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Number of divisors of repunit 111...111 (with n digits).
+10
9
1, 2, 4, 4, 4, 32, 4, 16, 12, 16, 4, 128, 8, 16, 64, 64, 4, 384, 2, 128, 128, 96, 2, 1024, 32, 64, 64, 256, 32, 8192, 8, 2048, 64, 64, 128, 3072, 8, 8, 64, 2048, 16, 24576, 16, 1536, 768, 64, 4, 8192, 16, 1024, 256, 512, 16, 8192, 256, 4096
EXAMPLE
a(9) = 12 since the divisors of 111111111 are 1, 3, 9, 37, 111, 333, 333667, 1001001, 3003003, 12345679, 37037037, 111111111.
CROSSREFS
Cf. A000005, A002275, A070528, A051064, A004023, A046413, A003020, A095370, A176973, A046053, A046412, A102380.
a(n) = sigma(10^n - 1), where sigma(n) is the sum of positive divisors of n.
+10
9
13, 156, 1520, 15912, 148512, 2042880, 14508000, 162493344, 1534205464, 16203253248, 144451398000, 2063316971520, 14903272088640, 158269280832000, 1614847741624320, 17205180696931968, 144444514193267496
CROSSREFS
Cf. A000203, A001270, A002283, A003020, A005422, A046053, A046107, A046412, A046415, A046416, A046417, A046418, A046419, A046420, A057951, A059892, A061075, A070528, A070529, A081317, A081318, A085035, A095370, A095413, A095414, A095417, A095418, A102347, A102380, A112505, A147556, A295503, A366669.
AUTHOR
Jun Mizuki (suzuki32(AT)sanken.osaka-u.ac.jp), Feb 14 2005
Squares arising as a concatenation of k and 9's complement of k.
+10
6
36, 81, 1089, 4356, 9801, 110889, 443556, 998001, 11108889, 44435556, 99980001, 1111088889, 4444355556, 9999800001, 111110888889, 444443555556, 999998000001, 11111108888889, 44444435555556, 99999980000001
COMMENTS
Numbers of the form j^2*x*(10^k-1) where x = A007913(10^k-1) and 10^(k-1)+1 <= j^2*x <= 10^k-1. If k >= 2 is not in A046412, there are only three terms with 2*k digits, namely (10^k-1)^2/9, 4*(10^k-1)^2/9, and 9*(10^k-1)^2/9. The first term not of one of those three forms is a(25)=197530863802469136.
(End)
EXAMPLE
1089 = 33^2 is a concatenation of 10 and 89, 10+89 = 99.
MAPLE
f:= proc(k) local F, x, p, t;
p:= 10^k-1;
F:= select(t -> t[2]::odd, ifactors(p)[2]);
x:= mul(t[1], t=F);
seq(j^2*x*p, j=ceil(sqrt((10^(k-1)+1)/x))..floor(sqrt(p/x)))
end proc:
AUTHOR
Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 23 2003
a(n) = smallest k such that (10^k-1)/9 == 0 mod prime(n)^2, or 0 if no such k exists.
+10
4
0, 9, 0, 42, 22, 78, 272, 342, 506, 812, 465, 111, 205, 903, 2162, 689, 3422, 3660, 2211, 2485, 584, 1027, 3403, 3916, 9312, 404, 3502, 5671, 11772, 12656, 5334, 17030, 1096, 6394, 22052, 11325, 12246, 13203, 27722, 7439, 31862, 32580, 18145, 37056, 19306
COMMENTS
For a given a(n)>0, all of the values of k such that (10^k-1)/9=0 mod prime(n)^2 is given by the sequence a(n)* A000027, i.e. integral multiples of a(n). For example, for n=2, prime(2)=3, a(n)=9, the set of values of k for which (10^k-1)/9=0 mod 3^2 is 9* A000027=9,18,27,36,45,... The union of the collection of sequences formed from the nonzero terms of a(n)* A000027, gives the values of k for which (10^k-1)/9 is not squarefree, see A046412. All of terms of the sequence a(n) are integer multiples of prime(n) for primes <1000 except for a(93)=486 where prime(93)=487. Conjecture: there are no 0 terms after a(3). That conjecture is easily proved, for a(n) is just the multiplicative order of 10 modulo (prime(n))^2 for n>3. - Jeppe Stig Nielsen, Dec 28 2015
EXAMPLE
a(2)=9 since 9 is least value of k for which (10^k-1)/9=0 mod 3^2.
MAPLE
0, 9, 0, seq(numtheory:-order(10, ithprime(i)^2), i=4..100); # Robert Israel, Dec 30 2015
PROG
(PARI) a(n)=p=prime(n); 10%p==0 && return(0); for(k=1, p^2, ((10^k-1)/9) % p^2 == 0 && return(k)); error() \\ Jeppe Stig Nielsen, Dec 28 2015 (PARI) a(n)=p=prime(n); if(10%p==0, 0, 10%p==1, 9, znorder(Mod(10, p^2))) \\ Jeppe Stig Nielsen, Dec 28 2015
111111111, 111111111111111111, 1111111111111111111111, 111111111111111111111111111, 111111111111111111111111111111111111, 111111111111111111111111111111111111111111, 11111111111111111111111111111111111111111111, 111111111111111111111111111111111111111111111
PROG
(PARI) lista(nn) = {for (n=1, nn, if (! issquarefree(r = sum(i=0, n, 10^i)), print1(r, ", ")); ); } \\ Michel Marcus, Nov 24 2013
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