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Preperiodic part of the decimal expansion of 1/k as k runs through A065502.
+20
4
5, 25, 2, 1, 125, 1, 8, 0, 0, 625, 0, 5, 0, 41, 4, 0, 3, 0, 3125, 0, 0, 2, 0, 25, 0, 2, 0, 0, 208, 2, 1, 0, 0, 17, 0, 1, 0, 15625, 0, 0, 1, 0, 13, 0, 1, 1, 0, 125, 0, 1, 0, 0, 11, 0, 1, 0, 0, 1041, 0, 1
COMMENTS
Multiples of 2 or 5 generate a quotient with a preperiodic sequence of digits, for example 1/24 = 0.041666666..., and 41 is the decimal form of the preperiodic part.
Usually a(n) = A114205( A065502(n)), but the convention in A114205 that leading zeros in the periodic part are attached to the preperiodic part seems not to be used here. - R. J. Mathar, Jul 20 2012
EXAMPLE
a(14)=4 is in the sequence because 1/25 = 0.040000... and 4 is the prefix.
208 is in the sequence because 1/48 = 2083333.... and 208 is the prefix.
MAPLE
local k, s, al ;
for s from 1 do
for al from 0 to s-1 do
if (10^s-10^al) mod k = 0 then
return floor(10^al/k) ;
end if;
end do:
end do:
Prime preperiodic part of the decimal expansion of 1/k as k runs through A065502.
+20
1
5, 2, 5, 41, 3, 2, 2, 2, 17, 13, 11, 89, 7, 5, 5, 5, 41, 3, 3, 347, 3, 3, 3, 29, 2, 2, 2, 2, 26041, 2, 2, 2, 23, 2, 2, 2, 2, 2, 17, 13, 13, 1201, 11, 11, 107, 919, 89, 7, 7, 7, 7, 7, 7, 61, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 41, 4111, 3
COMMENTS
Primes in A175555 in the order of appearance. Multiples of 2 or 5 generate a quotient with a preperiodic sequence of digits, for example 1/24 = 0.041666666..., and 41 is the decimal form of the preperiodic part.
The corresponding values of n are: 2, 5, 20, 24, 28, 36, 44, 50, 56, 72, 88, 112, 136, 168, 184, ...
REFERENCES
H. Rademacher and O. Toeplitz, Von Zahlen und Figuren (Springer 1930, reprinted 1968), ch. 19, 'Die periodischen Dezimalbrueche'.
EXAMPLE
The prime 347 is in the sequence because 1/288 = .00347222222222222222...
The prime 1201 is in the sequence because 1/832 =.001201 923076 923076 ...
MAPLE
for n from 1 do
if isprime(p) then
print(p) ;
end if;
Numbers that are odd but not divisible by 5.
+10
78
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
COMMENTS
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 k such that k^(4*j) mod 10 = 1, for any j. - Gary Detlefs, Jan 03 2012 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 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
FORMULA
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
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)
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 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 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
EXAMPLE
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 + ...
MATHEMATICA
Flatten[Table[10n + {1, 3, 7, 9}, {n, 0, 19}]] (* Alonso del Arte, Jan 13 2012 *) Map[(1/2*(5*# + Mod[3*# + 2, 4] - 4))&, Range[10^3]] (* Mikk Heidemaa, Nov 23 2017 *)
PROG
(GNU bc) scale=0; for(n=1; n<=100; n++) 10*((n-1)/4)+2*((4*((n-1)%4)+1)/3)+1 /* Carl R. White, Feb 06 2008 */ (PARI) {a(n) = 2*n - 1 + (n+1) \ 4 * 2}; /* Michael Somos, Jun 15 2014 */ (Magma) [ 2*n + 2*Floor((n-3)/4) + 1: n in [1..70] ]; // Vincenzo Librandi, Aug 01 2011 (Haskell)
a045572 n = a045572_list !! (n-1)
a045572_list = filter ((/= 0) . (`mod` 5)) a005408_list
(Python)
CROSSREFS
Cf. A000035, A000042, A001065, A001589, A002275, A005114, A045797, A045798, A065502, A078923, A079998, A082768 (numbers that begin with 1, 3, 7 or 9), A085820, A099679.
Numbers that cannot be either prefixed or followed by one digit to form a prime.
+10
8
20, 32, 62, 84, 114, 126, 134, 135, 146, 150, 164, 168, 176, 185, 192, 196, 204, 210, 218, 232, 236, 240, 248, 256, 258, 282, 294, 298, 305, 314, 315, 324, 326, 328, 342, 348, 350, 356, 366, 368, 374, 375, 378, 395, 406, 408, 410, 414, 416, 418
COMMENTS
Prefixing by 0 gives the number itself, implying that a(n) is not prime.
All integers of the form 100*(21*n)^3 belong to the sequence, so it is infinite. - Mauro Fiorentini, Jan 05 2023
EXAMPLE
If you prefix 20 with any digit you will get an even number. Also 201, 203, 207 and 209 are all composite.
MATHEMATICA
okQ[n_]:=If[EvenQ[n]||Divisible[n, 5], Union[PrimeQ[10 n+{1, 3, 7, 9}]] == {False}, !PrimeQ[n]&&Union[PrimeQ[10 n+{1, 3, 7, 9}]]=={False} && Union[ PrimeQ[Table[FromDigits[Join[{i}, IntegerDigits[n]]], {i, 9}]]] == {False}]; Select[Range[500], okQ] (* Harvey P. Dale, Jul 15 2011 *)
PROG
(PARI) is(n)=my(N=10*n, D=10^#Str(n)); forstep(k=n, n+9*D, D, if(isprime(k), return(0))); !(isprime(N+1)||isprime(N+3)||isprime(N+7)||isprime(N+9)) \\ Charles R Greathouse IV, Jul 15 2011 (Python)
from sympy import isprime
def ok(n):
s = str(n)
if any(isprime(int(s+c)) for c in "1379"): return False
return not any(isprime(int(c+s)) for c in "0123456789")
Numbers k such that 3^k - 1 is not squarefree.
+10
4
2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20, 22, 24, 25, 26, 28, 30, 32, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 50, 52, 54, 55, 56, 58, 60, 62, 64, 65, 66, 68, 70, 72, 74, 75, 76, 78, 80, 82, 84, 85, 86, 88, 90, 92, 94, 95, 96, 98, 100, 102, 104, 105, 106
COMMENTS
All even numbers are present (odd square - 1 == 0 mod 4). All multiples of 5 are present, since we can factorize 3^5k - 1 as (3^5-1)*[3^5(k-1) + ... + 1], and 3^5-1=121. Similarly all multiples of 39 are present since 3^39-1 = 405255515301=3^2*7*13^2*41^2*22643. - Jon Perry, Nov 09 2014
MAPLE
select(t -> igcd(t, 10) > 1 or not numtheory:-issqrfree(3^t-1), [$1..150]); # Robert Israel, Mar 16 2017
PROG
(PARI) for(k=1, 1e3, if(!issquarefree(3^k-1), print1(k, ", ")))
(Magma) [n: n in [1..110]| not IsSquarefree(3^n-1)]; // Vincenzo Librandi, Oct 25 2014
Start with n, concatenate its trivial divisors, and repeat until a prime is reached. a(n) = 0 if no prime is ever reached.
+10
1
0, 0, 13, 0, 0, 0, 17, 0, 19, 0, 1111111111111111111, 0, 113, 0, 0, 0, 1117, 0, 11119, 0, 111121, 0, 1123, 0, 0, 0, 127, 0, 1129, 0, 131, 0
COMMENTS
a(33) has 715 digits and is too large to include.
a( A065502(n)) = 0. There are other integers for which a(n) = 0 (i.e., n = 221). The number (10^270343 - 1)/9 appears 161046 times in this sequence.
All odd primes from A096497 are in the sequence.
EXAMPLE
17 -> {1, 17} = 117 (composite) -> {1, 117} = 1117 (prime), so a(17) = 1117.
MATHEMATICA
lst = {}; Do[If[DivisorSigma[0, n] == 1 || Divisible[n, 5] || EvenQ[n], AppendTo[lst, 0], If[PrimeQ[n], n = 10^Length[IntegerDigits[n]] + n]; While[True, If[PrimeQ[n], Break[]]; n = FromDigits[Flatten[IntegerDigits[{1, n}]]]]; AppendTo[lst, n]], {n, 32}]; lst
Numbers n not coprime to 10 such that there exists an integer k > 1 where n^k contains n as its last digits in base 10.
+10
1
2, 4, 5, 6, 8, 12, 16, 24, 25, 28, 32, 36, 44, 48, 52, 56, 64, 68, 72, 75, 76, 84, 88, 92, 96, 104, 112, 125, 128, 136, 144, 152, 168, 176, 184, 192, 208, 216, 224, 232, 248, 256, 264, 272, 288, 296, 304, 312, 328
COMMENTS
Intersection of A065502 (numbers not coprime with 10) and A072495 (k-morphic numbers). Also defined as all n not coprime with 10 where there exists k > 1 such that n^k mod 10^floor(log_10(n)) = n.
For n > 1, a(n) <= a(n-1) + 2^(ceiling(log_10(a(n))) + 1) (conjectured).
For a(n) >= 10^k where k >= 1, there exists a(m) = a(n) mod 10^j where m < n and j < k.
n with d digits is in the sequence if and only if n is either divisible by 2^d but not by 5, or divisible by 5^d but not by 2.
For d >= 2 the number of terms with d digits is 4*5^(d-1) + 2^(d-1) - 4*floor(5^d/50) - floor(2^d/20) - x(d) where x(d) = 3 if d == 2 or 3 mod 4, 2 otherwise.
(End)
FORMULA
For n > 6, a(n) > 2n since no term is divisible by 10 (but all are divisible by either 2 or 5). - Charles R Greathouse IV, May 13 2015
EXAMPLE
For n = 2, we have n^5 = 2^5 = 32, whose last digit is 2 = n, so 2 is in the sequence.
For n = 3, we have n^5 = 3^5 = 243, so 3 is in the sequence.
For n = 4, we have n^3 = 4^3 = 64, so 4 is in the sequence.
...
As a counterexample, n = 41 is not in the sequence because it is coprime with 10, even though we have 41^6 = 4750104241, whose last 2 digits are 41.
MAPLE
F:= d -> (seq(seq(2^d*(5*j+i), i=1..4), j=0..5^(d-1)-1), seq(5^d*(2*j+1), j=0..2^(d-1)-1)):
sort(convert({seq(F(d), d=1..4)}, list)); # Robert Israel, May 14 2015
PROG
(Tcl) See a255577.tcl in LINKS.
a(n) = the sum of all the multiples of 2 or 5 less than or equal to 10^n.
+10
1
35, 3050, 300500, 30005000, 3000050000, 300000500000, 30000005000000, 3000000050000000, 300000000500000000, 30000000005000000000, 3000000000050000000000, 300000000000500000000000, 30000000000005000000000000, 3000000000000050000000000000
FORMULA
a(n) = 3*10^(2*n-1) + 5*10^(n-1).
G.f.: 5*x*(7 - 160*x)/(1 - 110*x + 1000*x^2). - Stefano Spezia, Apr 21 2021
EXAMPLE
a(1) = 2 + 5 + 4 + 6 + 8 + 10 = 3*10^(2*1-1) + 5*10^(1-1) = 35.
Sum of positive integers <= n which are multiples of 2 or 5.
+10
0
0, 2, 2, 6, 11, 17, 17, 25, 25, 35, 35, 47, 47, 61, 76, 92, 92, 110, 110, 130, 130, 152, 152, 176, 201, 227, 227, 255, 255, 285, 285, 317, 317, 351, 386, 422, 422, 460, 460, 500, 500, 542, 542, 586, 631, 677, 677, 725, 725, 775, 775, 827, 827, 881, 936
LINKS
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,2,-2,0,0,0,0,0,0,0,0,-1,1).
FORMULA
a(n) = sn(n,2) + sn(n,5) - sn(n,10) where sn(n,d) = (an(n,d) * (an(n,d) + d))/(2*d) and an(n,d) = d * floor(n/d).
a(n) = Sum_{k=2..n} {k if gcd(k,10) > 1}.
MATHEMATICA
Accumulate[Table[n * Boole[MemberQ[Mod[n, {2, 5}], 0]], {n, 0, 55}]] (* Amiram Eldar, Aug 09 2023 *)
PROG
(Python)
sn = lambda k, n: ((n // k)*((n // k) + 1) * k) // 2
a = lambda n: sn(2, n) + sn(5, n) - sn(10, n)
print([a(n) for n in range(1, 56)])
(PARI) a(n) = vecsum(select(x->(!(x%2) || !(x%5)), [1..n])); \\ Michel Marcus, Aug 09 2023
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