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A350812
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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)).
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0
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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
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OFFSET
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1,7
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COMMENTS
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Blocks of consecutive numbers, duplicates, gaps and irregularities in the sequence explain the separated segments with small oscillations as shown by the graphs.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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