A099679 -id:A099679 - cp's OEIS frontend

A045572 Numbers that are odd but not divisible by 5.

1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, 31, 33, 37, 39, 41, 43, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 77, 79, 81, 83, 87, 89, 91, 93, 97, 99, 101, 103, 107, 109, 111, 113, 117, 119, 121, 123, 127, 129, 131, 133, 137, 139, 141, 143, 147, 149, 151, 153

Offset: 1

Jeff Burch, Dec 11 1999

Contains the repunits R_n, (A000042 or A002275): For any m in the sequence (divisible by neither 2 nor 5), Euler's theorem (i.e., m | 10^m - 1 = 9*R_n) guarantees that R_n is always some multiple of m (see A099679) and thus forms a subsequence. - Lekraj Beedassy, Oct 26 2004
Inverse formula: n = 4*floor(a(n)/10) + floor((a(n) mod 10)/3) + 1. - Carl R. White, Feb 06 2008
Numbers ending with 1, 3, 7 or 9. - Lekraj Beedassy, Apr 04 2009
Complement of A065502. - Reinhard Zumkeller, Nov 15 2009
Union of evenish and oddish numbers, cf. A045797, A045798. - Reinhard Zumkeller, Dec 10 2011
Numbers k such that k^(4*j) mod 10 = 1, for any j. - Gary Detlefs, Jan 03 2012
Numbers coprime to 10. - Charles R Greathouse IV, Sep 05 2013
This is also the sequence of numbers such that all their divisors are the sum of the proper divisors of some number (see A001065 (sum of proper divisors) and A078923 (possible values of sigma(n)-n)). This is due to the fact that in the set of untouchable numbers (A005114) there are only 2 prime numbers (2 and 5) and all other terms are even composite. - Michel Marcus, Jun 14 2014
Numbers n for which A001589(n) is divisible by 5. - Bruno Berselli, Jun 18 2014
For a(n) > 1, positive integers x such that the decimal representation of 1/x is purely periodic after the decimal point (1/x is a repeating decimal with no non-repeating portion). - Doug Bell, Aug 05 2015
The asymptotic density of this sequence is 2/5. - Amiram Eldar, Oct 18 2020

			a(18) = 10*floor(17/4) + 2*floor( (4*(17 mod 4) + 1)/3 ) + 1
      = 10*4 + 2*floor( (4*(1)+1)/3 ) + 1
      = 40 + 2*floor(5/3) + 1
      = 40 + 2*1 + 1
      = 43.
G.f. = x + 3*x^2 + 7*x^3 + 9*x^4 + 11*x^5 + 13*x^6 + 17*x^7 + 19*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Relative complement of A017329 in A005408.

Cf. A000035, A000042, A001065, A001589, A002275, A005114, A045797, A045798, A065502, A078923, A079998, A082768 (numbers that begin with 1, 3, 7 or 9), A085820, A099679.

a045572 n = a045572_list !! (n-1)
a045572_list = filter ((/= 0) . (`mod` 5)) a005408_list
-- Reinhard Zumkeller, Dec 10 2011

[ 2*n + 2*Floor((n-3)/4) + 1: n in [1..70] ]; // Vincenzo Librandi, Aug 01 2011

A045572:=n->2*n + 2*floor((n-3)/4) + 1; seq(A045572(n), n=1..50); # Wesley Ivan Hurt, Jun 14 2014

Flatten[Table[10n + {1, 3, 7, 9}, {n, 0, 19}]] (* Alonso del Arte, Jan 13 2012 *)
Select[2Range@78 -1, Mod[#, 5] > 0 &] (* Robert G. Wilson v, Apr 02 2017 *)
Map[(1/2*(5*# + Mod[3*# + 2, 4] - 4))&, Range[10^3]] (* Mikk Heidemaa, Nov 23 2017 *)

a(n)=10*(n>>2)+[-1,1,3,7][n%4+1] \\ Charles R Greathouse IV, Jul 31 2011

is(n)=gcd(n,10)==1 \\ Charles R Greathouse IV, Sep 05 2013

{a(n) = 2*n - 1 + (n+1) \ 4 * 2}; /* Michael Somos, Jun 15 2014 */

def A045572(n): return 2*(n+(n+1)//4) - 1 # Chai Wah Wu, Jan 08 2022

a(n) = 10*floor((n-1)/4) + 2*floor( (4*((n-1) mod 4) + 1)/3 ) + 1; a(n) = a(n-1) + 2 + 2*floor(((x+6) mod 10)/9). - Carl R. White, Feb 06 2008
a(n) = 2*n + 2*floor((n-3)/4) + 1. - Kenneth Hammond (weregoose(AT)gmail.com), Mar 07 2008
a(n) = -1 + 2*n + 2*floor((n+1)/4). - Kenneth Hammond (weregoose(AT)gmail.com), Mar 25 2008
From R. J. Mathar, Sep 22 2009: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5).
G.f.: x*(1 + 2*x + 4*x^2 + 2*x^3 + x^4)/((1+x) * (x^2+1) * (x-1)^2). (End)
A000035(a(n))*(1 - A079998(a(n))) = 1. - Reinhard Zumkeller, Nov 15 2009
a(n) = (10*n + 2*(-1)^(n*(n+1)/2) - (-1)^n - 5)/4. - Bruno Berselli, Nov 06 2011
G.f.: x * (1 + 2*x + 4*x^2 + 2*x^3 + x^4) / ((1 - x) * (1 - x^4)). - Michael Somos, Jun 15 2014
a(1 - n) = -a(n) for all n in Z. - Michael Somos, Jun 15 2014
0 = (a(n) - 2*a(n+1) + a(n+2)) * (a(n) - 4*a(n+2) + 3*a(n+3)) for all n in Z. - Michael Somos, Jun 15 2014
From Mikk Heidemaa, Nov 22 2017: (Start)
a(n) = (1/2)*(5*n + ((3*n + 2) mod 4) - 4);
a(n) = (1/4)*((-1)^(n + 1) + 10*n + 2*cos((n*Pi)/2) - 2*sin((n*Pi)/2) - 5);
a(n) = (1/4)*((-1)^(1 + n) + (1 - i)*exp(-(1/2)*i*n*Pi) + (1 + i)*exp(i*n*Pi/2) + 10*n - 5) (for n > 0), where i is the imaginary unit. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(10-2*sqrt(5))*Pi/10. - Amiram Eldar, Dec 12 2021
E.g.f.: (2 + cos(x) + (5*x - 3)*cosh(x) - sin(x) + (5*x - 2)*sinh(x))/2. - Stefano Spezia, Dec 07 2022

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

A340204 a(n) is the smallest proper multiple of n whose digit product is the same as the digit product of n; 0 if no such number exists.

Original entry on oeis.org

11, 12, 1113, 212, 15, 132, 11711, 24, 11133, 20, 1111, 11112, 1131, 21112, 11115, 32, 71111, 11124, 133, 40, 11111121, 1122, 161, 14112, 125, 1612, 11111111172, 224, 3132, 60, 11111113, 1312, 11111133, 612, 315, 1332, 11137, 342, 11193, 80, 1111141, 11214, 11223

Offset: 1

Bernard Schott, Jan 15 2021

Every odd integer k not ending with 5 has a multiple that is a repunit (see A099679), hence a(n) <= the concatenation of this repunit with this odd number (see example a(33)).

			a(16) = 32 because 32 is the smallest proper multiple of 16 such that 1*6 = 3*2.
a(33) = 11111133 is the concatenation of 111111 (that is the smallest repunit multiple of 33) with 33.

Cf. A007954, A069035, A087304.

prodig[n_] := Times @@ IntegerDigits[n]; a[n_] := Module[{k = 2*n, p = prodig[n]}, While[prodig[k] != p, k += n]; k]; Array[a, 20] (* Amiram Eldar, Jan 15 2021 *)

f(n) = vecprod(digits(n)); \\ A007954
a(n) = my(x = f(n), k = 2); while(f(k*n) != x, k++); k*n; \\ Michel Marcus, Jan 15 2021

from math import prod
def pd(n): return prod(map(int, str(n)))
def a(n):
  pdn, f = pd(n), 2
  while pd(f*n) != pdn: f += 1
  return f*n
print([a(n) for n in range(1, 27)]) # Michael S. Branicky, Jan 16 2021

a(10*k) = 20*k.

More terms from Amiram Eldar, Jan 15 2021

A343447 Smallest m such that alternating integer 101...101 = A094028(m) is a multiple of A045572(n), (i.e., integers coprime with 10).

Original entry on oeis.org

0, 2, 2, 8, 10, 2, 7, 8, 2, 10, 26, 13, 14, 32, 2, 2, 4, 20, 22, 20, 23, 12, 8, 28, 29, 8, 32, 32, 34, 3, 32, 12, 80, 40, 41, 21, 2, 14, 47, 98, 1, 16, 52, 53, 2, 55, 8, 23, 120, 14, 20, 20, 64, 8, 3, 22, 68, 32, 20, 73, 74, 71, 38, 38, 32, 80, 82, 38, 8, 42

Offset: 1

Bernard Schott, Apr 15 2021

Every number coprime with 10 has a smallest multiple that is repunit (A099679).
Every positive number has a smallest multiple consisting of a succession of 1's followed by a succession of 0's (A052983).
Every number coprime with 10 has a smallest multiple that is alternating of the form 1010...0101 (this sequence).

			A045572(3) = 7, the smallest alternating multiple of 7 in A094028 is A094028(2) = 10101 because 1443*7 = 10101, as 1 and 101 are not divisible by 7, so a(3) = 2.

Cf. A030141, A045572, A052983, A094028, A099679.

a[n_] := Module[{k = (5*n + (Mod[3*n + 2, 4] - 4))/2, m = 0}, While[! Divisible[1 + 100*(100^m - 1)/99, k], m++]; m]; Array[a, 100] (* Amiram Eldar, Apr 15 2021 *)

a045572(n)=10*(n>>2)+[-1,1,3,7][n%4+1] \\ after Charles R Greathouse IV in A045572
a094028(n) = 1+100*(100^n-1)/99
a(n) = for(m=0, oo, if(a094028(m)%a045572(n)==0, return(m))) \\ Felix Fröhlich, Apr 15 2021

More terms from Felix Fröhlich, Apr 15 2021

cp's OEIS Frontend

A045572 Numbers that are odd but not divisible by 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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A340204 a(n) is the smallest proper multiple of n whose digit product is the same as the digit product of n; 0 if no such number exists.

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A343447 Smallest m such that alternating integer 101...101 = A094028(m) is a multiple of A045572(n), (i.e., integers coprime with 10).

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