pari and p-adic numbers

[devarajkandadai]

On page 19 of " number theory" by Borevich and Shefarevich you will find a sequence that leads to 7-adic numbers. The sequence depends on the solution of a linear equation at every stage. An alternate method using pari is illustrated below:

10 = = 3 (mod 7)
108== 10(mod 7^2)
451 = = 108 (mod 7^3)
we now use code: {is(n) = Mod((108 + 343*n),7^4)^2 = = 2};
select (is,[1..100]). we get a sequence of numbers satisfying the code. The smallest
value of n obtained is 6 leading to the next member of above sequence: 2166.
Thus we have obtained a sequence ( using pari) without having to solve any linear
equation.This can be continued indefinitely.

[VBCurtis]

Why did you use 108 rather than 59 in the second congruence?

[CRGreathouse]

Originally Posted by devarajkandadai

On page 19 of " number theory" by Borevich and Shefarevich you will find a sequence that leads to 7-adic numbers. The sequence depends on the solution of a linear equation at every stage. An alternate method using pari is illustrated below:

10 = = 3 (mod 7)
108== 10(mod 7^2)
451 = = 108 (mod 7^3)
we now use code: {is(n) = Mod((108 + 343*n),7^4)^2 = = 2};
select (is,[1..100]). we get a sequence of numbers satisfying the code. The smallest
value of n obtained is 6 leading to the next member of above sequence: 2166.
Thus we have obtained a sequence ( using pari) without having to solve any linear
equation.This can be continued indefinitely.

So instead of solving a linear equation, you had gp solve 100 linear equations.

[Dr Sardonicus]

Originally Posted by CRGreathouse

So instead of solving a linear equation, you had gp solve 100 linear equations.

Why not use polhensellift() or factorpadic()?

[CRGreathouse]

Originally Posted by Dr Sardonicus

Why not use polhensellift() or factorpadic()?

The GP code in A034945 just uses the built-in p-adic type. It takes 9 milliseconds to find the 100,000-th term on my machine. devaraj, feel free to compare to your code, I'm not sure how to extend it to that case. As a quick check my answer has 84,510 decimal digits and is of the form

259265345916500277712186481134963754311965201010586807205594069157787134361146417423434312945701993552539489...9746222274903009641714981399405417166669775352016495125084837567091444060752850366048748957882425104

where the ... represents about 84,300 digits.

In fairness, it errors out if I ask for the millionth term:

> a(10^6)
  *** _+_: Warning: increasing stack size to 40000000.
  *** _+_: Warning: increasing stack size to 80000000.
  ***   at top-level: a(10^6)
  ***                 ^-------
  ***   in function a: truncate(sqrt(2+O(7^n)))
  ***                                 ^---------
  *** _+_: overflow in precp().
  ***   Break loop: type 'break' to go back to GP prompt

So maybe you can improve on this!

[devarajkandadai]

Originally Posted by VBCurtis

Why did you use 108 rather than 59 in the second congruence?

Because 59^2 not congruent to 2 (mod 7^3)

[devarajkandadai]

Originally Posted by CRGreathouse

So instead of solving a linear equation, you had gp solve 100 linear equations.

I cannot pretend to be a good pari code writer; even the little knowledge i picked up from you.

[science_man_88]

Originally Posted by devarajkandadai

I cannot pretend to be a good pari code writer; even the little knowledge i picked up from you.

solve() may come in handy. In my PARI there's also padicfields, padicappr, and padicprec.