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A286367
Compound filter: a(n) = P(A001511(n), A286364(n)), where P(n,k) is sequence A000027 used as a pairing function.
3
1, 3, 2, 6, 4, 5, 2, 10, 22, 8, 2, 9, 4, 5, 11, 15, 4, 30, 2, 13, 121, 5, 2, 14, 46, 8, 407, 9, 4, 17, 2, 21, 121, 8, 11, 39, 4, 5, 11, 19, 4, 138, 2, 9, 67, 5, 2, 20, 22, 57, 11, 13, 4, 437, 11, 14, 121, 8, 2, 24, 4, 5, 2212, 28, 211, 138, 2, 13, 121, 17, 2, 49, 4, 8, 92, 9, 121, 17, 2, 26, 7261, 8, 2, 156, 211, 5, 11, 14, 4, 80, 11, 9, 121, 5, 11, 27, 4, 30
OFFSET
1,2
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 the highest power of 2 dividing n. Note that this is more information than A286365, as it stores only the parity of the exponent of 2.
For all i, j: a(i) = a(j) => A286161(i) = A286161(j).
LINKS
Eric Weisstein's World of Mathematics, Pairing Function
FORMULA
a(n) = (1/2)*(2 + ((A001511(n)+A286364(n))^2) - A001511(n) - 3*A286364(n)).
PROG
(Scheme) (define (A286367 n) (* (/ 1 2) (+ (expt (+ (A001511 n) (A286364 n)) 2) (- (A001511 n)) (- (* 3 (A286364 n))) 2)))
(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 a001511(n): return 2 + bin(n - 1)[2:].count("1") - bin(n)[2:].count("1")
def a(n): return T(a001511(n), a286364(n)) # Indranil Ghosh, May 09 2017
KEYWORD
nonn
AUTHOR
Antti Karttunen, May 08 2017
STATUS
approved