

A210852


Approximations up to 7^n for one of the three 7adic integers (1)^(1/3).


16



0, 3, 31, 325, 1354, 1354, 34968, 740862, 2387948, 25447152, 146507973, 1276408969, 9185715941, 78392151946, 272170172760, 950393245609, 10445516265495, 43678446835096, 974200502783924, 10744682090246618, 22143577275619761
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,2


COMMENTS

The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 3^(7^(n1)) (mod 7^n), n >= 1, which satisfies a(n)^3 + 1 == 0 (mod 7^n), n >= 1. a(0) = 0 satisfies this congruence as well. The proof can be done by showing that each term in the binomial expansion of (3^(7^(n1)))^3 +1 = (28 1)^(7^(n1)) + 1 has a factor 7^n.
a(n) == 3 (mod 7), n >= 1. This follows from the formula given above, and 3^(7^(n1)) == 3 (mod 7), n >= 1 (proof by induction).
The digit t(n), n >= 0, multiplying 7^n in the 7adic integer (1)^(1/3) corresponding to this sequence is obtained from the (unique) solution of the linear congruence 3*a(n)^2*t(n) + b(n) == 0 (mod 7), n >= 1, with b(n):= (a(n)^3+1)/7^n = A210853(n). t(0):=3, one of the three solutions of X^3 + 1 == 0 (mod 7). For these digits see A212152. The 7adic number is, read from right to left, ...3143214516604202226653431432053116412125443426203643 =: u.
a(n) is obtained from reading u in base 7, and adding the first n terms.
One can show directly that a(n) = 7^n + 1  y(n), n >= 1, with y(n) = A212153(n) and z(n) = 7^n  1 = 6*A023000(n), n >= 0.
Iff a(n+1) = a(n) then t(n) = A212152(n) = 0.
See the Nagell reference given in A210848 for theorems 50 and 52 on p. 87, and formula (6) on page 86, adapted to this case. Because X^3 + 1 = 0 (mod 7) has the three simple roots 3, 5 and 6, one has for X(n)^3 + 1 == 0 (mod 7^n) exactly three solutions for each n >= 1, which can be chosen as a(n) == 3 (mod 7), y(n) == 5 (mod 7) and z(n) == 6 (mod 7) == 1 (mod 7). The y and z sequences are given in A212153 and 6*A023000, respectively.
For n > 0, a(n)  1 (== a(n)^2 (mod 7^n)) and 7^n  a(n) (== a(n)^4 (mod 7^n)) are the two primitive cubic roots of unity in Z/(7^n Z).  Álvar Ibeas, Feb 20 2017


LINKS

Kenny Lau, Table of n, a(n) for n = 0..1183


FORMULA

Recurrence: a(n) = a(n1)^7 (mod 7^n), n >= 2, a(0)=0, a(1)=3.
a(n) == 3^(7^(n1)) (mod 7^n) == 3 (mod 7), n >= 1.
a(n+1) = a(n) + A212152(n)*7^n, n >= 1.
a(n+1) = Sum_{k=0..n} A212152(k)*7^k, n >= 1.
a(n1)^2*a(n) + 1 == 0 (mod 7^(n1)), n >= 1 (from 3*a(n)^2* A212152(n) + A210853(n) == 0 (mod 7) and the secondtolast formula from above).
a(n) = 7^n + 1  A212153(n), n >= 1.


EXAMPLE

a(3) == 31^7 (mod 7^3) == 27512614111 (mod 343) = 325.
a(3) == 3^49 (mod 7^3) = 325.
a(3) = 31 + 6*7^2 = 325.
a(3) = 3*7^0 + 4*7^1 + 6*7^2 = 325.
a(3) = 7^3 +1  19 = 325.
a(5) = a(4) = 1354 because A212152(4) = 0.


MAPLE

a:=proc(n) option remember: if n=0 then 0 elif n=1 then 3
else modp(a(n1)^7, 7^n) fi end proc: [seq(a(n), n=0..30)];


MATHEMATICA

a[n_] := a[n] = Which[n == 0, 0, n == 1, 3, True, Mod[a[n1]^7, 7^n]]; Table[a[n], {n, 0, 30}] (* JeanFrançois Alcover, Mar 05 2014, after Maple *)


CROSSREFS

Cf. A048898 (case p=5). A212153, 6*A023000.
Sequence in context: A011545 A011546 A276195 * A152276 A136024 A051200
Adjacent sequences: A210849 A210850 A210851 * A210853 A210854 A210855


KEYWORD

nonn,easy


AUTHOR

Wolfdieter Lang, May 02 2012


STATUS

approved



