Displaying 1-10 of 11 results found.
Digits of one of the three 7-adic integers 6^(1/3) that is related to A319097.
+20
12
3, 3, 2, 2, 4, 6, 6, 1, 4, 4, 0, 3, 0, 5, 3, 5, 1, 5, 3, 6, 2, 2, 6, 4, 3, 3, 2, 0, 2, 1, 2, 3, 3, 4, 5, 6, 1, 5, 3, 0, 0, 3, 2, 6, 6, 0, 3, 5, 0, 6, 5, 1, 0, 3, 6, 4, 6, 6, 2, 4, 0, 4, 3, 3, 1, 4, 1, 5, 5, 6, 4, 4, 0, 1, 5, 2, 1, 1, 5, 0, 4, 1, 6, 5, 5, 5, 0, 4
COMMENTS
For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 13). If k is a cube in 7-adic field, then k has exactly three cubic roots.
EXAMPLE
The unique number k in [1, 7^3] and congruent to 3 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 122 = (233)_7, so the first three terms are 3, 3 and 2.
PROG
(PARI) a(n) = lift(sqrtn(6+O(7^(n+1)), 3) * (-1+sqrt(-3+O(7^(n+1))))/2)\7^n
CROSSREFS
Digits of p-adic cubic roots:
Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).
+10
16
0, 3, 31, 325, 1354, 1354, 34968, 740862, 2387948, 25447152, 146507973, 1276408969, 9185715941, 78392151946, 272170172760, 950393245609, 10445516265495, 43678446835096, 974200502783924, 10744682090246618, 22143577275619761
COMMENTS
The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 3^(7^(n-1)) (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^(n-1)))^3 +1 = (28 -1)^(7^(n-1)) + 1 has a factor 7^n.
a(n) == 3 (mod 7), n >= 1. This follows from the formula given above, and 3^(7^(n-1)) == 3 (mod 7), n >= 1 (proof by induction).
The digit t(n), n >= 0, multiplying 7^n in the 7-adic 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 7-adic 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 a(n) is the solution to x^2 - x + 1 == 0 (mod 7^n) that is congruent to 3 modulo 7 (if n>0).
A212153(n) is the multiplicative inverse of a(n) modulo 7^n. (End)
FORMULA
Recurrence: a(n) = a(n-1)^7 (mod 7^n), n >= 2, a(0)=0, a(1)=3.
a(n) == 3^(7^(n-1)) (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(n-1)^2*a(n) + 1 == 0 (mod 7^(n-1)), n >= 1 (from 3*a(n)^2* A212152(n) + A210853(n) == 0 (mod 7) and the second-to-last 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(n-1)^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[n-1]^7, 7^n]]; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Mar 05 2014, after Maple *)
PROG
(PARI) a(n) = lift((1-sqrt(-3+O(7^n)))/2) \\ Jianing Song, Aug 26 2022
CROSSREFS
Cf. A212152 (digits of (-1)^(1/3)), A212153 (approximations of another cube root of -1), 6* A023000 (approximations of -1).
Approximations up to 7^n for one of the three 7-adic integers (-1)^(1/3).
+10
15
0, 5, 19, 19, 1048, 15454, 82682, 82682, 3376854, 14906456, 135967277, 700917775, 4655571261, 18496858462, 406052900090, 3797168264335, 22787414304107, 188952067152112, 654213095126526, 654213095126526, 57648689021992241
COMMENTS
See A210852 for comments and the approximation of one of the other three 7-adic integers (-1)^(1/3), called there u. The numbers are computed from the recurrence given below in the formula field. This recurrence follows from the formula a(n) = 5^(7^(n-1)) (mod 7^n), n>= 1, which satisfies a(n)^3 + 1 == 0 (mod 7^n), n>=1. a(0) = 0 satisfies this congruence also. The proof can be done by showing that each term in the binomial expansion of (5^(7^(n-1)))^3 + 1 = (2*3^2*7 - 1)^(7^(n-1)) + 1 has a factor 7^n.
a(n) == 5 (mod 7), n>=1. This follows from the formula given above, and 5^(7^(n-1)) == 5 (mod 7), n>=1 (proof by induction).
The digit t(n), n>=0, multiplying 7^n in the 7-adic integer (-1)^(1/3) corresponding to the present 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 = A212154(n). t(0):=5. For these digits see A212155. The 7-adic number is, read from right to left, ...3452150062464440013235234613550254541223240463025 =: v.
a(n) is obtained from reading v in base 7, and adding the first n terms.
One can show directly that a(n) = 7^n + 1 - x(n), n>=1, with x(n) = A210852(n), and z(n) = 7^n - 1 = 6* A023000(n), n>=0. Iff a(n+1) = a(n) then t(n) = A212155(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 x(n) == 3 (mod 7), a(n) == 5 (mod 7) and z(n) == 6 (mod 7) == -1 (mod 7). The x- and z- sequences are given in A210852 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 a(n) is the solution to x^2 - x + 1 == 0 (mod 7^n) that is congruent to 5 modulo 7 (if n>0).
A210852(n) is the multiplicative inverse of a(n) modulo 7^n. (End)
FORMULA
Recurrence: a(n) = a(n-1)^7 (mod 7^n), n>=2, a(0):=0, a(1)=5.
a(n) == 5^(7^(n-1)) (mod 7^n) == 5 (mod 7), n>= 1.
a(n+1) = a(n) + A212155(n)*7^n, n>=1. a(n+1) = Sum_{k=0..n} A212155(k)*7^k, n>=1. a(n-1)^2*a(n) + 1 == 0 (mod 7^(n-1)), n>=1 (from 3*a(n)^2* A212155(n) + A212154(n) == 0 (mod 7) and the formula two lines above).
EXAMPLE
a(4) == 19^7 (mod 7^4) = 893871739 (mod 2401) = 1048.
a(4) == 5^343 (mod 7^4) = 1048.
a(4) = 19 + 3*7^3 = 1048.
a(4) = 5*7^0 + 2*7^1 + 0*7^2 + 3*7^3 = 1048.
a(4) = 7^4 + 1 - 1354 = 1048.
a(3) = a(2) = 19 because A212155(2) = 0.
MAPLE
a:=proc(n) option remember: if n=0 then 0 elif n=1 then 5
else modp(a(n-1)^7, 7^n) fi end proc:
PROG
(PARI) a(n) = lift((1+sqrt(-3+O(7^n)))/2) \\ Jianing Song, Aug 26 2022
CROSSREFS
Cf. A212155 (digits of (-1)^(1/3)), A210852 (approximations of another cube root of -1), 6* A023000 (approximations of -1).
Digits of one of the three 7-adic integers 6^(1/3) that is related to A319098.
+10
12
5, 5, 2, 2, 1, 5, 6, 0, 5, 4, 4, 3, 0, 0, 3, 3, 2, 4, 5, 2, 1, 1, 6, 1, 4, 6, 2, 2, 2, 6, 2, 4, 1, 0, 0, 2, 0, 3, 4, 2, 2, 1, 5, 2, 6, 1, 0, 4, 1, 6, 0, 6, 0, 1, 3, 0, 3, 4, 1, 1, 1, 0, 6, 5, 2, 4, 0, 2, 2, 5, 1, 1, 0, 4, 0, 0, 5, 6, 4, 1, 5, 2, 4, 3, 0, 1, 2, 5
COMMENTS
For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 13). If k is a cube in 7-adic field, then k has exactly three cubic roots.
EXAMPLE
The unique number k in [1, 7^3] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 138 = (255)_7, so the first three terms are 5, 5 and 2.
PROG
(PARI) a(n) = lift(sqrtn(6+O(7^(n+1)), 3) * (-1-sqrt(-3+O(7^(n+1))))/2)\7^n
CROSSREFS
Digits of p-adic cubic roots:
Digits of one of the three 7-adic integers 6^(1/3) that is related to A319199.
+10
12
6, 4, 1, 2, 1, 2, 0, 4, 4, 4, 1, 0, 6, 1, 0, 5, 2, 4, 4, 4, 2, 3, 1, 0, 6, 3, 1, 4, 2, 6, 1, 6, 1, 2, 1, 5, 4, 5, 5, 3, 4, 2, 6, 4, 0, 4, 3, 4, 4, 1, 0, 6, 5, 2, 4, 1, 4, 2, 2, 1, 5, 2, 4, 4, 2, 5, 4, 6, 5, 1, 0, 1, 6, 1, 1, 4, 0, 6, 3, 4, 4, 2, 3, 4, 0, 0, 4, 4
COMMENTS
For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 13). If k is a cube in 7-adic field, then k has exactly three cubic roots.
EXAMPLE
The unique number k in [1, 7^3] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 83 = (146)_7, so the first three terms are 6, 4 and 1.
PROG
(PARI) a(n) = lift(sqrtn(6+O(7^(n+1)), 3))\7^n
CROSSREFS
Digits of p-adic cubic roots:
One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 5 (mod 7) case (except for n = 0).
+10
11
0, 5, 40, 138, 824, 3225, 87260, 793154, 793154, 29617159, 191031587, 1320932583, 7252912812, 7252912812, 7252912812, 2041922131359, 16284606661188, 82750467800390, 1013272523749218, 9155340513301463, 31953130884047749, 111745397181659750, 670291261264943757
COMMENTS
For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 5 mod 7 such that k^3 - 6 is divisible by 7^n.
For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 7). If k is a cube in 7-adic field, then k has exactly three cubic roots.
EXAMPLE
The unique number k in [1, 7^2] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 40, so a(2) = 40.
The unique number k in [1, 7^3] and congruent to 5 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 138, so a(3) = 138.
PROG
(PARI) a(n) = lift(sqrtn(6+O(7^n), 3) * (-1-sqrt(-3+O(7^n)))/2)
CROSSREFS
Approximations of p-adic cubic roots:
One of the three successive approximations up to 7^n for 7-adic integer 6^(1/3). This is the 6 (mod 7) case (except for n = 0).
+10
11
0, 6, 34, 83, 769, 3170, 36784, 36784, 3330956, 26390160, 187804588, 470279837, 470279837, 83518003043, 180407013450, 180407013450, 23918214563165, 90384075702367, 1020906131651195, 7534560523292991, 53130141264785563, 212714673860009565, 1888352266109861586
COMMENTS
For n > 0, a(n) is the unique number k in [1, 7^n] and congruent to 6 mod 7 such that k^3 - 6 is divisible by 7^n.
For k not divisible by 7, k is a cube in 7-adic field if and only if k == 1, 6 (mod 7). If k is a cube in 7-adic field, then k has exactly three cubic roots.
EXAMPLE
The unique number k in [1, 7^2] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^2 is k = 34, so a(2) = 24.
The unique number k in [1, 7^3] and congruent to 6 modulo 7 such that k^3 - 6 is divisible by 7^3 is k = 83, so a(3) = 122.
PROG
(PARI) a(n) = lift(sqrtn(6+O(7^n), 3))
CROSSREFS
Approximations of p-adic cubic roots:
The successive approximations up to 2^n for 2-adic integer 3^(1/3).
+10
5
0, 1, 3, 3, 11, 27, 59, 123, 123, 379, 379, 379, 379, 4475, 12667, 29051, 61819, 127355, 127355, 127355, 127355, 127355, 2224507, 2224507, 2224507, 19001723, 52556155, 119665019, 253882747, 253882747, 253882747, 1327624571, 3475108219, 7770075515
COMMENTS
a(n) is the unique solution to x^3 == 3 (mod 2^n) in the range [0, 2^n - 1].
FORMULA
For n > 0, a(n) = a(n-1) if a(n-1)^3 - 3 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).
EXAMPLE
11^3 = 1331 = 83*2^4 + 3;
27^3 = 19683 = 615*2^5 + 3;
59^3 = 205379 = 3209*2^6 + 3.
PROG
(PARI) a(n) = lift(sqrtn(3+O(2^n), 3))
CROSSREFS
For the digits of 3^(1/3), see A323000. Approximations of p-adic cubic roots:
this sequence (2-adic, 3^(1/3));
The successive approximations up to 2^n for 2-adic integer 5^(1/3).
+10
5
0, 1, 1, 5, 13, 29, 29, 93, 93, 93, 605, 1629, 3677, 3677, 3677, 20061, 20061, 20061, 151133, 151133, 151133, 151133, 151133, 4345437, 4345437, 21122653, 54677085, 54677085, 188894813, 457330269, 457330269, 457330269, 2604813917, 6899781213, 6899781213
COMMENTS
a(n) is the unique solution to x^3 == 5 (mod 2^n) in the range [0, 2^n - 1].
FORMULA
For n > 0, a(n) = a(n-1) if a(n-1)^3 - 5 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).
EXAMPLE
13^3 = 2197 = 137*2^4 + 5;
29^3 = 24389 = 762*2^5 + 5 = 381*2^6 + 5;
93^3 = 804357 = 6284*2^7 + 5 = 3142*2^8 + 5 = 1571*2^9 + 5.
PROG
(PARI) a(n) = lift(sqrtn(5+O(2^n), 3))
CROSSREFS
For the digits of 5^(1/3), see A323045. Approximations of p-adic cubic roots:
this sequence (2-adic, 5^(1/3));
The successive approximations up to 2^n for 2-adic integer 7^(1/3).
+10
5
0, 1, 3, 7, 7, 23, 23, 23, 151, 407, 407, 1431, 3479, 3479, 11671, 11671, 44439, 109975, 241047, 503191, 1027479, 2076055, 2076055, 6270359, 6270359, 6270359, 6270359, 6270359, 6270359, 274705815, 811576727, 1885318551, 1885318551, 6180285847
COMMENTS
a(n) is the unique solution to x^3 == 7 (mod 2^n) in the range [0, 2^n - 1].
FORMULA
For n > 0, a(n) = a(n-1) if a(n-1)^3 - 7 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).
EXAMPLE
7^3 = 343 = 21*2^4 + 7;
23^3 = 12167 = 380*2^5 + 7 = 190*2^6 + 7 = 95*2^7 + 7;
151^3 = 3442951 = 13449*2^8 + 7.
PROG
(PARI) a(n) = lift(sqrtn(7+O(2^n), 3))
CROSSREFS
For the digits of 7^(1/3), see A323095. Approximations of p-adic cubic roots:
this sequence (2-adic, 7^(1/3));
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