A262114 -id:A262114 - cp's OEIS frontend

A262115 Irregular triangle read by rows: row b (b >= 2) gives periodic part of digits of the base-b expansion of 1/7.

0, 0, 1, 0, 1, 0, 2, 1, 2, 0, 2, 1, 0, 3, 2, 4, 1, 2, 0, 5, 1, 1, 1, 2, 5, 1, 4, 2, 8, 5, 7, 1, 6, 3, 1, 8, 6, 10, 3, 5, 1, 11, 2, 2, 2, 4, 9, 2, 7, 4, 14, 9, 12, 2, 10, 5, 2, 13, 10, 16, 5, 8, 2, 17, 3, 3, 3, 6, 13, 3, 10, 6, 20, 13, 17, 3, 14, 7, 3, 18, 14, 22, 7, 11, 3, 23, 4, 4, 4, 8, 17

Offset: 2

Michael De Vlieger, Sep 11 2015

The number of terms associated with a particular value of b are cyclical: 3, 5, 3, 5, 2, 1, 1, repeat. This is because the values are associated with b (mod 7), starting with 2 (mod 7).
The expansion of 1/7 either terminates after one digit when b == 0 (mod 7) or is purely recurrent in all other cases of b (mod 7), since 7 is prime and must either divide or be coprime to b.
The period for purely recurrent expansions of 1/7 must be a divisor of Euler's totient of 7 = 6, i.e., one of {1, 2, 3, 6}.
b == 0 (mod 7): 1 (terminating)
b == 1 (mod 7): 1 (purely recurrent)
b == 2 (mod 7): 3 (purely recurrent)
b == 3 (mod 7): 6 (purely recurrent)
b == 4 (mod 7): 3 (purely recurrent)
b == 5 (mod 7): 6 (purely recurrent)
b == 6 (mod 7): 2 (purely recurrent)
The expansion of 1/7 has a full-length period 6 when base b is a primitive root of p = 7.
Digits of 1/7 for the following bases:
2    0, 0, 1
3    0, 1, 0, 2, 1, 2
4    0, 2, 1
5    0, 3, 2, 4, 1, 2
6    0, 5
7*   1
8    1
9    1, 2, 5
10   1, 4, 2, 8, 5, 7
11   1, 6, 3
12   1, 8, 6, 10, 3, 5
13   1, 11
14*  2
15   2
16   2, 4, 9
17   2, 7, 4, 14, 9, 12
18   2, 10, 5
19   2, 13, 10, 16, 5, 8
20   2, 17
21*  3
...
Asterisks above denote terminating expansion; all other entries are digits of purely recurrent reptends.
Each entry associated with base b with more than one term has a second term greater than the first except for b = 2, where the first two terms are 0, 0.
Entries for b == 0 (mod 7) (i.e., integer multiples of 7) appear at 21, 43, 65, ..., every 22nd term thereafter.

			For b = 8, 1/7 = .111..., contributing the term 1 to the sequence.
For b = 9, 1/7 = .125125..., thus 1, 2, 5 are the next terms in the sequence.
For b = 10, 1/7 = .142857142857..., thus 1, 4, 2, 8, 5, 7 are terms that follow in the sequence.

U. Dudley, Elementary Number Theory, 2nd ed., Dover, 2008, pp. 119-126.
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 6th ed., Oxford Univ. Press, 2008, pp. 138-148.
Oystein Ore, Number Theory and Its History, Dover, 1988, pp. 311-325.

Michael De Vlieger, Table of n, a(n) for n = 2..10000
Eric Weisstein's World of Mathematics, Decimal Period.
Eric Weisstein's World of Mathematics, Repeating Decimal.

Cf. A004526 Digits of expansions of 1/2.
Cf. A026741 Full reptends of 1/3.
Cf. A130845 Digits of expansions of 1/3 (eliding first 2 terms).
Cf. A262114 Digits of expansions of 1/5.

F:= proc(N) # to get rows for bases 2 to N, flattened.
  local b, R, p, L;
  R:= NULL;
  for b from 2 to N do
    if b mod 7 = 0 then
      R:= R, b/7
    else
      p:= numtheory:-order(b, 7);
      L:= convert((b^p-1)/7, base, b);
      if nops(L) < p then L:= [op(L), 0$ (p - nops(L))] fi;
      R:= R, op(ListTools:-Reverse(L));
    fi
  od:
  R;
end proc:
F(100); # Robert Israel, Dec 04 2015

RotateLeft[Most@ #, Last@ #] &@ Flatten@ RealDigits[1/7, #] & /@ Range[2, 30] // Flatten (* Michael De Vlieger, Sep 11 2015 *)

Conjectures from Colin Barker, Oct 09 2015: (Start)
a(n) = 2*a(n-22) - a(n-44) for n>44.
G.f.: x^3*(x^39 +x^38 +x^37 +x^36 +2*x^35 +2*x^34 +2*x^33 +x^32 +x^31 +2*x^30 +x^29 +3*x^28 +3*x^27 +4*x^26 +2*x^25 +2*x^24 +x^23 +3*x^22 +2*x^21 +x^20 +x^19 +x^18 +5*x^17 +2*x^15 +x^14 +4*x^13 +2*x^12 +3*x^11 +x^9 +2*x^8 +2*x^6 +x^5 +2*x^4 +x^2 +1) / (x^44 -2*x^22 +1).
(End)
From Robert Israel, Dec 04 2015: (Start)
To prove the recursion, note that if a(n) is the k'th digit in the base-b expansion of 1/7, then a(n+22) and a(n+44) are the corresponding digits in the base-(b+7) and base-(b+14) expansions.
The one digit in the base-(7k) expansion of 1/7 is k.
For each d from 1 to 6, one can show that the digits in the base-(7k+d) expansion of ((7k+d)^p - 1)/7 where p is the order of d mod 7, and thus the digits of 1/7, are linear expressions in k.
Thus for d=3, these digits are [5k+2, 4k+1, 6k+2, 2k, 3k+1, k], since those are nonnegative integers < 7k+3 and (5k+2) + (4k+1)*(7k+3) + (6k+2)*(7k+3)^2 + (2k)*(7k+3)^3 + (3k+1)*(7k+3)^4 + k*(7k+3)^5 = ((7*k+3)^6 - 1)/7.
The g.f. follows from the recursion. (End)

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A262115 Irregular triangle read by rows: row b (b >= 2) gives periodic part of digits of the base-b expansion of 1/7.

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