login
A286365
Compound filter: a(n) = 2*A286364(n) + A000035(A007814(n)).
7
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
1,1
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
FORMULA
a(n) = (2*A286364(n)) + (1 - A035263(n)) = 2*A286364(n) + A000035(A007814(n)).
PROG
(Scheme) (define (A286365 n) (+ (* 2 (A286364 n)) (A000035 (A007814 n))))
(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
CROSSREFS
Cf. A286366, A286367 (similar, but contain more information).
Sequence in context: A091732 A299439 A109746 * A345061 A061020 A206369
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 08 2017
STATUS
approved