A074787 Sum of squares of the number of unitary divisors of k from 1 to n.

1, 5, 9, 13, 17, 33, 37, 41, 45, 61, 65, 81, 85, 101, 117, 121, 125, 141, 145, 161, 177, 193, 197, 213, 217, 233, 237, 253, 257, 321, 325, 329, 345, 361, 377, 393, 397, 413, 429, 445, 449, 513, 517, 533, 549, 565, 569, 585, 589, 605, 621, 637, 641, 657, 673

Offset: 1

Author

Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Sep 07 2002

Keywords

• nonn

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A034444, A064608, A069212, A285052, A319592.

Equals 4*A069811(n) + 1, for n <= 29.

Programs

Maple

with(numtheory): seq(add(2^(2*nops(ifactors(k)[2])),k=1..n),n=1..100);

Mathematica

Accumulate[Table[Count[Divisors[n],?(GCD[#,n/#]==1&)],{n,60}]^2] (* _Harvey P. Dale, Dec 06 2012 *)
Accumulate[Table[4^PrimeNu[n], {n, 1, 50}]] (* Amiram Eldar, Jul 02 2022 *)

Formula

a(n) = Sum_{k=1..n} ud(k)^2 = Sum_{k=1..n} A034444(k)^2 . a(n) = Sum_{k=1..n} 2^(2*omega(k)) = Sum_{k=1..n} 2^(2*A001221(k)).

a(n) ~ c * n * log(n)^3, where c = (1/6) * Product_{p prime} ((1-1/p)^3*(1+3/p)) = A319592 / 6. - Amiram Eldar, Jul 02 2022