[math-fun] n and n^3 have all digits odd
Suppose the positive integers n and n^3 both have all digits odd. Such numbers include 1, 11, 15, 33, 39, 71, 91, 173, 175, 179, 335, 3337, 5597, 7353. Is this list complete? This sequence is http://oeis.org/A085597 in OEIS. But there is no key word indicating that it is a finite sequence. Is this known to be a finite sequence?
I don't think it has been proven to be finite, though it surely is -- there are 5^n n-digit numbers with all digits odd, but the 'chance' that a 3n-digit number has all digits odd is 2^(-3n). (That the last few digits are OK doesn't matter, asymptotically.) The sum of (5/8)^n is of course finite. Charles Greathouse Analyst/Programmer Case Western Reserve University On Wed, Apr 18, 2012 at 11:09 AM, James Buddenhagen <jbuddenh@gmail.com> wrote:
Suppose the positive integers n and n^3 both have all digits odd. Such numbers include 1, 11, 15, 33, 39, 71, 91, 173, 175, 179, 335, 3337, 5597, 7353. Is this list complete? This sequence is http://oeis.org/A085597 in OEIS. But there is no key word indicating that it is a finite sequence. Is this known to be a finite sequence?
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James Buddenhagen:
Is this known to be a finite sequence?
I think the best we can ever really do with this sort of problem is conjecture it to be true and provide a large number up to which it is known to be true. There was another such question brought up recently on SeqFan asking if 34155 is the only odd number equal to the sum of its proper divisors greater than (or equal to) its square root.
I think there's hope in the case of the 34155 conjecture. There are powerful techniques available using sigma_{-1} which might be brought to bear. I feel like the 2e11 lower bound could be used to find a narrow range (2, 2.00...] in which sigma_{-1}(n) must reside, and then... well, I don't know or else I'd try to prove it myself. But in any case the odd digit problem looks much less amenable to solution. Charles Greathouse Analyst/Programmer Case Western Reserve University On Wed, Apr 18, 2012 at 12:17 PM, Hans Havermann <gladhobo@teksavvy.com> wrote:
James Buddenhagen:
Is this known to be a finite sequence?
I think the best we can ever really do with this sort of problem is conjecture it to be true and provide a large number up to which it is known to be true. There was another such question brought up recently on SeqFan asking if 34155 is the only odd number equal to the sum of its proper divisors greater than (or equal to) its square root.
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I'm doing some calculations pertaining to this. For all positive integers k define the set S_k := { 1 <= n < 10^k | OddDigits(n) & OddDigits((n^3) mod 10^k) } It looks like (log # S_k)/k --> 1 (where log is the natural log). Notice that 10^k-1 is always in S_k. The sequence for #S_k starts: [5,25,62,151,381,833,2163,5291,13317,33519,85179,213083,539212,1344272,3358571] and the sequence (log # S_k)/k starts: [1.6094379124341003,1.6094379124341003,1.3757114616816972,1.254319959203731,1.1885598750253403, 1.1208389403611405,1.097035917993294,1.0717203178630164,1.0551996326113493,1.0419867721215448, 1.0320463821421144,1.0224530866792414,1.0152203148865746,1.0079545114672392,1.0018017429237103] On Wed, Apr 18, 2012 at 11:09 AM, James Buddenhagen <jbuddenh@gmail.com>wrote:
Suppose the positive integers n and n^3 both have all digits odd. Such numbers include 1, 11, 15, 33, 39, 71, 91, 173, 175, 179, 335, 3337, 5597, 7353. Is this list complete? This sequence is http://oeis.org/A085597 in OEIS. But there is no key word indicating that it is a finite sequence. Is this known to be a finite sequence?
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That matches my modular calculations from earlier, though I didn't go quite as far. If accurate it suggests asymptotically (e/4)^n n-digit examples -- still finitely many, though not quite as few as the 5/8 the simpler heuristic suggested (since e/4 < 5/8). Charles Greathouse Analyst/Programmer Case Western Reserve University On Wed, Apr 18, 2012 at 10:57 PM, Victor Miller <victorsmiller@gmail.com> wrote:
I'm doing some calculations pertaining to this. For all positive integers k define the set
S_k := { 1 <= n < 10^k | OddDigits(n) & OddDigits((n^3) mod 10^k) }
It looks like (log # S_k)/k --> 1 (where log is the natural log). Notice that 10^k-1 is always in S_k.
The sequence for #S_k starts:
[5,25,62,151,381,833,2163,5291,13317,33519,85179,213083,539212,1344272,3358571]
and the sequence (log # S_k)/k starts:
[1.6094379124341003,1.6094379124341003,1.3757114616816972,1.254319959203731,1.1885598750253403, 1.1208389403611405,1.097035917993294,1.0717203178630164,1.0551996326113493,1.0419867721215448, 1.0320463821421144,1.0224530866792414,1.0152203148865746,1.0079545114672392,1.0018017429237103]
On Wed, Apr 18, 2012 at 11:09 AM, James Buddenhagen <jbuddenh@gmail.com>wrote:
Suppose the positive integers n and n^3 both have all digits odd. Such numbers include 1, 11, 15, 33, 39, 71, 91, 173, 175, 179, 335, 3337, 5597, 7353. Is this list complete? This sequence is http://oeis.org/A085597 in OEIS. But there is no key word indicating that it is a finite sequence. Is this known to be a finite sequence?
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On Thursday 19 April 2012 03:57:09 Victor Miller wrote:
I'm doing some calculations pertaining to this. For all positive integers k define the set
S_k := { 1 <= n < 10^k | OddDigits(n) & OddDigits((n^3) mod 10^k) }
It looks like (log # S_k)/k --> 1 (where log is the natural log). Notice that 10^k-1 is always in S_k.
What happens in other bases? (6 and 14 seem like the closest counterparts of 10.) -- g
Looks about the same in those two (though of course smaller bases gain digits faster and thus have fewer examples). 6 has 1, 7, 695, and no others up to 10^7. 14 has 1, 3, 15, 21, 183, 219, 2261, 3067, 3439, 68991, 146579, 222875, 431979, 3213973, and no others up to 10^7. In modular terms, 14 has 7, 16, 39, 95, 222, 671, 824 acceptable residues mod 14^n. Charles Greathouse Analyst/Programmer Case Western Reserve University On Thu, Apr 19, 2012 at 4:11 AM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Thursday 19 April 2012 03:57:09 Victor Miller wrote:
I'm doing some calculations pertaining to this. For all positive integers k define the set
S_k := { 1 <= n < 10^k | OddDigits(n) & OddDigits((n^3) mod 10^k) }
It looks like (log # S_k)/k --> 1 (where log is the natural log). Notice that 10^k-1 is always in S_k.
What happens in other bases? (6 and 14 seem like the closest counterparts of 10.)
-- g _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Charles, for the mod 14^n I get the sequence [7,37,147,502,1788,6287,22468] For mod 6^n I get [3,9,18,33,60,93,189,252,402,636] On Thu, Apr 19, 2012 at 9:09 AM, Charles Greathouse < charles.greathouse@case.edu> wrote:
Looks about the same in those two (though of course smaller bases gain digits faster and thus have fewer examples). 6 has 1, 7, 695, and no others up to 10^7. 14 has 1, 3, 15, 21, 183, 219, 2261, 3067, 3439, 68991, 146579, 222875, 431979, 3213973, and no others up to 10^7.
In modular terms, 14 has 7, 16, 39, 95, 222, 671, 824 acceptable residues mod 14^n.
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Thu, Apr 19, 2012 at 4:11 AM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Thursday 19 April 2012 03:57:09 Victor Miller wrote:
I'm doing some calculations pertaining to this. For all positive integers k define the set
S_k := { 1 <= n < 10^k | OddDigits(n) & OddDigits((n^3) mod 10^k) }
It looks like (log # S_k)/k --> 1 (where log is the natural log). Notice that 10^k-1 is always in S_k.
What happens in other bases? (6 and 14 seem like the closest counterparts of 10.)
-- g _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Certainly we can prove this theorem for base 2, can't we? How many cubes are there of the form 2^n-1? Something is telling me, "one," but I can't prove it. On Thu, Apr 19, 2012 at 9:57 AM, Victor Miller <victorsmiller@gmail.com>wrote:
Charles, for the mod 14^n I get the sequence [7,37,147,502,1788,6287,22468] For mod 6^n I get [3,9,18,33,60,93,189,252,402,636]
On Thu, Apr 19, 2012 at 9:09 AM, Charles Greathouse < charles.greathouse@case.edu> wrote:
Looks about the same in those two (though of course smaller bases gain digits faster and thus have fewer examples). 6 has 1, 7, 695, and no others up to 10^7. 14 has 1, 3, 15, 21, 183, 219, 2261, 3067, 3439, 68991, 146579, 222875, 431979, 3213973, and no others up to 10^7.
In modular terms, 14 has 7, 16, 39, 95, 222, 671, 824 acceptable residues mod 14^n.
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Thu, Apr 19, 2012 at 4:11 AM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Thursday 19 April 2012 03:57:09 Victor Miller wrote:
I'm doing some calculations pertaining to this. For all positive integers k define the set
S_k := { 1 <= n < 10^k | OddDigits(n) & OddDigits((n^3) mod 10^k) }
It looks like (log # S_k)/k --> 1 (where log is the natural log). Notice that 10^k-1 is always in S_k.
What happens in other bases? (6 and 14 seem like the closest counterparts of 10.)
-- g _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Hint: Mod 2^N, cubing permutes the odd residues. --Rich --- Quoting Allan Wechsler <acwacw@gmail.com>:
Certainly we can prove this theorem for base 2, can't we? How many cubes are there of the form 2^n-1? Something is telling me, "one," but I can't prove it.
On Thu, Apr 19, 2012 at 9:57 AM, Victor Miller <victorsmiller@gmail.com>wrote:
Charles, for the mod 14^n I get the sequence [7,37,147,502,1788,6287,22468] For mod 6^n I get [3,9,18,33,60,93,189,252,402,636]
On Thu, Apr 19, 2012 at 9:09 AM, Charles Greathouse < charles.greathouse@case.edu> wrote:
Looks about the same in those two (though of course smaller bases gain digits faster and thus have fewer examples). 6 has 1, 7, 695, and no others up to 10^7. 14 has 1, 3, 15, 21, 183, 219, 2261, 3067, 3439, 68991, 146579, 222875, 431979, 3213973, and no others up to 10^7.
In modular terms, 14 has 7, 16, 39, 95, 222, 671, 824 acceptable residues mod 14^n.
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Thu, Apr 19, 2012 at 4:11 AM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Thursday 19 April 2012 03:57:09 Victor Miller wrote:
I'm doing some calculations pertaining to this. For all positive integers k define the set
S_k := { 1 <= n < 10^k | OddDigits(n) & OddDigits((n^3) mod 10^k) }
It looks like (log # S_k)/k --> 1 (where log is the natural log). Notice that 10^k-1 is always in S_k.
What happens in other bases? (6 and 14 seem like the closest counterparts of 10.)
-- g _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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That's Catalan's conjecture, right? So the case b = 2 is solved thanks to Mihăilescu. I don't know if that's using a sledgehammer to crack a nut or if the problem is really that hard. Charles Greathouse Analyst/Programmer Case Western Reserve University On Thu, Apr 19, 2012 at 4:12 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Certainly we can prove this theorem for base 2, can't we? How many cubes are there of the form 2^n-1? Something is telling me, "one," but I can't prove it.
On Thu, Apr 19, 2012 at 9:57 AM, Victor Miller <victorsmiller@gmail.com>wrote:
Charles, for the mod 14^n I get the sequence [7,37,147,502,1788,6287,22468] For mod 6^n I get [3,9,18,33,60,93,189,252,402,636]
On Thu, Apr 19, 2012 at 9:09 AM, Charles Greathouse < charles.greathouse@case.edu> wrote:
Looks about the same in those two (though of course smaller bases gain digits faster and thus have fewer examples). 6 has 1, 7, 695, and no others up to 10^7. 14 has 1, 3, 15, 21, 183, 219, 2261, 3067, 3439, 68991, 146579, 222875, 431979, 3213973, and no others up to 10^7.
In modular terms, 14 has 7, 16, 39, 95, 222, 671, 824 acceptable residues mod 14^n.
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Thu, Apr 19, 2012 at 4:11 AM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Thursday 19 April 2012 03:57:09 Victor Miller wrote:
I'm doing some calculations pertaining to this. For all positive integers k define the set
S_k := { 1 <= n < 10^k | OddDigits(n) & OddDigits((n^3) mod 10^k) }
It looks like (log # S_k)/k --> 1 (where log is the natural log). Notice that 10^k-1 is always in S_k.
What happens in other bases? (6 and 14 seem like the closest counterparts of 10.)
-- g _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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On 4/19/2012 4:12 PM, Allan Wechsler wrote:
Certainly we can prove this theorem for base 2, can't we? How many cubes are there of the form 2^n-1? Something is telling me, "one," but I can't prove it.
If 2^n - 1 = m^3, then 2^n = m^3 + 1 = (m + 1)(m^2 - m + 1). The second factor is odd, so it's only a power of 2 if m^2 - m = 0. Thus 0 and 1 are the only cubes of the form 2^n - 1. (Of course Alan wanted n > 0, so only one solution is relevant to the original question.) By the way, the base-10 problem is easy if you replace n^3 with n^2. -- Fred W. Helenius fredh@ix.netcom.com
Well, OK then! Let us move on to ternary and see what we can establish! On Thu, Apr 19, 2012 at 4:43 PM, Fred W. Helenius <fredh@ix.netcom.com>wrote:
On 4/19/2012 4:12 PM, Allan Wechsler wrote:
Certainly we can prove this theorem for base 2, can't we? How many cubes are there of the form 2^n-1? Something is telling me, "one," but I can't prove it.
If 2^n - 1 = m^3, then 2^n = m^3 + 1 = (m + 1)(m^2 - m + 1). The second factor is odd, so it's only a power of 2 if m^2 - m = 0. Thus 0 and 1 are the only cubes of the form 2^n - 1. (Of course Alan wanted n > 0, so only one solution is relevant to the original question.)
By the way, the base-10 problem is easy if you replace n^3 with n^2.
-- Fred W. Helenius fredh@ix.netcom.com
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Wait a sec, I'm being stupid. Ternary odd-digit numbers also have only 1's, so they are of the form (3^n-1)/2. If (3^n-1)/2 is a cube, then 3^n = 2m^3 + 1. OK, I'm stuck again. Fred? On Thu, Apr 19, 2012 at 4:51 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Well, OK then! Let us move on to ternary and see what we can establish!
On Thu, Apr 19, 2012 at 4:43 PM, Fred W. Helenius <fredh@ix.netcom.com>wrote:
On 4/19/2012 4:12 PM, Allan Wechsler wrote:
Certainly we can prove this theorem for base 2, can't we? How many cubes are there of the form 2^n-1? Something is telling me, "one," but I can't prove it.
If 2^n - 1 = m^3, then 2^n = m^3 + 1 = (m + 1)(m^2 - m + 1). The second factor is odd, so it's only a power of 2 if m^2 - m = 0. Thus 0 and 1 are the only cubes of the form 2^n - 1. (Of course Alan wanted n > 0, so only one solution is relevant to the original question.)
By the way, the base-10 problem is easy if you replace n^3 with n^2.
-- Fred W. Helenius fredh@ix.netcom.com
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Cubes (mod 9) are always 0, 1, or 8. The low order ternary digits are 00, 01, or 22. --Rich ---- Quoting Allan Wechsler <acwacw@gmail.com>:
Wait a sec, I'm being stupid. Ternary odd-digit numbers also have only 1's, so they are of the form (3^n-1)/2. If (3^n-1)/2 is a cube, then 3^n = 2m^3 + 1. OK, I'm stuck again. Fred?
On Thu, Apr 19, 2012 at 4:51 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Well, OK then! Let us move on to ternary and see what we can establish!
On Thu, Apr 19, 2012 at 4:43 PM, Fred W. Helenius <fredh@ix.netcom.com>wrote:
On 4/19/2012 4:12 PM, Allan Wechsler wrote:
Certainly we can prove this theorem for base 2, can't we? How many cubes are there of the form 2^n-1? Something is telling me, "one," but I can't prove it.
If 2^n - 1 = m^3, then 2^n = m^3 + 1 = (m + 1)(m^2 - m + 1). The second factor is odd, so it's only a power of 2 if m^2 - m = 0. Thus 0 and 1 are the only cubes of the form 2^n - 1. (Of course Alan wanted n > 0, so only one solution is relevant to the original question.)
By the way, the base-10 problem is easy if you replace n^3 with n^2.
-- Fred W. Helenius fredh@ix.netcom.com
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On 4/19/2012 4:54 PM, Allan Wechsler wrote:
Wait a sec, I'm being stupid. Ternary odd-digit numbers also have only 1's, so they are of the form (3^n-1)/2. If (3^n-1)/2 is a cube, then 3^n = 2m^3 + 1. OK, I'm stuck again. Fred?
Cubes are -1, 0 or 1 mod 9, so the equation has no solutions for n >= 2. -- Fred W. Helenius fredh@ix.netcom.com
I've chased this out to n<=10^25 and found no more examples. The "chance" that we'll get a hit above this, assuming random distribution of digits, is pretty small. On Wed, Apr 18, 2012 at 8:09 AM, James Buddenhagen <jbuddenh@gmail.com> wrote:
Suppose the positive integers n and n^3 both have all digits odd. Such numbers include 1, 11, 15, 33, 39, 71, 91, 173, 175, 179, 335, 3337, 5597, 7353. Is this list complete? This sequence is http://oeis.org/A085597 in OEIS. But there is no key word indicating that it is a finite sequence. Is this known to be a finite sequence?
participants (9)
-
Allan Wechsler -
Charles Greathouse -
Fred W. Helenius -
Gareth McCaughan -
Hans Havermann -
James Buddenhagen -
rcs＠xmission.com -
Tom Rokicki -
Victor Miller