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

Author

Enoch Haga, Dec 11 1999

Keywords

• easy
• nonn
• nice

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

Examples

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

Links

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

Crossrefs

Cf. A047973.

Programs

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 *)

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

Formula

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