to my surprise that the sequence was missing.
Then I searched for "Straus", and quickly found A192787, which claims to be my A. The trouble is, A192787(4) = 4, and I say A(4) = 3.
Bear with me while I list my solutions, and then somebody tell me what I missed.
4/4 = 1, so the problem is to partition 1 into three reciprocals. I have the following solutions:
1/2 + 1/3 + 1/6 1/2 + 1/4 + 1/4 1/3 + 1/3 + 1/3
1/1 + 1/2 + 1/2
Brent Meeker
A192787 seems to be claiming that I missed one. Charles R. Greathouse IV was the sequence author, and I think he's a funster, so, Charles, if you're listening, can you tell me the missing dissection? ______________________________**_________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/**cgi-bin/mailman/listinfo/math-**fun<http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun>
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That's a good one, I hadn't thought of that. They're discussing A192787 over in the OEIS list. The problem was that the computer program was counting "1/2 + 1/3 + 1/6" and "1/3 + 1/2 + 1/6" as two different solutions. It was just a bug in the program. Regarding 1/1+1/-1+1/1, the definition of A192787: http://oeis.org/A192787 clearly says "Number of distinct solutions of 4/n = 1/a + 1/b + 1/c in positive integers.". So that's covered. On Tue, Feb 19, 2013 at 6:54 PM, Tom Rokicki <rokicki@gmail.com> wrote:
1/1 + 1/-1 + 1/1, perhaps? They say integer, not natural number.
On Tue, Feb 19, 2013 at 3:12 PM, meekerdb <meekerdb@verizon.net> wrote:
On 2/19/2013 1:36 PM, Allan Wechsler wrote:
Let A(n) be the number of ways of expressing 4/n as the sum of three integer reciprocals, where the mere permutation of a sum is regarded as not making a difference. [...]
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participants (6)
-
Allan Wechsler -
Charles Greathouse -
meekerdb -
Richard Guy -
Robert Munafo -
Tom Rokicki