A346528 Distance between first appearance and last appearance of n in A105224, counting both end points.

17, 33, 60, 67, 102, 103, 144, 162, 186, 204, 187, 246, 264, 288, 306, 273, 348, 307, 390, 408, 432, 450, 393, 492, 510, 534, 552, 477, 594, 513, 636, 654, 678, 696, 597, 738, 756, 780, 798, 683, 840, 858, 882, 900

Offset

1,1

Comments

When the sequence A105224 is plotted, we get horizontal lines which lie roughly on a parabola and correspond to repeated values in the sequence. This sequence is the length of these lines.

Empirically, all elements seem to roughly lie on either of the lines 20.5*n or 17*n.

Links

Elvar Wang Atlason, Table of n, a(n) for n = 1..5000

Formula

a(n) = A346527(n) - A346522(n) + 1.

Example

For n = 1, the number 1 appears in the sequence A105224 at indices ranging from 1 to 17. This is because the intervals [1,2] and [17,34] each contain 1 square, and are the smallest and largest intervals, respectively, of the form [k,2*k] that contain exactly 1 square. So a(1) = 17.

Prog

(Python)
k = 1
n = 1
while n<100:
    if math.isqrt(2*k)-math.isqrt(k-1) == n:
        k1 = k
        t = k
        k2 = k
        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-k1+1)
    k = k+1
(Python)
from math import isqrt
def A346528(n):
    if n == 1: return 17
    a, b, k, k2, m, r, s = -6*(n+1)**2, (n+1)**4, 2, 4, 1, 0, 0
    while 2*m+a < 0 or m*(m+a)+b < 0:
        if isqrt(2*m) - isqrt(m-1) == n:
            r = m
        if s == 0 and isqrt(2*m+2)-isqrt(m) == n:
            s = m
        k += 1
        k2 += 2*k-1
        m = (k2-1)//2
    return r-s # Chai Wah Wu, Sep 28 2021

Crossrefs

Cf. A105224, A346527, A346522.

Sequence in context: A158057 A249356 A329919 * A286679 A116523 A168579

Adjacent sequences: A346525 A346526 A346527 * A346529 A346530 A346531

Keyword

easy,nonn

Author

Hrólfur Eyjólfsson, Elvar Wang Atlason, Jul 22 2021

Status

approved