A047972 Distance of n-th prime to nearest square.

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

Offset

1,4

Comments

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

Links

T. D. Noe, Table of n, a(n) for n = 1..1000

Formula

For each prime, find the closest square (preceding or succeeding); subtract, take absolute value.

Example

For 13, 9 is the preceding square, 16 is the succeeding. 13-9 = 4, 16-13 is 3, so the distance is 3.

Mathematica

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 *)

Prog

(Python)
from sympy import integer_nthroot, prime
def A047972(n):
    p = prime(n)
    a = integer_nthroot(p, 2)[0]
    return min(p-a**2, (a+1)**2-p) # Chai Wah Wu, Apr 03 2021

Crossrefs

Cf. A047973.

Sequence in context: A088904 A241568 A335017 * A004595 A071681 A135621

Adjacent sequences: A047969 A047970 A047971 * A047973 A047974 A047975

Keyword

easy,nonn,nice

Author

Enoch Haga, Dec 11 1999

Status

approved