[seqfan] Re: I need a name for this sequence

Vladimir Shevelev shevelev at bgu.ac.il
Fri Jun 17 16:50:35 CEST 2011


I obtained the previous formulas thinking that it should be gcd(a,b,c)=1. In general , one can obtain by the same way the following formulas,  including case u:=gcd(a,b,c)>1 (here  "c"  I consider as a parameter):
1) a:=c+min((c-1)*u/v, c*v/u), b:=c+max((c-1)*u/v, c*v/u), c>1,v|c-1,u^2|c; 
2) a=c+min(c/(u*s), (c-1)*s*u), b=c+max(c/(u*s), (c-1)*s*u), c>1, u^2|c, s|(c/u^2) .   
It is easy to see that in both cases a=b is impossible.
          
 Example. Let in 1) c=k^2, u=k, v=k-1, k>1. Then  a=k^2+k*(k-1)=2*k^2-k,
 b= k^2+k*(k+1)=2*k^2+k and N=a*b+c=4*k^4, k=2,3,...

Regards,
Vladimir


----- Original Message -----
From: Vladimir Shevelev <shevelev at bgu.ac.il>
Date: Tuesday, June 14, 2011 13:42
Subject: [seqfan] Re: I need a name for this sequence
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>

> The set of all terms of the sequence is the set of different 
> numbers given by the following two formulas with positive u,v:
> 1) a=u*v+u, b=u*v^2+u*v-v, c=u*v, if v>1, and
>  a=2*u-1, b=2*u, c=u, if v=1.
> 2) a=u*v+u+1, b=u*v^2+u*v+v+1, c=u*v+1,
> such that in all cases we have (a+b)*c=a*b+c=N.
> Proof. Consider Diophantine equation (a+b)*c=a*b+c with the 
> condition 0<a<b. Then c>0.
> Put x-y=a, x+y=b. Since 2*x=a+b, then in order to not lose a 
> solution, we suppose that x,y are integers or half-integers. 
> Then we have 2*x*c=x^2-y^2+c, or (x-c)^2-y^2=c*(c-1). This 
> equation splits into two disjoint cases: 1) x-c-y=c/v, x-
> c+y=v*(c-1), where v is a divisor of c;  2) x-c-y=(c-1)/v, 
> x-c+y=v*c, where v is a divisor of c-1. Both cases are 
> considered quite analogously. Therefore, let us consider 1). Put 
> c=v*u. Then a=x-y=v*u+u, b=x+y=v*u+v^2*u-v and  1) follows.
> Note that the case v=1 is only  case when a=2*u>b=2*u-1. 
> Since the equation is (a,b)-symmetric, then  in this case 
> we put a=2*u-1,b=2*u with c=u.
>  <>
> Note that full description of odd terms of the sequence are 
> given by these formulas, if in 1) u,v are odd and in 2) u,v are 
> of different parities.
>  
> Regards.
> Vladimir
> 
> 
> ----- Original Message -----
> From: Maximilian Hasler <maximilian.hasler at gmail.com>
> Date: Friday, June 10, 2011 18:59
> Subject: [seqfan] Re: I need a name for this sequence
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> 
> > apologies for re-posting, but
> > 
> > is_abc(N)={fordiv(N,c,c*c>N&return;c>1&fordiv(N-c,a,(a+(N-
> > c)/a)*c==N&return(1)))}
> > is over 250 times faster, for N=1..9999
> > The "a>(N-c)/a & break" gave a significant speedup only in the 
> > other version.
> > 
> > One can also prove that always a>c.
> > (Assume a<=c, then N = ab+c <= cb+c < c(b+a) = N, 
> > impossible.)(But adding this test in the inner loop seems not 
> to 
> > give a
> > significant speedup.)
> > 
> > Maximilian
> > 
> > 
> > On Fri, Jun 10, 2011 at 11:09 AM, Maximilian Hasler
> > <maximilian.hasler at gmail.com> wrote:
> > > I agree with David on the missing requirement
> > > to exclude the "trivial" decomposition (1,N-1,1) for any N.
> > > (For example, to require c>1 is enough).
> > >
> > > So the definition of the sequence could be :
> > >
> > > Numbers N such that N=(a+b)*c=a*b+c for some a,b,c>1.
> > >
> > > and a more "playful" characterization would be to speak 
> about 
> > exchange> of + and * in the expression (a+b)*c.
> > >
> > > PARI code:
> > >
> > > is_abc(N)={for(c=2,N-2,fordiv(N-c,a,a>(N-c)/a&break;(a+(N-
> > c)/a)*c==N&return(1)))}>
> > > for(N=1,1999,is_abc(N)&print1(N","))
> > >
> > > Maximilian
> > >
> > >
> > >
> > >
> > > On Fri, Jun 10, 2011 at 9:45 AM, Vladimir Shevelev 
> > <shevelev at bgu.ac.il> wrote:
> > >>
> > >> An infinite set of  even N given by formula 
> > 8*u*v*(4*u*v^2+4*u*v+u-v) in view of the identity:
> > >> if x=4*u*v+2*u, y=8*u*v^2+4*u*v-2*v, z=4*u*v, then N=(x+y)z=xy+z.
> > >>
> > >> Regards,
> > >> Vladimir
> > >>
> > >>
> > >> ----- Original Message -----
> > >> From: Jose Miguel Ortiz Rodriguez <josemortiz at yahoo.com>
> > >> Date: Friday, June 10, 2011 9:26
> > >> Subject: [seqfan] Re: I need a name for this sequence
> > >> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> > >>
> > >>> Looks similar to the precedence rules for boolean numbers:
> > >>>
> > >>> Boolean Precedence:
> > >>> AB = A·B
> > >>> A·B+C = (A·B) + C
> > >>> A+B·C = A + (B·C)
> > >>>
> > >>> Your series seems to treat the numbers in a similar way to the
> > >>> boolean definition of De Morgan's Theorem, in the sense of
> > >>> (A*B+C)'=(A'+B')*C' {and then both =N in your case} or something
> > >>> like that.
> > >>>
> > >>> It's been a long time since my college years and my digital
> > >>> circuits theory courses, so this example that I wrote may be
> > >>> wrong. But I do remember something about flipping the *'s 
> by the
> > >>> +'s directly when boolean negation was in the equation in order
> > >>> to keep the equality in the logic circuit while using different
> > >>> gates for the implementation. This would be like your equivalent
> > >>> 'digital' output N in this case.
> > >>>
> > >>> Obviously you are not working just with 0's and 1's in your
> > >>> algorithm, but maybe looking at it from the variables 
> point of
> > >>> view, there could be some relationship, at least to help you
> > >>> come up with a name for it. Hope it helps :-)
> > >>>
> > >>>
> > >>> --- On Thu, 6/9/11, David Wilson 
> > <davidwwilson at comcast.net> wrote:
> > >>>
> > >>>
> > >>> From: David Wilson <davidwwilson at comcast.net>
> > >>> Subject: [seqfan] Re: I need a name for this sequence
> > >>> To: "Sequence Fanatics Discussion list" 
> <seqfan at list.seqfan.eu>> >>> Date: Thursday, June 9, 2011, 
> 6:38 PM
> > >>>
> > >>>
> > >>> If you allow a, b, c nonnegative, then (a, b, c) = (0, 1, N)
> > >>> implies every N >= 0 is one of your numbers.
> > >>>
> > >>> If you require a, b, c positive, then (a, b, c) = (1, N-1, 1)
> > >>> implies every N >= 2 is one of your numbers.
> > >>>
> > >>> Perhaps you meant to require a, b, c >= 2. This seems to
> > >>> generate the numbers you seek.
> > >>>
> > >>> I'm supposing this is the correct constraint, and that 
> your list
> > >>> below is meant to list all elements up
> > >>> to 10000. In that case, your list is incomplete. The first
> > >>> missing element I found was:
> > >>>
> > >>>     1694 = (16+105)*14 = (16*105)+14
> > >>>
> > >>> My Perl program below:
> > >>>
> > >>> my $N = 10000;
> > >>> my %seen = ();
> > >>> for (my $b = 2; 2*($b+1) <= $N; $b++) {
> > >>> print "b = $b\n";
> > >>>     my $v = $b*($b-1);
> > >>>     for (my $d = $b+1; $d <= $v; 
> $d++) {
> > >>>         next if $v%$d;
> > >>>         my $a = 
> ($d-$b+1);
> > >>>         last if $a 
> > $b;
> > >>>         my $c = 
> ($a*$b)/$d;> >>>         
> my $k = $a*$b+$c;
> > >>>         next if $k 
> > $N;
> > >>>         $seen{$k} 
> = 1;
> > >>>     }
> > >>> }
> > >>> my @k = sort {$a <=> $b} keys %seen;
> > >>> print map("$_\n", @k);
> > >>>
> > >>> generates the hopefully complete list
> > >>>
> > >>> 14 33 39 60 64 84 95 110 138 150 155 174 189 217 248 258 
> 259 272
> > >>> 315 324 360 368
> > >>> 390 399 405 410 430 473 504 530 539 564 584 624 663 670 
> 732 754
> > >>> 770 819 852 854
> > >>> 869 885 897 915 1005 1008 1024 1053 1056 1065 1104 1110 1120
> > >>> 1139 1155 1248 1278
> > >>> 1292 1360 1378 1422 1425 1463 1536 1545 1580 1615 1674 
> 1694 1743
> > >>> 1760 1785 1802
> > >>> 1806 1840 1869 1884 1914 1919 1974 2002 2055 2093 2134 
> 2280 2289
> > >>> 2369 2379 2420
> > >>> 2464 2475 2478 2500 2538 2544 2574 2625 2678 2751 2780 
> 2794 2800
> > >>> 2889 2924 2945
> > >>> 2954 2990 2997 3000 3108 3164 3171 3248 3267 3302 3325 
> 3335 3438
> > >>> 3472 3504 3570
> > >>> 3615 3668 3770 3807 3813 3885 3900 3990 4009 4064 4172 
> 4309 4323
> > >>> 4368 4375 4422
> > >>> 4488 4544 4560 4590 4640 4710 4779 4788 4794 4862 4865 
> 4884 5100
> > >>> 5115 5134 5148
> > >>> 5184 5219 5220 5249 5264 5334 5439 5478 5495 5508 5580 
> 5640 5709
> > >>> 5738 5760 5915
> > >>> 6045 6094 6142 6154 6156 6174 6272 6279 6325 6330 6336 
> 6360 6422
> > >>> 6480 6640 6683
> > >>> 6760 6798 6804 6825 6913 6923 6965 6972 7014 7120 7170 
> 7239 7289
> > >>> 7353 7488 7544
> > >>> 7685 7700 7824 7843 7854 7857 7956 7960 7995 8055 8145 
> 8184 8235
> > >>> 8299 8370 8418
> > >>> 8420 8520 8547 8624 8645 8708 8786 8789 8800 8874 8880 
> 9090 9168
> > >>> 9282 9324 9405
> > >>> 9430 9555 9604 9723 9810 9840 9847 9854 9917 9950
> > >>>
> > >>> On 6/9/2011 4:39 PM, Claudio Meller wrote:
> > >>> > Take a,b and c with a<b and a<>b,  b<>c
> > >>> >
> > >>> > I search for numbers N such N = (a+b) x c = (axb) + c
> > >>> >
> > >>> > For example :
> > >>> > a    b   
> c     N
> > >>> > (3 , 4 , 2) = 14
> > >>> > (5 , 6 , 3) = 33
> > >>> > (4 , 9 , 3) = 39
> > >>> >
> > >>> >
> > >>> > Sequence :
> > >>> > 14, 33, 39, 60, 64, 84, 95, 110, 138, 150, 155, 174, 
> 189, 217,
> > >>> 248, 258,
> > >>> > 259, 272, 315, 324, 360, 368, 390, 399, 405, 410, 430, 473,
> > >>> 504, 530, 539,
> > >>> > 564, 584, 624, 663, 670, 732, 754, 770, 819, 852, 854, 869,
> > >>> 885, 897, 915,
> > >>> > 1005, 1008, 1024, 1053, 1056, 1065, 1104, 1110, 1120, 1139,
> > >>> 1155, 1248,
> > >>> > 1278, 1292, 1360, 1378, 1422, 1425, 1536, 1545, 1580, 1615,
> > >>> 1674, 1743,
> > >>> > 1760, 1785, 1802, 1806, 1840, 1869, 1914, 1919, 1974, 2002,
> > >>> 2093, 2134,
> > >>> > 2280, 2289, 2369, 2420, 2475, 2478, 2500, 2544, 2574, 2625,
> > >>> 2678, 2794,
> > >>> > 2800, 2889, 2990, 2997, 3000, 3000, 3108, 3164, 3248, 3267,
> > >>> 3302, 3325,
> > >>> > 3335, 3472, 3570, 3668, 3770, 3813, 3990, 4064, 4309, 4323,
> > >>> 4422, 4488,
> > >>> > 4544, 4590, 4794, 4865, 4884, 5134, 5148, 5184, 5220, 5439,
> > >>> 5508, 5580,
> > >>> > 5738, 6045, 6279, 6360, 6480, 6640, 6683, 7014, 7353, 7700,
> > >>> 8055, 8145,
> > >>> > 8418, 8789, 9168, 9555, 9950
> > >>> >
> > >>> > How can I define this sequence?
> > >>> >
> > >>> > Thanks
> > >>>
> > >>>
> > >>> _______________________________________________
> > >>>
> > >>> Seqfan Mailing list - http://list.seqfan.eu/
> > >>>
> > >>> _______________________________________________
> > >>>
> > >>> Seqfan Mailing list - http://list.seqfan.eu/
> > >>>
> > >>
> > >>  Shevelev Vladimir‎
> > >>
> > >> _______________________________________________
> > >>
> > >> Seqfan Mailing list - http://list.seqfan.eu/
> > >>
> > >
> > 
> > _______________________________________________
> > 
> > Seqfan Mailing list - http://list.seqfan.eu/
> > 
> 
>  Shevelev Vladimir‎
> 
> _______________________________________________
> 
> Seqfan Mailing list - http://list.seqfan.eu/
> 

 Shevelev Vladimir‎



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