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A373114
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Cardinality of the largest subset of {1,...,n} such that no odd number of terms from this subset multiply to a square.
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5
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0, 1, 2, 2, 3, 3, 4, 5, 5, 5, 6, 7, 8, 9, 9, 9, 10, 11, 12, 12, 13, 13, 14, 15, 15, 16, 17, 18, 19, 19, 20, 20, 20, 21, 21, 21, 22, 23, 23, 24, 25, 26, 27, 28, 29, 30, 31, 31, 31, 31, 32, 33, 34, 34, 34, 35, 36, 37, 38, 39, 40, 41, 42, 42, 42, 43, 44, 45, 46, 46, 47
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OFFSET
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1,3
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LINKS
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FORMULA
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n-2*a(n) = A360659(n) (see Footnote 2 of the linked paper of Tao).
Asymptotically, a(n)/n converges to log(1+sqrt(e)) - 2*Integral_{t=1..sqrt(e)} log(t)/(t+1) dt = A246849 ~ 0.828499... (essentially due to Granville and Soundararajan).
a(n+1)-a(n) is either 0 or 1 for any n.
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EXAMPLE
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For n=6, {2,3,5} is the largest set without an odd product being a square, so a(6)=3.
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PROG
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(Python)
import itertools
import sympy
def generate_all_completely_multiplicative_functions(primes):
combinations = list(itertools.product([-1, 1], repeat=len(primes)))
functions = []
for combination in combinations:
func = dict(zip(primes, combination))
functions.append(func)
return functions
def evaluate_function(f, n):
if n == 1:
return 1
factors = sympy.factorint(n)
value = 1
for prime, exp in factors.items():
value *= f[prime] ** exp
return value
def compute_minimum_sum(N: int):
primes = list(sympy.primerange(1, N + 1))
functions = generate_all_completely_multiplicative_functions(primes)
min_sum = float("inf")
for func in functions:
total_sum = 0
for n in range(1, N + 1):
total_sum += evaluate_function(func, n)
if total_sum < min_sum:
min_sum = total_sum
return min_sum
results = [(N - compute_minimum_sum(N)) // 2 for N in range(1, 12)]
print(", ".join(map(str, results)))
(Python)
from itertools import product
from sympy import primerange, primepi, factorint
a = dict(zip(primerange(n+1), range(c:=primepi(n))))
return n-min(sum(sum(e for p, e in factorint(m).items() if b[a[p]])&1^1 for m in range(1, n+1)) for b in product((0, 1), repeat=c)) # Chai Wah Wu, May 31 2024
(PARI)
F(n, b)={vector(n, k, my(f=factor(k)); prod(i=1, #f~, if(bittest(b, primepi(f[i, 1])-1), 1, -1)^f[i, 2]))}
a(n)={my(m=oo); for(b=0, 2^primepi(n)-1, m=min(m, vecsum(F(n, b)))); (n-m)/2} \\ adapted from Andrew Howroyd, Feb 16 2023 at A360659 by David A. Corneth, May 25 2024
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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