cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A130507 First differences of A130845.

Original entry on oeis.org

0, 0, 1, 0, 0, 0, 2, -1, 0, 0, 3, -2, 0, 0, 4, -3, 0, 0, 5, -4, 0, 0, 6, -5, 0, 0, 7, -6, 0, 0, 8, -7, 0, 0, 9, -8, 0, 0, 10, -9, 0, 0, 11, -10, 0, 0, 12, -11, 0, 0, 13, -12, 0, 0, 14, -13, 0, 0, 15, -14, 0, 0, 16, -15, 0, 0, 17, -16, 0, 0, 18, -17, 0, 0, 19, -18, 0, 0, 20, -19, 0, 0, 21, -20, 0, 0, 22, -21, 0, 0, 23, -22, 0, 0, 24, -23, 0, 0, 25, -24, 0
Offset: 0

Views

Author

Paul Curtz, Aug 16 2007

Keywords

Links

Crossrefs

Cf. A130845.

Programs

  • PARI
    concat(vector(2), Vec(x^2*(1 + x + x^2 + x^3 + x^4) / ((1 - x)*(1 + x)^2*(1 + x^2)^2) + O(x^100))) \\ Colin Barker, Sep 25 2017

Formula

a(n) = (1/16)*(cos(n*Pi/2)+sin(n*Pi/2)-1)*((2n-1)*cos(n*Pi/2)-5*cos(n*Pi)+(2n-1)*sin(n*Pi/2))*(-1)^floor((n-1)/2). - Wesley Ivan Hurt, Sep 24 2017
From Colin Barker, Sep 25 2017: (Start)
G.f.: x^2*(1 + x + x^2 + x^3 + x^4) / ((1 - x)*(1 + x)^2*(1 + x^2)^2).
a(n) = -a(n-1) - a(n-2) - a(n-3) + a(n-4) + a(n-5) + a(n-6) + a(n-7) for n>6.
(End)

Extensions

One term corrected by Colin Barker, Sep 25 2017

A130717 Second differences of A130845.

Original entry on oeis.org

0, 1, -1, 0, 0, 2, -3, 1, 0, 3, -5, 2, 0, 4, -7, 3, 0, 5, -9, 4, 0, 6, -11, 5, 0, 7, -13, 6, 0, 8, -15, 7, 0, 9, -17, 8, 0, 10, -19, 9, 0, 11, -21, 10, 0, 12, -23, 11, 0, 13, -25, 12, 0, 14, -27, 13, 0, 15, -29, 14, 0, 16, -31, 15, 0, 17, -33, 16, 0, 18, -35, 17, 0, 19, -37, 18, 0, 20, -39, 19, 0, 21, -41, 20
Offset: 0

Views

Author

Paul Curtz, Aug 16 2007

Keywords

A262114 Irregular triangle read by rows: row b (b >= 2) gives periodic part of digits of the base-b expansion of 1/5.

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 2, 1, 0, 3, 1, 1, 1, 2, 5, 4, 1, 4, 6, 3, 1, 7, 2, 2, 2, 4, 9, 7, 2, 7, 10, 5, 2, 11, 3, 3, 3, 6, 13, 10, 3, 10, 14, 7, 3, 15, 4, 4, 4, 8, 17, 13, 4, 13, 18, 9, 4, 19, 5, 5, 5, 10, 21, 16, 5, 16, 22, 11, 5, 23, 6, 6, 6, 12, 25, 19, 6, 19, 26, 13, 6, 27, 7, 7, 7, 14, 29, 22, 7, 22
Offset: 2

Views

Author

Michael De Vlieger, Sep 11 2015

Keywords

Comments

The number of terms associated with a particular value of b are cyclical: 4, 4, 2, 1, 1, repeat. This is because the values are associated with b (mod 5), starting with 2 (mod 5).
The expansion of 1/5 either terminates after one digit when b == 0 (mod 5) or is purely recurrent in all other cases of b (mod 5), since 5 is prime and must either divide or be coprime to b.
The period for purely recurrent expansions of 1/5 must be a divisor of Euler's totient of 5 = 4, i.e., one of {1, 2, 4}.
b == 0 (mod 5): 1 (terminating)
b == 1 (mod 5): 1 (purely recurrent)
b == 2 (mod 5): 4 (purely recurrent)
b == 3 (mod 5): 4 (purely recurrent)
b == 4 (mod 5): 2 (purely recurrent)
The expansion of 1/5 has a full-length period 4 when base b is a primitive root of p = 5.
Digits of 1/5 for the following bases:
2 0, 0, 1, 1
3 0, 1, 2, 1
4 0, 3
5* 1
6 1
7 1, 2, 5, 4
8 1, 4, 6, 3
9 1, 7
10* 2
11 2
12 2, 4, 9, 7
13 2, 7, 10, 5
14 2, 11
15* 3
16 3
17 3, 6, 13, 10
18 3, 10, 14, 7
19 3, 15
20* 4
...
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 5) (i.e., integer multiples of 5) appear at 11, 23, 35, ..., every 12th term thereafter.

Examples

			For b = 8, 1/5 = .14631463..., thus 1, 4, 6, 3 are terms in the sequence.
For b = 10, 1/5 = .2, thus 2 is a term in the sequence.
For b = 13, 1/5 = .27a527a5..., thus 2, 7, 10, 5 are terms in the sequence.

References

  • 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.

Links

Crossrefs

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. A262115 Digits of expansions of 1/7.

Programs

  • Mathematica
    RotateLeft[Most@ #, Last@ #] &@ Flatten@ RealDigits[1/5, #] & /@ Range[2, 38] // Flatten (* Michael De Vlieger, Sep 11 2015 *)

Formula

Conjectures from Colin Barker, Oct 09 2015: (Start)
a(n) = 2*a(n-12) - a(n-24) for n>24.
G.f.: x^3*(x^19 +x^18 +x^17 +2*x^16 +2*x^15 +x^14 +2*x^13 +3*x^12 +2*x^11 +x^10 +x^9 +x^8 +3*x^7 +x^5 +2*x^4 +x^3 +x +1) / (x^24 -2*x^12 +1).
(End)

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

Original entry on oeis.org

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

Views

Author

Michael De Vlieger, Sep 11 2015

Keywords

Comments

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.

Examples

			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.

References

  • 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.

Links

Crossrefs

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.

Programs

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

Formula

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)
Showing 1-4 of 4 results.

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