# duplicate hunting cont., pt3

Martin Fuller martin_n_fuller at btinternet.com
Tue Apr 24 19:27:40 CEST 2007

%I A119892
%C A119892 a(2322)=389999 is the first value different from A106766

(PARI) - should be three lines:
DigitSum(n,b=10)=local(x);x=0;while(n,x+=n%b;n\=b);x
PrimeEnsemble(n,b=10)=local(x);x=1;while(ispseudoprime(n),if(n<b,return(x));n=DigitSum(n,b);x++);0
forprime(p=2, 16000, if(PrimeEnsemble(p)>=4, print1(p", ")));

Martin Fuller

--- Joshua Zucker <joshua.zucker at gmail.com> wrote:

> On 4/23/07, Andrew Plewe <aplewe at sbcglobal.net> wrote:
> > Possible duplicates:
> > Will the sequences A068310 and A086485 ever differ? Glancing at the problem
> > I think they will at some value for n, but n would probably have to be very
> > large. If not, then these are duplicates.
>
> I don't understand.  Their definitions look identical to me!
>
> > On a similar note, A106766 and
> > A119892 will probably differ at some point, too. It is kind of interesting
> > that they match as well as they do.
>
> Well, the digit sum must be the largest member of a prime trio, so the
> first number where the sequences differ must be with digit sum 47, and
> thus have at least 6 digits -- so until then you get all the primes
> with 4 or 5 digits that have digit sum 29, and that's a lot of terms.
> Similarly A119891 has all its entries having digit sum 11 for a long
> while until the first 23 comes in.
>
> --Joshua Zucker
>

Several people have emailed me off-list to point out that A106766 and
A119892 are trivially the same and why (the largest square divisor of n^2 -
1 will be the product of all the square divisors of n^2 - 1, hence division
by the largest square divisor guarantees that the result will be
today:

A098769 and A098900
A105184 and A121607
A111480 and A111713
A098768 and A098898
A061150 and A127369
A061421 and A100360
A028888 and A082627
A108165 and A108170
A092767 and A108311
A063976 and A064000

Sequences A087695 and A087749 match, but it appears that A087749 is wrong.
It is defined as "Primes n such that 6^n-5^n is composite." Here are two
examples of terms that are invalid:

A087749(1) = 2, 6^2 - 5^2 = 11, 11 is prime
A087749(2) = 5, 6^5 - 5^5 = 4651, 4651 is prime

This is the Pari code included with the sequence:

apmb(a, b, n) = ( forprime(x=2, n, y=a^x-b^x; if(!ispseudoprime(y),
print1(x", "); ) ) )

Running abpm(6,5,400) I get:

3, 7, 17, 19, 29, 31, 37, 41, 43, 47, 53, 59, 67, 71, 73, 79, 89, 97, 101,
103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179,
181, 191,193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263,
269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353,
359, 367, 373, 379, 383, 389, 397

Doing a lookup on the first fifteen or so terms of that sequence returns no
results. Can someone confirm that this is indeed the correct sequence?
Thanks!

-----Original Message-----

%I A119892
%C A119892 a(2322)=389999 is the first value different from A106766

(PARI) - should be three lines:
DigitSum(n,b=10)=local(x);x=0;while(n,x+=n%b;n\=b);x
PrimeEnsemble(n,b=10)=local(x);x=1;while(ispseudoprime(n),if(n<b,return(x));
n=DigitSum(n,b);x++);0
forprime(p=2, 16000, if(PrimeEnsemble(p)>=4, print1(p", ")));

Martin Fuller

--- Joshua Zucker <joshua.zucker at gmail.com> wrote:

> On 4/23/07, Andrew Plewe <aplewe at sbcglobal.net> wrote:
> > Possible duplicates:
> > Will the sequences A068310 and A086485 ever differ? Glancing at the
problem
> > I think they will at some value for n, but n would probably have to be
very
> > large. If not, then these are duplicates.
>
> I don't understand.  Their definitions look identical to me!
>
> > On a similar note, A106766 and
> > A119892 will probably differ at some point, too. It is kind of
interesting
> > that they match as well as they do.
>
> Well, the digit sum must be the largest member of a prime trio, so the
> first number where the sequences differ must be with digit sum 47, and
> thus have at least 6 digits -- so until then you get all the primes
> with 4 or 5 digits that have digit sum 29, and that's a lot of terms.
> Similarly A119891 has all its entries having digit sum 11 for a long
> while until the first 23 comes in.
>
> --Joshua Zucker
>