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A346527 a(n) is the largest number such that there are precisely n squares between a(n) and 2*a(n) inclusive. 2
17, 40, 84, 127, 199, 264, 364, 449, 577, 684, 799, 967, 1104, 1300, 1457, 1624, 1860, 2047, 2311, 2520, 2812, 3041, 3280, 3612, 3871, 4231, 4512, 4801, 5201, 5512, 5940, 6271, 6727, 7080, 7441, 7937, 8320, 8844, 9247, 9660, 10224, 10657, 11249, 11704, 12324 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Note that the number of squares between k and 2*k is floor(sqrt(2*k))-floor(sqrt(k-1)), which may be bounded below by (sqrt(2)-1)*sqrt(k-1)-1, so we know a(n) < (3+2*sqrt(2))*(n+1)^2.
Either 2*a(n)+1 or 2*a(n)+2 is square.
Neither a(n) nor a(n)+1 is square.
LINKS
FORMULA
The sequence grows like (3+2*sqrt(2))*n^2, or around 5.8*n^2 (see A156035).
EXAMPLE
The first term is 17 as there is only one square between 17 and 34 inclusive, namely 5^2. Any number k larger than 17 will have at least 2 squares between k and 2*k inclusive.
The third term is 84 as there are 3 squares between 84 and 168, namely 10^2, 11^2, and 12^2. Any number k larger than 84 will have at least 4 squares between k and 2*k inclusive.
PROG
(Python)
k = 1
n = 1
while n<100:
if math.isqrt(2*k)-math.isqrt(k-1) == n:
t = k
k2 = t
while t < 6*(n+1)**2:
if math.isqrt(2*t)-math.isqrt(t-1) == n:
k2 = t
t = t+1
n = n+1
print(k2)
k = k+1
(Python)
from math import isqrt
def A346527(n):
a, b, k, k2, m, r = -6*(n+1)**2, (n+1)**4, 2, 4, 1, 0
while 2*m+a < 0 or m*(m+a)+b < 0:
if isqrt(2*m) - isqrt(m-1) == n:
r = m
k += 1
k2 += 2*k-1
m = (k2-1)//2
return r # Chai Wah Wu, Sep 28 2021
CROSSREFS
Largest index where A105224 takes value n.
Sister sequence of A346522.
Sequence in context: A070143 A274555 A140679 * A152817 A259683 A078553
KEYWORD
easy,nonn
AUTHOR
STATUS
approved

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Last modified May 19 14:26 EDT 2024. Contains 372697 sequences. (Running on oeis4.)