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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, 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 (list; graph; refs; listen; history; text; internal format)
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
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
KEYWORD
nonn,base,easy
AUTHOR
STATUS
approved

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Last modified May 23 02:40 EDT 2024. Contains 372758 sequences. (Running on oeis4.)