A350812 a(n) = ceiling((n-R(ceiling(n^(1/2))))^2/(n+R(ceiling(n^(1/2))))), where R(ceiling(n^(1/2))) is the digit reversal of ceiling(n^(1/2)).

0, 0, 1, 1, 1, 1, 2, 3, 3, 3, 4, 4, 5, 6, 7, 8, 7, 8, 9, 9, 10, 11, 12, 13, 14, 13, 14, 15, 16, 16, 17, 18, 19, 20, 21, 22, 21, 22, 23, 24, 25, 25, 26, 27, 28, 29, 30, 31, 32, 31, 32, 33, 34, 35, 36, 36, 37, 38, 39, 40, 41, 42, 43, 44, 43, 44, 45, 46, 47, 48, 49, 49, 50

Offset

1,7

Comments

Blocks of consecutive numbers, duplicates, gaps and irregularities in the sequence explain the separated segments with small oscillations as shown by the graphs.

Links

Table of n, a(n) for n=1..73.

Example

For n = 1, R(ceiling(n^(1/2))) = 1, thus a(1) = ceiling((1-1)^2/(1+1)) = 0.
For n = 16, R(ceiling(n^(1/2))) = 4, thus a(16) = ceiling((16-4)^2/(16+4)) = 8.
For n = 21, R(ceiling(n^(1/2))) = 5, thus a(21) = ceiling((21-5)^2/(21+5)) = 10.

Mathematica

Table[Ceiling[(n-FromDigits[Reverse[IntegerDigits[Ceiling[n^(1/2)]]]])^2/(n+FromDigits[Reverse[IntegerDigits[Ceiling[n^(1/2)]]]])], {n, 73}] (* Stefano Spezia, Jan 18 2022 *)

Prog

(PARI) a(n)=my(x=fromdigits(Vecrev(digits(ceil(sqrt(n)))))); r=ceil((n-x)^2/(n+x));
for(n=1, 2000, print1(a(n)", "))
(Python)
from math import isqrt
def R(n): return int(str(n)[::-1])
def a(n):
    root = isqrt(n)
    Rcroot = R(root) if root**2 ==n else R(root+1)
    q, r = divmod((n-Rcroot)**2, n+Rcroot)
    return q if r == 0 else q + 1
print([a(n) for n in range(1, 94)]) # Michael S. Branicky, Jan 17 2022

Crossrefs

Cf. A004086.

Sequence in context: A189674 A156250 A029108 * A181550 A134841 A071112

Adjacent sequences: A350809 A350810 A350811 * A350813 A350814 A350815

Keyword

nonn,base,easy

Author

Claude H. R. Dequatre, Jan 17 2022

Status

approved