A045572 -id:A045572 - cp's OEIS frontend

A110435 Beginning with 11, least number such that every partial concatenation is a prime.

11, 3, 11, 1, 3, 3, 53, 13, 39, 9, 3, 21, 53, 79, 11, 19, 59, 27, 49, 21, 23, 211, 153, 189, 3, 161, 121, 167, 183, 193, 77, 21, 349, 107, 129, 343, 119, 241, 143, 37, 77, 31, 159, 183, 531, 1517, 7, 59, 159, 123, 9, 1513, 203, 343, 59, 9, 999

Offset: 1

Amarnath Murthy, Aug 03 2005

As n tends to infinity, does everyy term in A045572 arise infinitely many often and with same frequency? - Stefan Steinerberger, Feb 05 2006

			11,113,11311,113111,1131113,11311133 are all prime.

a(n)=A092528(n+1), n>1. [From R. J. Mathar, Aug 18 2008]

More terms from Stefan Steinerberger, Feb 05 2006

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

A178836 Numbers n such that the period of 1/n equals the period of 1/R(n), where R(n) (A004086) is the reversal of n.

Original entry on oeis.org

3, 7, 9, 11, 33, 77, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 707, 717, 727, 737, 747, 757, 767, 777, 787, 797, 909, 919, 929, 939, 949, 959, 969, 979, 989, 999, 1001, 1111, 1221, 1331, 1441, 1551

Offset: 1

Michel Lagneau, Jun 16 2010

Non-palindromic numbers are included in this sequence :

{3267, 3927, 7293, 7632,...}

			3267 is in the sequence because period (1/3267) = 66 and also period(1/7623) = 66.
3927 is in the sequence because period (1/3927) = 48 and also period(1/7293) = 48.

Cf. A004086, A045572, A002329.

with(numtheory): nn:=8000:for n from 3 to nn do: s:=0:l:=length(n):for q from 0 to l-1 do:x:=iquo(n,10^q):y:=irem(x,10):s:=s+y*10^(l-1-q): od: indic1:=0:for p from 1 to nn do:if irem(10^p, n) = 1 and gcd(n, 5) = 1 and indic1=0 then pp:=p: indic1:=1:else fi:od: indic2:=0:for p from 1 to nn do:if irem(10^p, s) = 1 and gcd(s, 5) = 1 and indic2=0 then ppp:=p:indic2:=1:else fi:od: if pp=ppp and indic1=1 and indic2=1 then print(n):else fi:od:

A241662 Numbers of the form m * 10^k where gcd(10, m) = 1 and k >= 0 and m > 0.

Original entry on oeis.org

1, 3, 7, 9, 10, 11, 13, 17, 19, 21, 23, 27, 29, 30, 31, 33, 37, 39, 41, 43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 70, 71, 73, 77, 79, 81, 83, 87, 89, 90, 91, 93, 97, 99, 100, 101, 103, 107, 109, 110, 111, 113, 117, 119, 121, 123, 127, 129, 130, 131, 133

Offset: 1

Jeppe Stig Nielsen, Apr 26 2014

Also natural numbers x such that fractions of the form x/10^r, after reduction to lowest terms, still have a power of ten in the denominator.

Superset of A045572, and every member of a(n) arises from a member of A045572 by appending zero or more '0' digits at the right.

			19/10 is already fully reduced, and has a power of 10 in its denominator, so 19 is in the sequence;
30/100 reduces to 3/10, which has a power of 10 in its denominator, so 30 is in the sequence;
12/10, 15/100, and 20/100 reduce to 6/5, 3/20, and 1/5, respectively (none of which has a power of 10 in its denominator), so 12, 15, and 20 are not in the sequence.

Cf. A064615 (uses 6 instead of 10).

for(i=1,400,if(valuation(i,2)==valuation(i,5),print1(i,", ")))

is(n)=gcd(n/10^valuation(n,10),10)==1 \\ Charles R Greathouse IV, May 14 2014

a(n) = 9n/4 + O(log n). - Charles R Greathouse IV, May 14 2014

A279535 Triangle read by rows: The number of digits in the smallest 9-repdigit that is a multiple of n and m, where n and m are coprime to 2 and 5.

Original entry on oeis.org

1, 1, 1, 6, 6, 42, 1, 3, 6, 9, 2, 2, 6, 2, 22, 6, 6, 6, 6, 6, 78, 16, 16, 48, 16, 16, 48, 272, 18, 18, 18, 18, 18, 18, 144, 342, 6, 6, 42, 6, 6, 6, 48, 18, 42, 22, 22, 66, 22, 22, 66, 176, 198, 66, 506, 3, 9, 6, 27, 6, 6, 48, 18, 18, 66, 81, 28, 28, 84, 28, 28, 84, 112, 252, 84, 308, 84, 812

Offset: 1

R. J. Mathar, Dec 14 2016

The number of digits of the smallest member of A002283 divisible by n and m, as n and m run through A045572. Numbers of the form 10^d-1 are not divisible through 5 or 2, so these are excluded in the table. Losely related to A278588.

			The 3rd smallest number coprime to 2 and 5 is A045572(3)=7. The smallest 9-repdigit divisible by 7*7=49 is 10^42-1 = A002283(42), to T(3,3)=42.
The triangle starts
   1;
   1,  1;
   6,  6, 42;
   1,  3,  6,  9;
   2,  2,  6,  2, 22;
   6,  6,  6,  6,  6, 78;
  16, 16, 48, 16, 16, 48, 272;

Cf. A002283, A002329, A045572

T(i,j) = min{d: (n*m) | (10^d-1)} where n=A045572(i) and m=A045572(j).

A306544 Any positive integer n has a smallest multiple consisting of a succession of 1's followed by a succession of 0's (A052983). This multiple is regarded as a binary number and a(n) is its conversion to base 10.

Original entry on oeis.org

2, 2, 14, 4, 2, 14, 126, 8, 1022, 2, 6, 28, 126, 126, 14, 16, 131070, 1022, 524286, 4, 126, 6, 8388606, 56, 4, 126, 268435454, 252, 536870910, 14, 65534, 32, 126, 131070, 126, 2044, 14, 524286, 126, 8, 62, 126, 4194302, 12, 1022, 8388606, 140737488355326, 112, 8796093022206

Offset: 1

Bernard Schott, Feb 22 2019

For any odd number m not divisible by 5 (A045572), Euler's theorem (lcm(9*m,10) = 1, so 10^phi(9*m) == 1 (mod 9*m); i.e., 9*m | 10^d - 1 = 9*R_d with d = phi(9*m)) guarantees that the repunit R_d is always some multiple of m.
The numbers of the form 2^i*5^j with i, j >= 0 (A003592) clearly have a multiple equal to 10^r, for r = max(i,j).
These multiples of n end in a string of one or more 0's, so all the terms of this sequence are even.
The powers 2^k are fixed points of this sequence: the smallest multiple of 2^k consisting of a succession of 1's followed by a succession of 0's is 10^k, and 10^k in base 2 is 2^k in base 10.

			The smallest multiple of the integer 7 consisting of a succession of 1's followed by a succession of 0's is 1111110, and 1111110_2 = 126_10, so a(7) = 126. This is also the case for n=13, 14, 21, 26, 33, 35, 37, ...

Makoto Kamada, Factorization of 11...11 (Repunit).

Cf. A000079, A052983, A045572, A099679, A002275, A007088, A276349, A003592, A011557, A276348.

More terms from Michel Marcus, Feb 28 2019

A344734 a(n) is the smallest divisibility checking number for the n-th number coprime to 10.

Original entry on oeis.org

1, 2, 2, 8, 1, 9, 5, 17, 2, 16, 8, 26, 3, 23, 11, 35, 4, 30, 14, 44, 5, 37, 17, 53, 6, 44, 20, 62, 7, 51, 23, 71, 8, 58, 26, 80, 9, 65, 29, 89, 10, 72, 32, 98, 11, 79, 35, 107, 12, 86, 38, 116, 13, 93, 41, 125, 14, 100, 44, 134, 15, 107, 47, 143, 16, 114, 50

Offset: 1

Zach Pollard, May 27 2021

nonn

Conjecture: the subsequence of divisibility checking numbers for numbers that end in 1 form an arithmetic sequence. This is also the case for numbers ending in 3, 7, and 9.

			For n = 6, the sixth number coprime to 10 is 13. The divisibility checking number for 13 is 9, as demonstrated below:
  13 divides 12961
  1296 - (1 * 9) = 1287
  13 divides 1287
  128 - (7 * 9) = 65
  13 divides 65.
This is useful in quickly checking divisibility by hand. 13 divides 65 implies that 13 divides 12961.

Cf. A045572 (numbers coprime to 10).

def divisibility_checking_number(a):
    if a%5 == 0 or a % 2 == 0:
        return(0)
    x = 1
    while (x*10 + 1)%a != 0 :
        x += 1
    return(x)
h = []
for i in range(1,250):
    if divisibility_checking_number(i) != 0:
        h.insert(len(h),divisibility_checking_number(i))
print(h)

Conjectures from Chai Wah Wu, Dec 29 2021: (Start)
a(n) = 2*a(n-4) - a(n-8) for n > 9.
G.f.: x*(x^8 + x^7 + x^6 + 5*x^5 - x^4 + 8*x^3 + 2*x^2 + 2*x + 1)/(x^8 - 2*x^4 + 1). (End)

cp's OEIS Frontend

A110435 Beginning with 11, least number such that every partial concatenation is a 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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A178836 Numbers n such that the period of 1/n equals the period of 1/R(n), where R(n) (A004086) is the reversal of n.

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A241662 Numbers of the form m * 10^k where gcd(10, m) = 1 and k >= 0 and m > 0.

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A279535 Triangle read by rows: The number of digits in the smallest 9-repdigit that is a multiple of n and m, where n and m are coprime to 2 and 5.

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A306544 Any positive integer n has a smallest multiple consisting of a succession of 1's followed by a succession of 0's (A052983). This multiple is regarded as a binary number and a(n) is its conversion to base 10.

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A344734 a(n) is the smallest divisibility checking number for the n-th number coprime to 10.

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