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A047972
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Distance of n-th prime to nearest square.
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5
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1, 1, 1, 2, 2, 3, 1, 3, 2, 4, 5, 1, 5, 6, 2, 4, 5, 3, 3, 7, 8, 2, 2, 8, 3, 1, 3, 7, 9, 8, 6, 10, 7, 5, 5, 7, 12, 6, 2, 4, 10, 12, 5, 3, 1, 3, 14, 2, 2, 4, 8, 14, 15, 5, 1, 7, 13, 15, 12, 8, 6, 4, 17, 13, 11, 7, 7, 13, 14, 12, 8, 2, 6, 12, 18, 17, 11, 3, 1, 9, 19, 20, 10, 8, 2, 2, 8, 16, 20, 21
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OFFSET
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1,4
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COMMENTS
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Conjecture: a(n) < sqrt(prime(n)) and lim_{n->infinity} a(n)/sqrt(prime(n)) = 1, where prime(n) is the n-th prime. - Ya-Ping Lu, Nov 29 2021
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LINKS
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FORMULA
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For each prime, find the closest square (preceding or succeeding); subtract, take absolute value.
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EXAMPLE
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For 13, 9 is the preceding square, 16 is the succeeding. 13-9 = 4, 16-13 is 3, so the distance is 3.
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MATHEMATICA
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a[n_] := (p = Prime[n]; ns = p+1; While[ !IntegerQ[ Sqrt[ns++]]]; ps = p-1; While[ !IntegerQ[ Sqrt[ps--]]]; Min[ ns-p-1, p-ps-1]); Table[a[n], {n, 1, 90}] (* Jean-François Alcover, Nov 18 2011 *)
Table[Apply[Min, Abs[p - Through[{Floor, Ceiling}[Sqrt[p]]]^2]], {p, Prime[Range[90]]}] (* Jan Mangaldan, May 07 2014 *)
Min[Abs[#-Through[{Floor, Ceiling}[Sqrt[#]]]^2]]&/@Prime[Range[100]] (* More concise version of program immediately above *) (* Harvey P. Dale, Dec 04 2019 *)
Rest[Table[With[{s=Floor[Sqrt[p]]}, Abs[p-Nearest[Range[s-2, s+2]^2, p]]], {p, Prime[ Range[ 100]]}]//Flatten] (* Harvey P. Dale, Apr 27 2022 *)
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PROG
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(Python)
from sympy import integer_nthroot, prime
p = prime(n)
a = integer_nthroot(p, 2)[0]
return min(p-a**2, (a+1)**2-p) # Chai Wah Wu, Apr 03 2021
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CROSSREFS
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KEYWORD
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easy,nonn,nice
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AUTHOR
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STATUS
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approved
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