OFFSET
1,1
COMMENTS
Or numbers n >= 36 having a divisor t^2 > 1, where t=k/m, 1 <= m < k, such that n == n/t^2 (mod 11).
Or positive numbers n such that if n == 0 (mod 11), then n is divisible by 11^3 or by the square of some other prime; otherwise n is divisible by k^2, such that there is a k_1, 0 < k_1 < k with k_1^2 == k^2 (mod 11) (or, according to the comment in A130290, n is divisible by some k^2 >= 36).
For a generalization, see the Sequence Fans mailing list for Jun 13 2016 (correction Jun 14 2016).
From David A. Corneth, Jun 26 2016: (Start)
If k is a term then m * k is a term for m > 0. Hence closed under multiplication. For k > 11, k^2 is in the sequence. So k^t is as well for t > 2.
Summarizing, k is a term iff
- k is of the form k^2 for floor(11/2) < k except k = 11.
- k is of the form 11 * p^2 for p < floor(11/2)
- of the form k * t for k of one of the forms above and integer t > 0. (End)
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000
David A. Corneth, n, a(n) and j as described in name for n = 1..10000
EXAMPLE
49 is member, since 16 == 49 (mod 11) and 16*49 is a square.
108 is member, since 75 == 108 (mod 11) and 75*108 is a square.
MATHEMATICA
Select[Range@500, Function[n, AnyTrue[Range[n - 1], And[Mod[#, 11] == Mod[n, 11], IntegerQ@ Sqrt[# n]] &]]] (* Michael De Vlieger, Jun 23 2016, Version 10 *)
PROG
(PARI) is(n) = for(j=1, n-1, if(Mod(j, 11)==n && issquare(j*n), return(1))); return(0) \\ Felix Fröhlich, Jun 15 2016
(PARI) is(n)=my(f=factor(n)); f[, 2]=f[, 2]%2; t=prod(i=1, matsize(f)[1], f[i, 1] ^ f[i, 2]); for(i=1, sqrtint((n-1)\t), if(Mod(t*i^2, 11)==n, return(1))); 0 \\ David A. Corneth, Jun 26 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Shevelev, Jun 15 2016
EXTENSIONS
More terms from Felix Fröhlich, Jun 15 2016
STATUS
approved