

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
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OFFSET

1,1


COMMENTS

Note that the number of squares between k and 2*k is floor(sqrt(2*k))floor(sqrt(k1)), which may be bounded below by (sqrt(2)1)*sqrt(k1)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(k1) == n:
t = k
k2 = t
while t < 6*(n+1)**2:
if math.isqrt(2*t)math.isqrt(t1) == n:
k2 = t
t = t+1
n = n+1
print(k2)
k = k+1
(Python)
from math import isqrt
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(m1) == n:
r = m
k += 1
k2 += 2*k1
m = (k21)//2


CROSSREFS

Largest index where A105224 takes value n.


KEYWORD

easy,nonn


AUTHOR



STATUS

approved



