A178505 -id:A178505 - cp's OEIS frontend

A175545 Numbers n (relatively prime to 10) such that the decimal form of the period of 1/n is prime.

3, 27, 33, 333, 369, 909, 2151, 2439, 2997, 3333, 27027, 33333, 37683, 41841, 76923, 90909, 142857, 194841, 243603, 333333

Offset: 1

Michel Lagneau, Jun 24 2010

This sequence is infinite because the numbers 3, 33, 333, ... generate the decimal form 3. The correspondant primes of this sequence such that :
{3, 37, 3, 3, 271, 11, 4649, 41, 333667, 3} are included in the sequence A178505.
The Maple program below is very slow for the numbers > 3333.

			27 is in the sequence because 1/27 = 0.037 037 ... and 37 is prime.
2997 is in the sequence because 1/2997 = 0.000333667 000333667 ... and 333667 is prime.

H. Rademacher and O. Toeplitz, Von Zahlen und Figuren (Springer 1930, reprinted 1968), ch. 19, 'Die periodischen Dezimalbrueche'.

Index entries for sequences related to decimal expansion of 1/n.

Cf. A178505, A045572, A002329.

with(numtheory): Digits:=4000:nn:=4000:for n from 3 by 2 to nn do:z:=evalf(1/n): indic:=0:for p from 1 to nn do:if irem(10^p, n) = 1 and gcd(n, 5) = 1 and indic=0 then pp:=p:indic:=1:z1:=floor(z*10^pp): else fi:od:if indic=1 and type(z1,prime)=true then print(n):else fi:od:

Extended and name corrected by T. D. Noe, Nov 18 2010

a(17)-a(20) from Ray Chandler, Apr 17 2017

A175550 Period of the decimal expansion of 1/F as F runs through the Fibonacci numbers greater than 1 and not divisible by 2 or 5.

Original entry on oeis.org

1, 6, 6, 44, 232, 84, 138, 133, 336, 396, 28656, 3016, 84, 514228, 335824, 152214, 67830, 4440, 261744, 504628, 108373609, 47124, 3295440, 2971215072, 49349664, 45240, 4438362040, 203028, 3599596, 10841042784, 104340657248, 252736776688

Offset: 1

Michel Lagneau, Jun 26 2010

The Fibonacci numbers contributing to this sequence are {3, 13, 21, 89, 233, ...}, i.e., Fibonacci(k) for k = 4, 7, 8, 11, 13, ... (A229829, starting with A229829(3)).

			For n = 1, the 1st Fibonacci number > 1 and coprime to 2 and 5 is Fibonacci(4) = 3, and period(1/3) = 1, so a(1) = 1.
For n = 2, the 2nd Fibonacci number > 1 and coprime to 2 and 5 is Fibonacci(7) = 13, and period (1/13) = 6, so a(2) = 6.

Cf. A000045, A175561, A178505, A045572, A002329, A229829.

with(combinat, fibonacci):nn:= 50:for q from 1 to nn do:n:=fibonacci(q):indic:=0:for p from 1 to n do:if irem(10^p, n) = 1 and gcd(n, 5) = 1 and indic=0 then printf(`%d, `, p):indic:=1:else fi:od:od:

Table[MultiplicativeOrder[10, n/Times @@ ({2, 5}^IntegerExponent[n, {2, 5}])], {n, Select[Fibonacci[Range[3, 70]], CoprimeQ[#, 10] &]}] (* Amiram Eldar, May 27 2024 *)

a(15) onwards from Robert G. Wilson v, Jun 29 2010

A179192 Numbers n, not relatively prime to 10, such that the decimal form of the period of 1/n is prime.

Original entry on oeis.org

12, 18, 30, 36, 45, 48, 75, 120, 180, 192, 198, 270, 288, 300, 330, 360, 450, 480, 495, 750, 768, 1152, 1200, 1584, 1800, 1875, 1920, 1980, 1998, 2304, 2700, 2880, 3000, 3072, 3300, 3330, 3600, 3690, 4500, 4800, 4950, 4995, 5625, 7500, 7680, 9090, 11520, 12000, 12288, 15840, 18000, 18432, 18750, 19200, 19800, 19980, 19998

Offset: 1

Michel Lagneau, Jul 01 2010

The sequence A175545 (numbers n such that the decimal form of the period of 1/n is prime) is only concerned with numbers n such that gcd(n,10)=1. Each number n such that gcd(n,10)<>1 generates a quotient where there exist a sequence of digits which is periodic after a finite sequence of digits, for example 1/36 = .0277777.... and 7 is periodic.
The prime numbers corresponding to this sequence are :
3, 5, 3, 7, 2, 3, 3, 3, 5, 3, 5, 37, 2, 3, 3, 7, 2, 3, 2,...

			1584 is in the sequence because 1/1584 = .0006313131313131313131... and 31 is prime.

H. Rademacher and O. Toeplitz, Von Zahlen und Figuren (Springer 1930, reprinted 1968), ch. 19, 'Die periodischen Dezimalbrueche'.

Cf. A175557, A175555, A178505, A045572, A002329

Reap[Do[p=RealDigits[1/n][[1,-1]]; If[GCD[10,n]>1 && Head[p] === List, While[p[[-1]] == 0, p=Most[p]]; If[PrimeQ[FromDigits[p]], Sow[n]]], {n, 20000}]][[2,1]]

Union of A179192 and A175545 is A061564.

Sequence corrected by T. D. Noe, Nov 18 2010

A175551 Decimal form of the period of 1/Fibonacci(n) for n such that gcd(10,Fibonacci(n)) = 1.

Original entry on oeis.org

3, 76923, 47619, 1123595505617977528089887640449438202247191

Offset: 1

Michel Lagneau, Jun 26 2010

A curiosity: the first six digits (with the first digit zero) of the 4th number, {0,1,1,2,3,5}, are the first six Fibonacci numbers!

The next number of this sequence contains 230 digits (decimal form of the period of 1/233 = 0.004291845493562231759656652360515021459227...).

			3 is in the sequence because Fibonacci(4) = 3 and 1/3 = 0.3333 => 3;
76923 is in the sequence because Fibonacci(7) = 13, and 1/13 = 0.076923 076923 => 76923.

Cf. A000045, A175545, A178505, A045572, A002329.

with(combinat, fibonacci):Digits:=100:nn:=10000:for m from 1 to nn do:n:=fibonacci(m):l:=length(n):z:=evalf(1/n): indic:=0:for p from 1 to nn do:if irem(10^p, n) = 1 and gcd(n, 5) = 1 and indic=0 then pp:=p:indic:=1:z1:=floor(z*10^pp): else fi:od:if indic=1 then printf(`%d, `, z1):else fi:od:

Name corrected by T. D. Noe, Jul 06 2010

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.

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A175545 Numbers n (relatively prime to 10) such that the decimal form of the period of 1/n is prime.

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A175550 Period of the decimal expansion of 1/F as F runs through the Fibonacci numbers greater than 1 and not divisible by 2 or 5.

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A179192 Numbers n, not relatively prime to 10, such that the decimal form of the period of 1/n is prime.

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A175551 Decimal form of the period of 1/Fibonacci(n) for n such that gcd(10,Fibonacci(n)) = 1.

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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.

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