A101776 Smallest k such that k^2 is equal to the sum of n not-necessarily-distinct primes plus 1.

1, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13

Offset: 0

Author

Ray Chandler, Jan 10 2005

Keywords

• nonn

Comments

Pattern appears to be: one 1, one 2, three 3's, three 4's, ..., (2k+1) (2k+1)'s, (2k+1) (2k+2)'s.

It appears that a(n) is also the number of pixels in C_{n}, a pixelated arc of circle x^2 + y^2 = n, defined as the set of the (x, y), ordered pairs of nonnegative integers, such that (x^2 + y^2 = n) or ((x^2 + y^2 < n) and ((x+1)^2 + y^2 > n or x^2 + (y+1)^2 > n)). - Luc Rousseau, Dec 30 2019

Links

• Jinyuan Wang, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A100555, A101777, A101778.

Programs

Mathematica

iMax[k_,n_]:=PrimePi[k^2-2*n+1]
f[k_,n_]:=IntegerPartitions[k^2-1,{n},Table[Prime[i],{i,1,iMax[k,n]}]]
a[n_]:=Module[{k=1},While[f[k,n]=={},k++];k]
Table[a[n],{n,0,100}]
(* Luc Rousseau, Dec 30 2019 *)

PARI

a(n) = ceil(sqrt(2*n+1)); \\ Jinyuan Wang, Jan 28 2020

Formula

a(n) = sqrt(A100555(n)).

a(n) = ceiling(sqrt(2*n+1)). - Mohammad K. Azarian, Jun 15 2016 [Proof: for any k > 1 and 1 <= m <= 2*k, a(2*k^2-2*k+m) = 2*k because (2*k-1)^2 < 2*(2*k^2-2*k+m) + 1 and (2*k)^2 = 2*(2*k^2-6*k+3*m+1) + 3*(4*k-2*m-1) + 1; a(2*k^2+m) = 2*k + 1 because (2*k)^2 < 2*(2*k^2+m) + 1 and (2*k+1)^2 = 2*(2*k^2-4*k+3*m) + 3*(4*k-2*m) + 1. Therefore, a(n) = ceiling(sqrt(2*n+1)) for n >= 5. Note that the formula is also correct for n < 5, hence a(n) = ceiling(sqrt(2*n+1)). - Jinyuan Wang, Jan 28 2020]