A286366 Compound filter: a(n) = 2*A286365(n) + floor(A072400(n)/4).

4, 6, 8, 4, 13, 11, 9, 6, 28, 14, 8, 8, 13, 11, 21, 4, 12, 30, 8, 13, 65, 11, 9, 11, 40, 14, 116, 9, 13, 23, 9, 6, 64, 14, 20, 28, 13, 11, 21, 14, 12, 66, 8, 8, 49, 11, 9, 8, 28, 42, 20, 13, 13, 119, 21, 11, 64, 14, 8, 21, 13, 11, 269, 4, 84, 66, 8, 12, 65, 23, 9, 30, 12, 14, 56, 8, 65, 23, 9, 13, 484, 14, 8, 65, 85, 11, 21, 11, 12, 50, 20, 9, 65, 11, 21, 11

Offset

1,1

Comments

Each term of this sequence contains, in addition to the information contained in A286365 (which packs the values of A286361(n) and A286363(n) and parity of the exponent of the highest power of 2 dividing n) also the bit-2 of A072400(n) (its third least significant bit), which is here stored as the least significant bit of a(n). Note that the whole A072400(n) can be recovered based on the other information contained in a(n). Together all this information is enough - by Lagrange's "Four Squares theorem" - to determine what is the least number of squares that add up to n. Thus it follows that for all i, j: a(i) = a(j) => A002828(i) = A002828(j).

A286369 is similar, but without the parity of the 2-adic value present.

Links

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

Formula

a(n) = 2*A286365(n) + floor(A072400(n)/4).

Prog

(Scheme) (define (A286366 n) (+ (* 2 (A286365 n)) (floor->exact (/ (A072400 n) 4))))
(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 a286365(n): return 2*a286364(n) + a007814(n)%2
def a072400(n): return int(str(int(''.join(map(str, digits(n, 4)[1:]))[::-1]))[::-1], 4)%8
def a(n): return 2*a286365(n) + int(a072400(n)/4) # Indranil Ghosh, May 09 2017

Crossrefs

Cf. A000035, A002828, A007814, A072400, A286361, A286363, A286364, A286365, A286368, A286369, A286372, A286373.

Sequence in context: A071851 A083256 A083257 * A331059 A201336 A244850

Adjacent sequences: A286363 A286364 A286365 * A286367 A286368 A286369

Keyword

nonn

Author

Antti Karttunen, May 08 2017

Status

approved