Voozh

A323275

Let f(p, q) denote the pair (p + q, wt(p) + wt(q)); a(n) is obtained by iterating f starting at (n, 1) until p/q is an integer (and then a(n) is that integer), or if no integer is ever reached then a(n) = -1. (Here wt is binary weight, A000120.)

1, 2, 2, 2, 2, 2, 2, 10, 7, 10, 3, 7, 5, 10, 7, 22, 6, 22, 5, 7, 8, 22, 10, 10, 8, 8, 10, 10, 6, 8, 22, 8, 22, 8, 9, 8, 10, 8, 8, 22, 10, 22, 22, 22, 22, 22, 8, 15, 22, 11, 15, 15, 22, 11, 16, 16, 22, 15, 10, 16, 15, 22, 15, 14, 22, 14, 17, 23, 40, 15, 22, 22, 40, 12, 22, 22, 16, 12, 18, 27, 18, 40, 40, 40, 22, 40, 40, 14, 18, 34

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OFFSET

1,2

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..10000

Jeffrey C. Lagarias, Wild and Wooley numbers, The American Mathematical Monthly, Vol. 113, No. 2 (2006), pp. 97-108; arXiv preprint, arXiv:math/0411141 [math.NT], 2004-2005.

EXAMPLE

(8, 1) -> (9, 2) -> (11, 3) -> (14, 5) -> (19, 5) -> (24, 5) -> (29, 4) -> (33, 5) -> (38, 4) -> (42, 4) -> (46, 4) -> (50, 5). 50/5 is an integer, so a(8) = 50/5 = 10.

MATHEMATICA

a[n_] := Divide@@ NestWhile[{Total[#], Total[DigitCount[#, 2, 1]]}&, {n, 1}, Last[#] == 1 || !Divisible@@# &]; Array[a, 100] (* Amiram Eldar, Jul 29 2025 *)

PROG

(PARI) f(v) = return([v[1]+v[2], hammingweight(v[1])+hammingweight(v[2])]);

a(n) = {my(nb = 0, v = [n, 1]); while (1, v = f(v); nb++; if (frac(q=v[1]/v[2]) == 0, return (q))); } \\ Michel Marcus, Jan 13 2019

(Python)

from itertools import count

def f(p, q): return (p+q, p.bit_count() + q.bit_count())

def a(n):

t = n, 1

return next(t[0]//t[1] for i in count(1) if (t:=f(*t))[0]%t[1] == 0)

print([a(n) for n in range(1, 91)]) # Michael S. Branicky, Dec 23 2025

CROSSREFS

Cf. A000120, A059175, A323375.