Problem 282

[ukimiku]

Dear forum members,

I have difficulties understanding problem 282.
I know the Ackermann function, but what does the existence quantor E 0<=n<=6 A(n,n) mean?
Should my program calculate all seven values for A(0,0), A(1,1), A(2,2), A(3,3), A(4,4), A(5,5), and A(6,6) (mod 14^8)?

Thanks for taking the time to clarify.

Regards

[harryh]

Problem 282 (View Problem) That's not E (for existence); it's a &sum; meaning the sum for n=0 to n=6.

[ukimiku]

(blush)

thanks for clearing up this mis-reading.
Regards,

[Raman]

At 227/350 problem solving number estimates
Breaking my head as for this problem at as such throughout this day

Which is wrong enough?

each value being given as (mod 14⁸) as well
A(0,0) = skip
A(1,1) = skip
A(2,2) = skip
A(3,3) = skip
A(4,4) = skip
A(5,5) = skip
A(6,6) = skip

For total value for
being = skip being

I believe everything is being correct itself
rather within that way

it is being up so thus t within
down that way sez within
zyx tuch tay tush stz within
z zearry of r ancy coursy linkagy factery knrgy stz lrmny ni yn y y y y y y y y y y y y y y y y y y y y z y x - - - - - - - - - - - - eht - - - - - - - - - - -
trfn tgrz dz_?=- - () - (-) ()- {} - {-} {}- [] - [-] []- - - - - - - - - - - - eht - - - - - - - - - - -

[ParadiceCity9]

I get a stack overflow exception when the numbers get high enough, but I can't think of any way of averting that. Any advice?

[Marcus_Andrews]

The Ackermann function gets very large very quickly as you ramp up m and n. It's too large to compute directly. Try taking advantage of the fact that the answer is asking for the modulus of each term, instead.

[Oliver1012]

Apologies for pushing this topic....

I'm trying to wrap my head around it. Might I ask if

A(4, 4) mod 13⁷ = 51262425
A(5, 5) mod 13⁷ = 58052009

is correct?