The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A319555 Digits of one of the three 7-adic integers 6^(1/3) that is related to A319199. 10
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 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. LINKS Seiichi Manyama, Table of n, a(n) for n = 0..10000 Wikipedia, p-adic number FORMULA Equals A319297*(A212152-1) = A319297*A212152^2, where each A-number represents a 7-adic number. Equals A319305*(A212155-1) = A319305*A212155^2. 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 Cf. A319097, A319098, A319199. Digits of p-adic cubic roots: A290566 (5-adic, 2^(1/3)); A290563 (5-adic, 3^(1/3)); A309443 (5-adic, 4^(1/3)); A319297, A319305, this sequence (7-adic, 6^(1/3)); A321106, A321107, A321108 (13-adic, 5^(1/3)). Sequence in context: A106333 A104748 A117335 * A244980 A021863 A259620 Adjacent sequences:  A319552 A319553 A319554 * A319556 A319557 A319558 KEYWORD nonn,base AUTHOR Jianing Song, Aug 27 2019 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified June 14 20:21 EDT 2021. Contains 345038 sequences. (Running on oeis4.)