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• Note for below: negatives included as primes. Given positive a,b, a
Message 1 of 10 , May 5, 2006
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Note for below: negatives included as primes.

Given positive a,b, a<b, we want b+a and b-a both prime. First
solution: (a,b): (1,4)

Given positive a,b,c, a<b<c, we want c+b+a, c+b-a, c-b+a,-c+b+a all
prime. First solution: (a,b,c) :(1,4,8)

Given positive a,b,c,d, a<b<c<d we want d+c+b+a, d+c+b-a, d+c-b+a, d-
c+b+a, -d+c+b+a, d+c-b-a, d-c+b-a,-d+c+b-a, d-c-b+a, -d+c-b+a, -d-c+b+a
all prime. (11 combinations.) First solution: (a,b,c,d): (3,5,8,13)
(No "fibbing"!)

Given positive a,b,c,d,e, a<b<c<d<e, we want all 16 combinations to be
prime. First solution: (a,b,c,d,e): (3,10,12,15,27)

What about for 6 terms? That would be 1 + 6 + 15 + 20 = 42
combinations. All prime? Eeek! If it is solved it won't be on my
computer....

Mark
• ... -d+c+b-a and d-c-b+a are the same prime -d+c-b+a and d-c+b-a are the same prime -d-c+b+a and d+c-b-a are the same prime ... Just line up the
Message 2 of 10 , May 5, 2006
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--- Mark Underwood <mark.underwood@...> wrote:
> Note for below: negatives included as primes.
>
> Given positive a,b, a<b, we want b+a and b-a both prime. First
> solution: (a,b): (1,4)
>
> Given positive a,b,c, a<b<c, we want c+b+a, c+b-a, c-b+a,-c+b+a all
> prime. First solution: (a,b,c) :(1,4,8)
>
> Given positive a,b,c,d, a<b<c<d we want d+c+b+a, d+c+b-a, d+c-b+a, d-
> c+b+a, -d+c+b+a, d+c-b-a, d-c+b-a,-d+c+b-a, d-c-b+a, -d+c-b+a, -d-c+b+a
> all prime. (11 combinations.) First solution: (a,b,c,d): (3,5,8,13)
> (No "fibbing"!)

-d+c+b-a and d-c-b+a are the same prime
-d+c-b+a and d-c+b-a are the same prime
-d-c+b+a and d+c-b-a are the same prime

> Given positive a,b,c,d,e, a<b<c<d<e, we want all 16 combinations to be
> prime. First solution: (a,b,c,d,e): (3,10,12,15,27)
>
> What about for 6 terms? That would be 1 + 6 + 15 + 20 = 42
> combinations. All prime? Eeek! If it is solved it won't be on my
> computer....

Just line up the sums/differences, so that many combinations evaluate to the
same thing.

I think your count of combinations is incorrect. There are 2^(n-1) ways to
chose the signs. See above examples of miscounting.

Phil

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• ... all ... b+a, d- ... c+b+a ... (3,5,8,13) ... to be ... evaluate to the ... ways to ... You re right Phil! As it turns out my method of counting works only
Message 3 of 10 , May 6, 2006
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--- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...>
wrote:
>
> --- Mark Underwood <mark.underwood@...> wrote:
> > Note for below: negatives included as primes.
> >
> > Given positive a,b, a<b, we want b+a and b-a both prime. First
> > solution: (a,b): (1,4)
> >
> > Given positive a,b,c, a<b<c, we want c+b+a, c+b-a, c-b+a,-c+b+a
all
> > prime. First solution: (a,b,c) :(1,4,8)
> >
> > Given positive a,b,c,d, a<b<c<d we want d+c+b+a, d+c+b-a, d+c-
b+a, d-
> > c+b+a, -d+c+b+a, d+c-b-a, d-c+b-a,-d+c+b-a, d-c-b+a, -d+c-b+a, -d-
c+b+a
> > all prime. (11 combinations.) First solution: (a,b,c,d):
(3,5,8,13)
> > (No "fibbing"!)
>
> -d+c+b-a and d-c-b+a are the same prime
> -d+c-b+a and d-c+b-a are the same prime
> -d-c+b+a and d+c-b-a are the same prime
>
> > Given positive a,b,c,d,e, a<b<c<d<e, we want all 16 combinations
to be
> > prime. First solution: (a,b,c,d,e): (3,10,12,15,27)
> >
> > What about for 6 terms? That would be 1 + 6 + 15 + 20 = 42
> > combinations. All prime? Eeek! If it is solved it won't be on my
> > computer....
>
> Just line up the sums/differences, so that many combinations
evaluate to the
> same thing.
>
> I think your count of combinations is incorrect. There are 2^(n-1)
ways to
> chose the signs. See above examples of miscounting.
>

You're right Phil! As it turns out my method of counting works only
when the number of terms is odd. If there are five terms:

1 + 5 + (5*4)/2 = 16 = 2^(5-1) additive combinations.

If there are seven terms:

1 + 7 + (7*6)/2 + (7*6*5)/(3*2) = 64 = 2^(7-1) additive combinations.

Mark
• Well surprise of surprises: The six numbers (5,30,33,42,60,63) yield primes for all 32 additive combinations of the six numbers.(!) I thought it would have
Message 4 of 10 , May 6, 2006
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Well surprise of surprises:

The six numbers (5,30,33,42,60,63) yield primes for all 32 additive
combinations of the six numbers.(!) I thought it would have been
waaay higher.

So we have as first cases:

(1,4) has all 2 additive combinations yielding 2 primes.
(1,4,8) has all 4 additive combos yielding 4 primes.
(3,5,8,13) has all 8 additive combos yielding 8 primes.
(3,10,12,15,27) has all 16 additive combos yielding 16 primes.
(5,30,33,42,60,63) has all 32 additive combos yielding 32 primes.

Interesting that thus far the greatest number in each set does not
quite exceed 2^n. Surely that can't hold up for long...or can it?

Wonders never cease!
Mark
• ... Something interesting with the product of the numbers ? (1,4)-- 4 = 2^2 (1,4,8)-- 32 = 2^5 (3,5,8,13)-- 1560 = 2^3 * 3 * 5 * 13 (3,10,12,15,27)--
Message 5 of 10 , May 7, 2006
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--- "Mark Underwood" <mark.underwood@...> wrote:
>
> Well surprise of surprises:
>
> The six numbers (5,30,33,42,60,63) yield primes for all 32 additive
> combinations of the six numbers.(!) I thought it would have been
> waaay higher.
>
> So we have as first cases:
>
> (1,4) has all 2 additive combinations yielding 2 primes.
> (1,4,8) has all 4 additive combos yielding 4 primes.
> (3,5,8,13) has all 8 additive combos yielding 8 primes.
> (3,10,12,15,27) has all 16 additive combos yielding 16 primes.
> (5,30,33,42,60,63) has all 32 additive combos yielding 32 primes.
>
> Interesting that thus far the greatest number in each set does not
> quite exceed 2^n. Surely that can't hold up for long...or can it?
>
> Wonders never cease!
> Mark
----------------------------------------------------------------------

Something interesting with the product of the numbers ?

(1,4)--> 4 = 2^2
(1,4,8)--> 32 = 2^5
(3,5,8,13)--> 1560 = 2^3 * 3 * 5 * 13
(3,10,12,15,27)--> 145800 = 2^3 * 3^6 * 5^2
(5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11

In each case the product of the n numbers gives a number whose number
of different prime factors (in the factorization) is smaller or equal
to n.
Is it always the case ?

Patrick Capelle.
• ... number ... equal ... Perhaps. Anyways, the numbers are relatively factor rich. A prime guru on very quick order came up with 7 more examples where all 32
Message 6 of 10 , May 8, 2006
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<patrick.capelle@...> wrote:
>
> Something interesting with the product of the numbers ?
>
> (1,4)--> 4 = 2^2
> (1,4,8)--> 32 = 2^5
> (3,5,8,13)--> 1560 = 2^3 * 3 * 5 * 13
> (3,10,12,15,27)--> 145800 = 2^3 * 3^6 * 5^2
> (5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11
>
> In each case the product of the n numbers gives a number whose
number
> of different prime factors (in the factorization) is smaller or
equal
> to n.
> Is it always the case ?
>

Perhaps. Anyways, the numbers are relatively factor rich.

A prime guru on very quick order came up with 7 more examples where
all 32 additive combinations yield primes (!):

(30, 33, 35, 60, 63, 72)
(30, 33, 47, 60, 72, 75)
(30, 42, 47, 60, 63, 75)
(15, 42, 48, 57, 70, 75)
(30, 33, 42, 60, 75, 77)
(15, 33, 42, 57, 70, 90)
(15, 35, 42, 57, 72, 90)

What is curious to me is that each set has three numbers with factors
of two. Also each set has three numbers with factors of five.
But it is easy to see why each set has five numbers with factors of
three.

So far, no luck finding 7 numbers with all 64 additive combinations
yielding primes.

Mark
• ... Something interesting with the product of the numbers ? (1,4)-- 4 = 2^2 (1,4,8)-- 32 = 2^5 (3,5,8,13)-- 1560 = 2^3 * 3 * 5 * 13 (3,10,12,15,27)--
Message 7 of 10 , May 8, 2006
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--- "Patrick Capelle" <patrick.capelle@> wrote:
Something interesting with the product of the numbers ?
(1,4)--> 4 = 2^2
(1,4,8)--> 32 = 2^5
(3,5,8,13)--> 1560 = 2^3 * 3 * 5 * 13
(3,10,12,15,27)--> 145800 = 2^3 * 3^6 * 5^2
(5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11

In each case the product of the n numbers gives a number whose
number of different prime factors (in the factorization) is smaller
or equal to n. Is it always the case ?
--------------------------------------------------------------------
--- "Mark Underwood" <mark.underwood@...> wrote:
Perhaps. Anyways, the numbers are relatively factor rich.

A prime guru on very quick order came up with 7 more examples where
all 32 additive combinations yield primes (!)
...
--------------------------------------------------------------------

It holds for n = 6 :
( 5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11
(30,33,35,60,63,72)--> 9430344000 = 2^6 * 3^7 * 5^3 * 7^2 * 11
(30,33,47,60,72,75)--> 15075720000 = 2^6 * 3^6 * 5^4 * 11^ * 47
(30,42,47,60,63,75)--> 16788870000 = 2^4 * 3^6 * 5^4 * 7^2 * 47
(15,42,48,57,70,75)--> 9049320000 = 2^6 * 3^5 * 5^4 * 7^2 * 19
(30,33,42,60,75,77)--> 14407470000 = 2^4 * 3^5 * 5^4 * 7^2 * 11^2
(15,33,42,57,70,90)--> 7465689000 = 2^3 * 3^6 * 5^3 * 7^2 * 11 * 19
(15,35,42,57,72,90)--> 8144388000 = 2^5 * 3^7 * 5^3 * 7^2 * 19

Patrick Capelle.
• ... Something interesting with the product of the numbers ? (1,4)-- 4 = 2^2 (1,4,8)-- 32 = 2^5 (3,5,8,13)-- 1560 = 2^3 * 3 * 5 * 13 (3,10,12,15,27)--
Message 8 of 10 , May 9, 2006
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--- "Patrick Capelle" <patrick.capelle@...> wrote:
Something interesting with the product of the numbers ?
(1,4)--> 4 = 2^2
(1,4,8)--> 32 = 2^5
(3,5,8,13)--> 1560 = 2^3 * 3 * 5 * 13
(3,10,12,15,27)--> 145800 = 2^3 * 3^6 * 5^2
(5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11

In each case the product of the n numbers gives a number whose
number of different prime factors (in the factorization) is smaller
or equal to n. Is it always the case ?
---------------------------------------------------------------------
--- "Patrick Capelle" <patrick.capelle@...> wrote:
It holds for n = 6 :
( 5,30,33,42,60,63)--> 785862000 = 2^4 * 3^6 * 5^3 * 7^2 * 11
(30,33,35,60,63,72)--> 9430344000 = 2^6 * 3^7 * 5^3 * 7^2 * 11
(30,33,47,60,72,75)--> 15075720000 = 2^6 * 3^6 * 5^4 * 11^ * 47
(30,42,47,60,63,75)--> 16788870000 = 2^4 * 3^6 * 5^4 * 7^2 * 47
(15,42,48,57,70,75)--> 9049320000 = 2^6 * 3^5 * 5^4 * 7^2 * 19
(30,33,42,60,75,77)--> 14407470000 = 2^4 * 3^5 * 5^4 * 7^2 * 11^2
(15,33,42,57,70,90)--> 7465689000 = 2^3 * 3^6 * 5^3 * 7^2 * 11 * 19
(15,35,42,57,72,90)--> 8144388000 = 2^5 * 3^7 * 5^3 * 7^2 * 19
---------------------------------------------------------------------
--- "Mark Underwood" <mark.underwood@...> wrote:
I assumed you meant it would hold for first cases. Afterall it
certainly doesn't hold for some cases after the first case. For
instance

14 + 9 = 23
14 - 9 = 5

And 14*9 has three different prime factors.

I'm going to try to look for solutions for n=7 and all 64 additive
combinations yielding primes. My computer is so slow however it does
not look promising.

kind regards,
Mark
---------------------------------------------------------------------

Important precision.
Thank you Mark.
I only started with the examples that you gave, without thinking to
other cases.
When we look at each n it is possible that it holds only for the
smallest set(s).
Or for all the sets when n is not too small ?

Best regards,
Patrick Capelle.
• ... Took a nice sabbatical from numbers, namely the primes. The things drive me crazy. But in a momement of weakness/boredom, perhaps abit of seasonal
Message 9 of 10 , Feb 5, 2008
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<mark.underwood@...> wrote:
>
> Well surprise of surprises:
>
> The six numbers (5,30,33,42,60,63) yield primes for all 32 additive
> combinations of the six numbers.(!) I thought it would have been
> waaay higher.
>
> So we have as first cases:
>
> (1,4) has all 2 additive combinations yielding 2 primes.
> (1,4,8) has all 4 additive combos yielding 4 primes.
> (3,5,8,13) has all 8 additive combos yielding 8 primes.
> (3,10,12,15,27) has all 16 additive combos yielding 16 primes.
> (5,30,33,42,60,63) has all 32 additive combos yielding 32 primes.
>
> Interesting that thus far the greatest number in each set does not
> quite exceed 2^n. Surely that can't hold up for long...or can it?
>
> Wonders never cease!
> Mark
>

Took a nice sabbatical from numbers, namely the primes. The things
drive me crazy. But in a momement of weakness/boredom, perhaps abit of
seasonal affective disorder, I looked back over some of my
investigations.

Very shortly after this post from about a year and a half ago, I found
a way to greatly speed up the search for additive combinations which
yeild all primes. For instance: for six numbers, each not exceeding
500, I found on the order of about 70 different solutions, each
yeilding (in their 32 additive combinations) all primes. But I noted
only one which yielded 32 *distinct* primes:

42,60,75,77,105,108

in its 32 additive combinations yields the 32 unique primes

13,17,19,23,41,43,47,53,79,97,
101,103,107,109,113,131,137,163,167,173,193,197,
229,233,251,257,263,313,317,347,383,467.

I tried a similar technique to get 7 numbers with all 64 additive
combinations yielding only primes, but with no success. I don't think
there is a solution if all the numbers are below 1000, but it wasn't
quite an exhaustive search.

Just throwing this out in case anyone is interested, before I entirely

Mark
• ... [snip] ... [snip] I meant each number not exceeding 150 (not 500). Mark
Message 10 of 10 , Feb 5, 2008
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--- In primenumbers@yahoogroups.com, "Mark Underwood" <mark.underwood@...> wrote:

[snip]
> For instance: for six numbers, each not exceeding
> 500, I found on the order of about 70 different solutions, each
> yeilding (in their 32 additive combinations) all primes.
[snip]

I meant each number not exceeding 150 (not 500).

Mark
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