A188169 -id:A188169 - cp's OEIS frontend

1, 9, 81, 153, 891, 1377, 8019, 3825, 11025, 15147, 88209, 31977, 354375, 99225, 121275, 95931, 7144929, 187425, 893025, 287793, 1403325, 1499553, 1715175, 675675, 1091475, 6024375, 1576575, 1686825, 72335025, 2027025, 2264802453041139, 2297295, 11609325, 121463793, 9823275

Offset: 1

Jianing Song, Apr 05 2021

nonn

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).
If for some k, A188169(k) > m then A188169(k*t) > m for all t > 0. This can be used to trim searches when looking for some a(m).
If gcd(k, m) = 1 then A188169(k) * A188169(m) <= A188169(k*m) (End)

			a(4) = 153 since it is the smallest number with exactly 4 divisors congruent to 1 modulo 8, namely 1, 9, 17 and 153.

Smallest number having exactly n divisors of the form 8*k + i: this sequence (i=1), A343105 (i=3), A343106 (i=5), A188226 (i=7).
Cf. A188169.
Cf. A007519, A007520, A007521, A007522.

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))))

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.

More terms from David A. Corneth, Apr 06 2021

cp's OEIS Frontend

A343104 Smallest number having exactly n divisors of the form 8*k + 1.

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