OFFSET
1,2
COMMENTS
Smallest index of n in A188169.
a(n) exists for all n, since 3^(2n-2) has exactly n divisors of the form 8*k + 1, namely 3^0, 3^2, ..., 3^(2n-2). This actually gives an upper bound for a(n).
From David A. Corneth, Apr 05 2021: (Start)
All terms are odd since if a term is even then the odd part has the same number of such divisors.
No a(2*k + 1) is divisible by a prime congruent to 1 (mod 8).
FORMULA
a(2n-1) <= 3^(2n-2) * 11, since 3^(2n-2) * 11 has exactly 2n-1 divisors congruent to 1 modulo 8: 3^0, 3^2, ..., 3^(2n-2), 3^1 * 11, 3^3 * 11, ..., 3^(2n-3) * 11.
a(2n) <= 3^(n-1) * 187, since 3^(n-1) * 187 has exactly 2n divisors congruent to 1 modulo 8: 3^0, 3^2, ..., 3^b, 3^0 * 17, 3^2 * 17, ..., 3^b * 17, 3^1 * 11, 3^3 * 11, ..., 3^a * 11, 3^1 * 187, 3^3 * 187, ... 3^a * 187, where a is the largest odd number <= n-1 and b is the largest even number <= n-1.
EXAMPLE
a(4) = 153 since it is the smallest number with exactly 4 divisors congruent to 1 modulo 8, namely 1, 9, 17 and 153.
PROG
(PARI) res(n, a, b) = sumdiv(n, d, (d%a) == b)
a(n) = if(n>0, for(k=1, oo, if(res(k, 8, 1)==n, return(k))))
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Apr 05 2021
EXTENSIONS
More terms from David A. Corneth, Apr 06 2021
STATUS
approved