A286364 Compound filter: a(n) = P(A286361(n), A286363(n)), where P(n,k) is sequence A000027 used as a pairing function.

1, 1, 2, 1, 3, 2, 2, 1, 7, 3, 2, 2, 3, 2, 5, 1, 3, 7, 2, 3, 16, 2, 2, 2, 10, 3, 29, 2, 3, 5, 2, 1, 16, 3, 5, 7, 3, 2, 5, 3, 3, 16, 2, 2, 12, 2, 2, 2, 7, 10, 5, 3, 3, 29, 5, 2, 16, 3, 2, 5, 3, 2, 67, 1, 21, 16, 2, 3, 16, 5, 2, 7, 3, 3, 14, 2, 16, 5, 2, 3, 121, 3, 2, 16, 21, 2, 5, 2, 3, 12, 5, 2, 16, 2, 5, 2, 3, 7, 67, 10, 3, 5, 2, 3, 23, 3, 2, 29, 3, 5, 5, 2, 3

Offset: 1

Author

Antti Karttunen, May 08 2017

Keywords

• nonn

Comments

This sequence packs the values of A286361(n) and A286363(n) to a single value with the pairing function A000027. These two components essentially give the prime signature of 4k+1 part and the prime signature of 4k+3 part, and they can be accessed from a(n) with functions A002260 and A004736. For example, A004431 lists all such numbers that the first component is larger than one and the second component is a perfect square.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Pairing Function

Crossrefs

Cf. A000027, A002260, A004431, A004736, A038722, A267099, A286361, A286363, A286365.

Programs

Python

from sympy import factorint
from operator import mul
def P(n):
    f = factorint(n)
    return sorted([f[i] for i in f])
def a046523(n):
    x=1
    while True:
        if P(n) == P(x): return x
        else: x+=1
def A(n, k):
    f = factorint(n)
    return 1 if n == 1 else reduce(mul, [1 if i%4==k else i**f[i] for i in f])
def T(n, m): return ((n + m)**2 - n - 3*m + 2)/2
def a(n): return T(a046523(n/A(n, 1)), a046523(n/A(n, 3))) # Indranil Ghosh, May 09 2017

Scheme

(define (A286364 n) (* (/ 1 2) (+ (expt (+ (A286361 n) (A286363 n)) 2) (- (A286361 n)) (- (* 3 (A286363 n))) 2)))

Formula

a(n) = (1/2)*(2+((A286361(n)+A286363(n))^2) - A286361(n) - 3*A286363(n)).

Other identities. For all n >= 1:

a(A267099(n)) = A038722(a(n)).