A343105 Smallest number having exactly n divisors of the form 8*k + 3.

1, 3, 27, 99, 297, 891, 1683, 8019, 5049, 17325, 15147, 99225, 31977, 190575, 136323, 121275, 95931, 3189375, 225225, 64304361, 287793, 1289925, 1686825, 15526875, 675675, 1091475, 3239775, 1576575, 2590137, 251644717004571, 2027025, 15436575, 2297295, 28676025, 33350625, 9823275, 3828825, 42879375, 760816875

Offset: 0

Author

Jianing Song, Apr 05 2021

Keywords

• nonn

Comments

Smallest index of n in A188170.

a(n) exists for all n, since 3^(2n-1) has exactly n divisors of the form 8*k + 3, namely 3^1, 3^3, ..., 3^(2n-1). This actually gives an upper bound (which is too far from reality when n is large) for a(n).

Examples

			a(4) = 297 since it is the smallest number with exactly 4 divisors congruent to 3 modulo 8, namely 3, 11, 27 and 297.
		

Links

• Bert Dobbelaere, Table of n, a(n) for n = 0..200

Crossrefs

Smallest number having exactly n divisors of the form 8*k + i: A343104 (i=1), this sequence (i=3), A343106 (i=5), A188226 (i=7).

Cf. A188170.

Programs

PARI

res(n,a,b) = sumdiv(n, d, (d%a) == b)
a(n) = for(k=1, oo, if(res(k,8,3)==n, return(k)))

Formula

a(2n-1) <= 3^(2n-2) * 11, since 3^(2n-2) * 11 has exactly 2n-1 divisors congruent to 3 modulo 8: 3^1, 3^3, ..., 3^(2n-3), 3^0 * 11, 3^2 * 11, ..., 3^(2n-2) * 11.

a(2n) <= 3^(n-1) * 187, since 3^(n-1) * 187 has exactly 2n divisors congruent to 3 modulo 8: 3^1, 3^3, ..., 3^a, 3^1 * 17, 3^3 * 17, ..., 3^a * 17, 3^0 * 11, 3^2 * 11, ..., 3^b * 11, 3^0 * 187, 3^2 * 187, ... 3^b * 187, where a is the largest odd number <= n-1 and b is the largest even number <= n-1.

Extensions

More terms from Bert Dobbelaere, Apr 09 2021