A379474 Number of prime factors of the form p^e || n : [p == 1 (mod 8), e == 1 (mod 4)] or [p == 5 (mod 8), e == -1 (mod 4)].

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1

Offset: 1

Author

Antti Karttunen, Jan 07 2025

Keywords

• nonn

Comments

In the definition p^e || n indicates that p^e divides n, but p^(1+e) does not divide n.

In Prasad and Sunitha paper prime factors of the form {p^2e : [p == 1 (mod 8), e == 1 (mod 4)] or [p == 5 (mod 8), e == -1 (mod 4)]} are called "special factors", and it is shown there that a necessary condition for an odd primitive abundant square (A379949) to be a quasiperfect number (number x such that sigma(x) = 2x+1) is that it has an odd number of such factors.

Examples

			a(17) = 1 because 17 is of the form 8m+1 and its exponent 1 is of the form 4m+1.
a(697) = 2, as 697 = 17^1 * 41^1, a product of two primes of the form 8m+1 with exponents of the form 4m+1.
a(2125) = 2 because 2125 = 17^1 * 5^3, the first factor is a prime of the form 8m+1 with exponent of the form 4m+1, and the second factor is a prime of the form 8m+5 with exponent of the form 4m+3.
a(50881) = 3 as 50881 = 17^1 * 41^1 * 73^1, a product of three primes of the form 8m+1 with exponents of the form 4m+1.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..100000
• V. Siva Rama Prasad and C. Sunitha, On quasiperfect numbers, Notes on Number Theory and Discrete Mathematics, Vol. 23, 2017, No. 3, 73-78.

Crossrefs

Cf. A379949.

Programs

PARI

A379474(n) = if(n<=1, 0, my(f=factor(n)); sum(i=1, #f~, if(1==(f[i, 1]%8), 1==(f[i, 2]%4), if(5==(f[i, 1]%8), 3==(f[i, 2]%4), 0))));

PARI

A379474(n) = if(n<=1, 0, my(f=factor(n), is(p,e) = (((1==p) || (5==p)) && (2*e == (1+p)))); sum(i=1, #f~, is(f[i, 1]%8, f[i, 2]%4)));