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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 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

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