Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
Eric, Very nice! I just added A209674, which is the number of steps to reach 5. Neil On Sun, Mar 11, 2012 at 2:23 PM, Eric Angelini <Eric.Angelini@kntv.be>wrote:
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
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Thanks Neil -- I've omitted to say that integers ending in 00 (like 100, 1000...) are obviously not hitting 5... ________________________________________ De : math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] de la part de Neil Sloane [njasloane@gmail.com] Date d'envoi : dimanche 11 mars 2012 19:55 À : math-fun Objet : Re: [math-fun] Hitting 5 Eric, Very nice! I just added A209674, which is the number of steps to reach 5. Neil On Sun, Mar 11, 2012 at 2:23 PM, Eric Angelini <Eric.Angelini@kntv.be>wrote:
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
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Yes, so A209674 either needs to be defined on 1..99, 101..199, etc. or else have some special value for multiples of 100. Either way, the sequence is periodic. (The exact period depends on the convention chosen above, as well as how numbers 5 mod 100 are treated.) I do get different values for many terms, though, to wit: 4, 9, 9, 5, 0, 4, 3, 10, 11, 5, 3, 2, 6, 1, 5, 8, 7, 9, 6, 10, 7, 8, 1, 7, 4, 3, 10, 6, 4, 10, 2, 1, 8, 5, 8, 7, 6, 4, 3, 6, 1, 4, 7, 4, 3, 6, 4, 3, 7, 1, 9, 11, 5, 8, 6, 4, 3, 7, 2, 5, 8, 7, 4, 6, 4, 3, 7, 2, 6, 4, 10, 5, 6, 4, 3, 7, 2, 6, 9, 11, 7, 6, 4, 3, 7, 2, 6, 9, 8, 12, 6, 4, 3, 7, 2, 6, 9, 8, 10. using (PARI) a(n)=my(m=0,t,k);while(n!=5,t=if(n>9,n%100\10+n%10,n+m%10);m=n;n=t;k++);k Charles Greathouse Analyst/Programmer Case Western Reserve University On Sun, Mar 11, 2012 at 3:04 PM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Thanks Neil -- I've omitted to say that integers ending in 00 (like 100, 1000...) are obviously not hitting 5...
________________________________________ De : math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] de la part de Neil Sloane [njasloane@gmail.com] Date d'envoi : dimanche 11 mars 2012 19:55 À : math-fun Objet : Re: [math-fun] Hitting 5
Eric, Very nice! I just added A209674, which is the number of steps to reach 5. Neil
On Sun, Mar 11, 2012 at 2:23 PM, Eric Angelini <Eric.Angelini@kntv.be>wrote:
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
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Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
This brings up the point that OEIS should allow a value of infinity for elements of a sequence. Another possible value is uncomputable (there may be some sequences like that), or "not yet computed". The latter would especially important for sequences which have individual terms that are arduous to compute. Victor On Sun, Mar 11, 2012 at 3:17 PM, Charles Greathouse <charles.greathouse@case.edu> wrote:
Yes, so A209674 either needs to be defined on 1..99, 101..199, etc. or else have some special value for multiples of 100.
Either way, the sequence is periodic. (The exact period depends on the convention chosen above, as well as how numbers 5 mod 100 are treated.)
I do get different values for many terms, though, to wit: 4, 9, 9, 5, 0, 4, 3, 10, 11, 5, 3, 2, 6, 1, 5, 8, 7, 9, 6, 10, 7, 8, 1, 7, 4, 3, 10, 6, 4, 10, 2, 1, 8, 5, 8, 7, 6, 4, 3, 6, 1, 4, 7, 4, 3, 6, 4, 3, 7, 1, 9, 11, 5, 8, 6, 4, 3, 7, 2, 5, 8, 7, 4, 6, 4, 3, 7, 2, 6, 4, 10, 5, 6, 4, 3, 7, 2, 6, 9, 11, 7, 6, 4, 3, 7, 2, 6, 9, 8, 12, 6, 4, 3, 7, 2, 6, 9, 8, 10.
using
(PARI) a(n)=my(m=0,t,k);while(n!=5,t=if(n>9,n%100\10+n%10,n+m%10);m=n;n=t;k++);k
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Sun, Mar 11, 2012 at 3:04 PM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Thanks Neil -- I've omitted to say that integers ending in 00 (like 100, 1000...) are obviously not hitting 5...
________________________________________ De : math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] de la part de Neil Sloane [njasloane@gmail.com] Date d'envoi : dimanche 11 mars 2012 19:55 À : math-fun Objet : Re: [math-fun] Hitting 5
Eric, Very nice! I just added A209674, which is the number of steps to reach 5. Neil
On Sun, Mar 11, 2012 at 2:23 PM, Eric Angelini <Eric.Angelini@kntv.be>wrote:
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
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-- Dear Friends, I will soon be retiring from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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The convention in the OEIS is to use -1 in cases when a term would be infinite or undefined. Neil On Sun, Mar 11, 2012 at 3:52 PM, Victor Miller <victorsmiller@gmail.com>wrote:
This brings up the point that OEIS should allow a value of infinity for elements of a sequence. Another possible value is uncomputable (there may be some sequences like that), or "not yet computed". The latter would especially important for sequences which have individual terms that are arduous to compute.
Victor
On Sun, Mar 11, 2012 at 3:17 PM, Charles Greathouse <charles.greathouse@case.edu> wrote:
Yes, so A209674 either needs to be defined on 1..99, 101..199, etc. or else have some special value for multiples of 100.
Either way, the sequence is periodic. (The exact period depends on the convention chosen above, as well as how numbers 5 mod 100 are treated.)
I do get different values for many terms, though, to wit: 4, 9, 9, 5, 0, 4, 3, 10, 11, 5, 3, 2, 6, 1, 5, 8, 7, 9, 6, 10, 7, 8, 1, 7, 4, 3, 10, 6, 4, 10, 2, 1, 8, 5, 8, 7, 6, 4, 3, 6, 1, 4, 7, 4, 3, 6, 4, 3, 7, 1, 9, 11, 5, 8, 6, 4, 3, 7, 2, 5, 8, 7, 4, 6, 4, 3, 7, 2, 6, 4, 10, 5, 6, 4, 3, 7, 2, 6, 9, 11, 7, 6, 4, 3, 7, 2, 6, 9, 8, 12, 6, 4, 3, 7, 2, 6, 9, 8, 10.
using
(PARI) a(n)=my(m=0,t,k);while(n!=5,t=if(n>9,n%100\10+n%10,n+m%10);m=n;n=t;k++);k
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Sun, Mar 11, 2012 at 3:04 PM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Thanks Neil -- I've omitted to say that integers ending in 00 (like
100, 1000...) are obviously not hitting 5...
________________________________________ De : math-fun-bounces@mailman.xmission.com [
math-fun-bounces@mailman.xmission.com] de la part de Neil Sloane [ njasloane@gmail.com]
Date d'envoi : dimanche 11 mars 2012 19:55 À : math-fun Objet : Re: [math-fun] Hitting 5
Eric, Very nice! I just added A209674, which is the number of steps to reach 5. Neil
On Sun, Mar 11, 2012 at 2:23 PM, Eric Angelini <Eric.Angelini@kntv.be wrote:
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Dear Friends, I will soon be retiring from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Dear Friends, I will soon be retiring from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
From: Neil Sloane <njasloane@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Sunday, March 11, 2012 3:50 PM Subject: Re: [math-fun] RE : Hitting 5
The convention in the OEIS is to use -1 in cases when a term would be infinite or undefined. Neil
But can't -1 be a legitimate element of an integer sequence? -- Gene
Yes, of course if the sequence already contains negative terms then we do something else. It depends on the particular sequence. Neil On Sun, Mar 11, 2012 at 7:10 PM, Eugene Salamin <gene_salamin@yahoo.com>wrote:
From: Neil Sloane <njasloane@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Sunday, March 11, 2012 3:50 PM Subject: Re: [math-fun] RE : Hitting 5
The convention in the OEIS is to use -1 in cases when a term would be infinite or undefined. Neil
But can't -1 be a legitimate element of an integer sequence?
-- Gene _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Dear Friends, I will soon be retiring from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
="Eugene Salamin" <gene_salamin@yahoo.com> But can't -1 be a legitimate element of an integer sequence?
Yes, but a "canonical" OEIS sequence consists only of non-negative integers. Thus -1 is a convenient conventional marker for an exceptional value. When used the meaning is always documented in the entry, typically by adding text of the form "...or -1 if..." to the description. When a sequence truly is best defined with signed integers they are given as an addendum to the main canonical sequence consisting of their absolute values, and the entry is tagged with a special keyword. Supporting a variety of values might seem congenial, but it would complicate lookup and open Pandora's Box. See for example A193616, which would include "1/2", "?", "infinity" and "coffee cup"!
In fact the -1 standard is not adhered. Sequences whose terms would normally be positive often use 0 if the particular term is undefined. To wit: https://oeis.org/search?q=or+0+if+no+such&language=english&go=Search returns 1368 results, whereas: https://oeis.org/search?q=or+-1+if+no+such&sort=&language=english&go=Search returns only 720 results. On Sun, Mar 11, 2012 at 19:10, Eugene Salamin <gene_salamin@yahoo.com>wrote:
From: Neil Sloane
The convention in the OEIS is to use -1 in cases when a term would be infinite or undefined. Neil
But can't -1 be a legitimate element of an integer sequence?
-- Gene
-- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com
Ok, I see. But shouldn't we distinguish between a term which is *proved* to be infinite, and one that just hasn't been computed? Victor On Sun, Mar 11, 2012 at 6:50 PM, Neil Sloane <njasloane@gmail.com> wrote:
The convention in the OEIS is to use -1 in cases when a term would be infinite or undefined. Neil
On Sun, Mar 11, 2012 at 3:52 PM, Victor Miller <victorsmiller@gmail.com>wrote:
This brings up the point that OEIS should allow a value of infinity for elements of a sequence. Another possible value is uncomputable (there may be some sequences like that), or "not yet computed". The latter would especially important for sequences which have individual terms that are arduous to compute.
Victor
On Sun, Mar 11, 2012 at 3:17 PM, Charles Greathouse <charles.greathouse@case.edu> wrote:
Yes, so A209674 either needs to be defined on 1..99, 101..199, etc. or else have some special value for multiples of 100.
Either way, the sequence is periodic. (The exact period depends on the convention chosen above, as well as how numbers 5 mod 100 are treated.)
I do get different values for many terms, though, to wit: 4, 9, 9, 5, 0, 4, 3, 10, 11, 5, 3, 2, 6, 1, 5, 8, 7, 9, 6, 10, 7, 8, 1, 7, 4, 3, 10, 6, 4, 10, 2, 1, 8, 5, 8, 7, 6, 4, 3, 6, 1, 4, 7, 4, 3, 6, 4, 3, 7, 1, 9, 11, 5, 8, 6, 4, 3, 7, 2, 5, 8, 7, 4, 6, 4, 3, 7, 2, 6, 4, 10, 5, 6, 4, 3, 7, 2, 6, 9, 11, 7, 6, 4, 3, 7, 2, 6, 9, 8, 12, 6, 4, 3, 7, 2, 6, 9, 8, 10.
using
(PARI) a(n)=my(m=0,t,k);while(n!=5,t=if(n>9,n%100\10+n%10,n+m%10);m=n;n=t;k++);k
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Sun, Mar 11, 2012 at 3:04 PM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Thanks Neil -- I've omitted to say that integers ending in 00 (like
100, 1000...) are obviously not hitting 5...
________________________________________ De : math-fun-bounces@mailman.xmission.com [
math-fun-bounces@mailman.xmission.com] de la part de Neil Sloane [ njasloane@gmail.com]
Date d'envoi : dimanche 11 mars 2012 19:55 À : math-fun Objet : Re: [math-fun] Hitting 5
Eric, Very nice! I just added A209674, which is the number of steps to reach 5. Neil
On Sun, Mar 11, 2012 at 2:23 PM, Eric Angelini <Eric.Angelini@kntv.be wrote:
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Dear Friends, I will soon be retiring from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Dear Friends, I will soon be retiring from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Sun, Mar 11, 2012 at 2:55 PM, Neil Sloane <njasloane@gmail.com> wrote:
Eric, Very nice! I just added A209674, which is the number of steps to reach 5. Neil
Every number also reaches 4, and 1, as well as 5. Any reason the time to reach 5 is the thing that seems interesting, rather than the time to reach 1 or 4? I think everything ultimately ends up in one of two cycles: 358134711235 or 459145. Isn't the time to reach a cycle really the interesting thing, rather than the time to reach the 5 in the cycle? Andy
On Sun, Mar 11, 2012 at 2:23 PM, Eric Angelini <Eric.Angelini@kntv.be>wrote:
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Dear Friends, I will soon be retiring from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
14 -> 5 -> 9 -> 14 -> 5 -> 9 -> ... 5 is the only number common to the trajectories of the numbers 1 through 99 Neil On Sun, Mar 11, 2012 at 5:38 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Sun, Mar 11, 2012 at 2:55 PM, Neil Sloane <njasloane@gmail.com> wrote:
Eric, Very nice! I just added A209674, which is the number of steps to reach 5. Neil
Every number also reaches 4, and 1, as well as 5. Any reason the time to reach 5 is the thing that seems interesting, rather than the time to reach 1 or 4?
I think everything ultimately ends up in one of two cycles: 358134711235 or 459145. Isn't the time to reach a cycle really the interesting thing, rather than the time to reach the 5 in the cycle?
Andy
On Sun, Mar 11, 2012 at 2:23 PM, Eric Angelini <Eric.Angelini@kntv.be wrote:
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Dear Friends, I will soon be retiring from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
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-- Dear Friends, I will soon be retiring from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
On Sun, Mar 11, 2012 at 6:33 PM, Neil Sloane <njasloane@gmail.com> wrote:
14 -> 5 -> 9 -> 14 -> 5 -> 9 -> ...
5 is the only number common to the trajectories of the numbers 1 through 99
Ah. I was thinking of it as a sequence of digits, rather than a sequence of numbers, which is why I had 1 and 4 as well as 5 in my list of digits that are always reached. Andy
Neil
On Sun, Mar 11, 2012 at 5:38 PM, Andy Latto <andy.latto@pobox.com> wrote:
On Sun, Mar 11, 2012 at 2:55 PM, Neil Sloane <njasloane@gmail.com> wrote:
Eric, Very nice! I just added A209674, which is the number of steps to reach 5. Neil
Every number also reaches 4, and 1, as well as 5. Any reason the time to reach 5 is the thing that seems interesting, rather than the time to reach 1 or 4?
I think everything ultimately ends up in one of two cycles: 358134711235 or 459145. Isn't the time to reach a cycle really the interesting thing, rather than the time to reach the 5 in the cycle?
Andy
On Sun, Mar 11, 2012 at 2:23 PM, Eric Angelini <Eric.Angelini@kntv.be wrote:
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Dear Friends, I will soon be retiring from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
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-- Dear Friends, I will soon be retiring from AT&T. New coordinates:
Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
Could you sketch a bit more of what's going on here? Suppose S begins with 321. How does this hit 5? Rich ________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] on behalf of Eric Angelini [Eric.Angelini@kntv.be] Sent: Sunday, March 11, 2012 12:23 PM To: math-fun Subject: [EXTERNAL] [math-fun] Hitting 5 Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
321,3,4,7,11,2,3,5 Envoyé d'un aPhone Le 11 mars 2012 à 20:56, "Schroeppel, Richard" <rschroe@sandia.gov> a écrit :
Could you sketch a bit more of what's going on here? Suppose S begins with 321. How does this hit 5?
Rich
________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] on behalf of Eric Angelini [Eric.Angelini@kntv.be] Sent: Sunday, March 11, 2012 12:23 PM To: math-fun Subject: [EXTERNAL] [math-fun] Hitting 5
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
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I got confused too. I looked at A209674, which has the description -------- For each n >= 1, define a sequence S(n) by S(0)=n, S(i) = sum of last two digits of (S(0), S(1), ..., S(n-1)); a(n) = smallest m such that S(m) = 5, or -1 if 5 is never reached. -------- So at first glance it SEEMS that you're to sum the last two digits of ALL the previous S(n). However the example for S4 is given as S4(0) = 4 S4(1) = 4 S4(2) = 8 S4(3) = 12 S4(4) = 3 S4(5) = 5 But if you summed everything then S4(3) would be 16 instead. What I think is actually going on is more like you append digits onto a string, and then sum the last two and append them, and so on 4 44 448 44812 448123 4481235. S1 would then apparently go 1 11 112 1123 11235. Which checks So S321 would go 321 3213 32134 321347 32134711 321347112 3213471123 32134711235. Giving a(321)=7. I would suggest that the description would be a lot clearer if A. a better notation than (S(0), S(1), ..., S(n-1)) was used for string concatenation, perhaps S(0):S(1):...:S(n-1) B. the index variable n wasn't reused to mean two different things, perhaps by writing Sn(k) for the k-th element of the auxiliary digit string sequence used to compute a(n).
="Schroeppel, Richard" <rschroe@sandia.gov>
Could you sketch a bit more of what's going on here? Suppose S begins with 321. How does this hit 5?
Rich
________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] on behalf of Eric Angelini [Eric.Angelini@kntv.be] Sent: Sunday, March 11, 2012 12:23 PM To: math-fun Subject: [EXTERNAL] [math-fun] Hitting 5
Start sequence S with any integer. If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
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Start sequence T with any integer (except an integer ending with 3 zeroes); if a(n) is the sum of the last three digits of T, T will always hit 11 at some point. Best, É. Le 11 mars 2012 à 19:23, "Eric Angelini" <Eric.Angelini@kntv.be<mailto:Eric.Angelini@kntv.be>> a écrit : Start sequence S with any integer (except an integer ending with two zeroes). If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com<mailto:math-fun@mailman.xmission.com> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Eric, I have entered that one as A209686. Neil On Sun, Mar 11, 2012 at 4:35 PM, Eric Angelini <Eric.Angelini@kntv.be>wrote:
Start sequence T with any integer (except an integer ending with 3 zeroes); if a(n) is the sum of the last three digits of T, T will always hit 11 at some point. Best, É.
Le 11 mars 2012 à 19:23, "Eric Angelini" <Eric.Angelini@kntv.be<mailto: Eric.Angelini@kntv.be>> a écrit :
Start sequence S with any integer (except an integer ending with two zeroes). If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
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-- Dear Friends, I will soon be retiring from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
What about 59,14,14,9,14,14,9,.....? On Sun, Mar 11, 2012 at 4:35 PM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Start sequence T with any integer (except an integer ending with 3 zeroes); if a(n) is the sum of the last three digits of T, T will always hit 11 at some point. Best, É.
Le 11 mars 2012 à 19:23, "Eric Angelini" <Eric.Angelini@kntv.be<mailto:Eric.Angelini@kntv.be>> a écrit :
Start sequence S with any integer (except an integer ending with two zeroes). If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
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You are right - and indeed the sequence correctly had a(59) = -1, although the comment was certainly wrong. What IS the nasc for n never to reach 11? Neil On Mon, Mar 12, 2012 at 12:09 AM, Victor Miller <victorsmiller@gmail.com>wrote:
What about 59,14,14,9,14,14,9,.....?
On Sun, Mar 11, 2012 at 4:35 PM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Start sequence T with any integer (except an integer ending with 3 zeroes); if a(n) is the sum of the last three digits of T, T will always hit 11 at some point. Best, É.
Le 11 mars 2012 à 19:23, "Eric Angelini" <Eric.Angelini@kntv.be<mailto:
Eric.Angelini@kntv.be>> a écrit :
Start sequence S with any integer (except an integer ending with two zeroes). If a(n) is the sum of the last two digits of S, S will always hit 5 at some point. Best, É.
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-- Dear Friends, I will soon be retiring from AT&T. New coordinates: Neil J. A. Sloane, President, OEIS Foundation 11 South Adelaide Avenue, Highland Park, NJ 08904, USA Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
participants (9)
-
Andy Latto -
Charles Greathouse -
Eric Angelini -
Eugene Salamin -
Marc LeBrun -
Neil Sloane -
Robert Munafo -
Schroeppel, Richard -
Victor Miller