A132740 -id:A132740 - cp's OEIS frontend

A132726 Triangle read by rows: T(n,k) = length of period in decimal representation of k/n, 1<=k<=n.

0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 6, 6, 6, 6, 6, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 0, 6, 6, 6, 6, 6, 6, 0, 6, 6, 6, 6, 6, 6, 0

Offset: 1

Reinhard Zumkeller, Aug 27 2007

T(n,1) = A051626(n); T(n,n) = 0;
T(n,k) = T(1,k/A050873(n,k));
T(n,k) = T(n,A132740(k)), 1<=k<=n;
T(A003592(n),k) = 0, 1<=k<=A003592(n).

Eric Weisstein's World of Mathematics, Decimal expansion
Index entries for sequences related to decimal expansion of 1/n.

A173484 a(n) = the smallest number ending in n+1 zeros divisible by n.

Original entry on oeis.org

100, 1000, 30000, 100000, 1000000, 30000000, 700000000, 1000000000, 90000000000, 100000000000, 11000000000000, 30000000000000, 1300000000000000, 7000000000000000, 30000000000000000

Offset: 1

Jaroslav Krizek, Mar 04 2010

See A173478 - A173486.

Robert Israel, Table of n, a(n) for n = 1..995

Cf. A132740.

f:= n -> 10^(n+1)*n/2^padic:-ordp(n,2)/5^padic:-ordp(n,5):
map(f, [$1..30]); # Robert Israel, Jun 28 2018

a(n) = 10^(n+1)*A132740(n). - Robert Israel, Jun 28 2018

A370757 a(n) is the least k > 0 such that 1/n and 1/k have equivalent repeating decimal digits.

Original entry on oeis.org

1, 1, 3, 1, 1, 6, 7, 1, 9, 1, 11, 3, 13, 7, 6, 1, 17, 18, 19, 1, 21, 22, 23, 6, 1, 26, 27, 7, 29, 3, 31, 1, 33, 17, 7, 36, 37, 19, 39, 1, 41, 42, 43, 44, 45, 23, 47, 3, 49, 1, 51, 13, 53, 54, 55, 7, 57, 29, 59, 6, 61, 31, 63, 1, 26, 66, 67, 17, 69, 7, 71, 72

Offset: 1

Rémy Sigrist, Feb 29 2024

In other words, a(n) is the least k > 0 such that the fractional parts of (10^i)/n and (10^j)/k are equal for some integers i, j.

a(n) is not always a divisor of n. For example, a(65) = 26 is not a divisor of 65. - David A. Corneth, Mar 01 2024

			The first terms, alongside the decimal expansion of 1/n with its repeating decimal digits in parentheses, are:
  n   a(n)  1/n
  --  ----  -----------
   1     1  1.(0)
   2     1  0.5(0)
   3     3  0.(3)
   4     1  0.25(0)
   5     1  0.2(0)
   6     6  0.1(6)
   7     7  0.(142857)
   8     1  0.125(0)
   9     9  0.(1)
  10     1  0.1(0)
  11    11  0.(09)
  12     3  0.08(3)
  13    13  0.(076923)
  14     7  0.07(142857)
  15     6  0.0(6)

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program
Index entries for sequences related to decimal expansion of 1/n

Cf. A000265 (base-2 analog), A038502 (base-3 analog), A132739 (base-5 analog), A242603 (base-7 analog).

Cf. A003592, A007732, A132740.

PARI
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\\ See Links section.
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from itertools import count
from sympy import multiplicity, n_order
def A370757(n):
    m2, m5 = (~n & n-1).bit_length(), multiplicity(5,n)
    r = max(m2,m5)
    w, m = 10**r, 10**(t:=n_order(10,n2) if (n2:=(n>>m2)//5**m5)>1 else 1)-1
    c = w//n
    s = str(m*w//n-c*m).zfill(t)
    l = len(s)
    for k in count(1):
        m2, m5 = (~k & k-1).bit_length(), multiplicity(5,k)
        r = max(m2,m5)
        w, m = 10**r, 10**(t:=n_order(10,k2) if (k2:=(k>>m2)//5**m5)>1 else 1)-1
        c = w//k
        if any(s[i:]+s[:i] == str(m*w//k-c*m).zfill(t) for i in range(l)):
            return k # Chai Wah Wu, Mar 03 2024

a(n) = 1 iff n belongs to A003592.
a(10*n) = a(n).
A007732(a(n)) = A007732(n).

A245570 Rectangular array A read by (upward) antidiagonals: A(n,k) = n/gcd(n,10^k), n,k >= 1.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 2, 3, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 7, 3, 1, 1, 3, 1, 1, 4, 7, 3, 1, 1, 3, 1, 1, 9, 2, 7, 3, 1, 1, 3, 1, 1, 1, 9, 1, 7, 3, 1, 1, 3, 1, 1, 11, 1, 9, 1, 7, 3, 1, 1, 3, 1, 1, 6, 11, 1, 9, 1, 7, 3, 1, 1, 3, 1, 1, 13, 3, 11, 1, 9, 1, 7, 3, 1, 1, 3, 1, 1

Offset: 1

L. Edson Jeffery, Jul 25 2014

Columns of A converge to A132740.

			Array A begins:
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
  3, 3, 3, 3, 3, 3, 3, 3, 3, 3, ...
  2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
  3, 3, 3, 3, 3, 3, 3, 3, 3, 3, ...
  7, 7, 7, 7, 7, 7, 7, 7, 7, 7, ...
  4, 2, 1, 1, 1, 1, 1, 1, 1, 1, ...
  9, 9, 9, 9, 9, 9, 9, 9, 9, 9, ...
  ...

Cf. A132740.

Grid[Table[n/GCD[n, 10^k], {n, 12}, {k, 12}]] (* Array *)
Flatten[Table[(n - k + 1)/GCD[n - k + 1, 10^k], {n, 13}, {k, n}]] (* Array antidiagonals flattened *)

A357756 a(n) is the least k > 0 such that A007953(nk) equals A007953((nk)^2), where A007953 is the sum of the digits.

Original entry on oeis.org

1, 1, 5, 3, 25, 2, 3, 27, 62, 1, 1, 5, 15, 27, 128, 3, 31, 17, 1, 1, 5, 9, 9, 2, 75, 4, 18, 7, 64, 5, 3, 16, 56, 3, 85, 17, 5, 27, 5, 9, 25, 9, 45, 13, 27, 1, 1, 27, 66, 54, 2, 9, 9, 18, 22, 1, 32, 15, 25, 135, 3, 18, 8, 3, 28, 9, 3, 43, 47, 72, 27, 8, 25, 126, 27

Offset: 0

Thomas Scheuerle, Oct 12 2022

A task in the German competition "Bundeswettbewerb Mathematik 2021" was to prove that for each positive integer n there exists a k such that A007953(n*k) = A007953((n*k)^2).
One of the proposed proofs uses the argument that numbers of the form m = (10^x-1)*(10^y) will have the desired property A007953(m) = A007953(m^2). Thus we need to prove that we can find for all n a k, x and y such that n*k = (10^x-1)*(10^y). Let n be of the form b*2^c*5^d with b odd and not divisible by 5, then we know that y = max(c, d). From Euler's totient theorem we know that 10^x-1 will be divisible by e if x = A000010(e) where A000010 is Euler's totient function. See the formula section for the corresponding resulting k.
a(n) will never be divisible by 10.
If n is divisible by 3 but not by 9, then a(n) is divisible by 3. - Robert Israel, Oct 13 2022

Johann Beurich, Wie kann man das beweisen? (Bundeswettbewerb Mathematik 2021), YouTube video, 2021 (in German).
Bundeswettbewerb Mathematik 2021, Die Aufgaben_und_Loesungen der 2. Runde 2021 (in German). The proof and solutions.

Cf. A000010, A007953, A051628, A058369, A060284, A132740, A178505.

f:= proc(n) local k;
   for k from 1 do if sd(n*k) = sd((n*k)^2) then return k fi od
end proc:
map(f, [$1..100]); # Robert Israel, Oct 13 2022

a(n) = {my(k = 1); while(sumdigits(n*k)!=sumdigits((n*k)^2),k++);k}

def sd(n): return sum(map(int, str(n)))
def a(n):
    k = 1
    while not sd(n*k) == sd((n*k)**2): k += 1
    return k
print([a(n) for n in range(75)]) # Michael S. Branicky, Oct 13 2022

a(A058369(n)) = 1.
a(a(n)) <= n.
a(n) <= A132740(n)*A060284(A132740(n))*10^A051628(n)/n.
or a(n) <= (10^A000010(A132740(n))-1)*10^A051628(n)/n.

cp's OEIS Frontend

A132726 Triangle read by rows: T(n,k) = length of period in decimal representation of k/n, 1<=k<=n.

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A173484 a(n) = the smallest number ending in n+1 zeros divisible by n.

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Maple

Formula

A370757 a(n) is the least k > 0 such that 1/n and 1/k have equivalent repeating decimal digits.

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PARI

Python

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A245570 Rectangular array A read by (upward) antidiagonals: A(n,k) = n/gcd(n,10^k), n,k >= 1.

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Mathematica

A357756 a(n) is the least k > 0 such that A007953(n*k) equals A007953((n*k)^2), where A007953 is the sum of the digits.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

PARI

Python

Formula

A357756 a(n) is the least k > 0 such that A007953(nk) equals A007953((nk)^2), where A007953 is the sum of the digits.