cp's OEIS Frontend

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)].

%I A379474 #16 Jan 07 2025 15:55:29
%S A379474 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,
%T A379474 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,
%U A379474 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
%N 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)].
%C A379474 In the definition p^e || n indicates that p^e divides n, but p^(1+e) does not divide n.
%C A379474 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.
%H A379474 Antti Karttunen, <a href="/A379474/b379474.txt">Table of n, a(n) for n = 1..100000</a>
%H A379474 V. Siva Rama Prasad and C. Sunitha, <a href="http://nntdm.net/papers/nntdm-23/NNTDM-23-3-073-078.pdf">On quasiperfect numbers</a>, Notes on Number Theory and Discrete Mathematics, Vol. 23, 2017, No. 3, 73-78.
%e A379474 a(17) = 1 because 17 is of the form 8m+1 and its exponent 1 is of the form 4m+1.
%e A379474 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.
%e A379474 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.
%e A379474 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.
%o A379474 (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))));
%o A379474 (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)));
%Y A379474 Cf. A379949.
%K A379474 nonn
%O A379474 1,697
%A A379474 _Antti Karttunen_, Jan 07 2025