A286365 Compound filter: a(n) = 2*A286364(n) + A000035(A007814(n)).

2, 3, 4, 2, 6, 5, 4, 3, 14, 7, 4, 4, 6, 5, 10, 2, 6, 15, 4, 6, 32, 5, 4, 5, 20, 7, 58, 4, 6, 11, 4, 3, 32, 7, 10, 14, 6, 5, 10, 7, 6, 33, 4, 4, 24, 5, 4, 4, 14, 21, 10, 6, 6, 59, 10, 5, 32, 7, 4, 10, 6, 5, 134, 2, 42, 33, 4, 6, 32, 11, 4, 15, 6, 7, 28, 4, 32, 11, 4, 6, 242, 7, 4, 32, 42, 5, 10, 5, 6, 25, 10, 4, 32, 5, 10, 5, 6, 15, 134, 20, 6, 11, 4, 7, 46, 7

Offset: 1

Author

Antti Karttunen, May 08 2017

Keywords

• nonn

Comments

This sequence contains, in addition to the information contained in A286364 (which packs the values of A286361(n) and A286363(n) to a single value with the pairing function A000027) also information whether the exponent of the highest power of 2 dividing n is even or odd, which is stored in the least significant bit of a(n). Thus, for example, all squares (A000290) can be obtained by listing such numbers n that a(n) is even and both A002260(a(n)/2) & A004736(a(n)/2) are perfect squares.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000035, A007814, A035263, A286361, A286362, A286363, A286364, A286369.

Cf. A286366, A286367 (similar, but contain more information).

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 a286364(n): return T(a046523(n/A(n, 1)), a046523(n/A(n, 3)))
def a007814(n): return 1 + bin(n - 1)[2:].count("1") - bin(n)[2:].count("1")
def a(n): return 2*a286364(n) + a007814(n)%2 # Indranil Ghosh, May 09 2017

Scheme

(define (A286365 n) (+ (* 2 (A286364 n)) (A000035 (A007814 n))))

Formula

a(n) = (2*A286364(n)) + (1 - A035263(n)) = 2*A286364(n) + A000035(A007814(n)).