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

0,2

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

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]

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 *)

Prog

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

Crossrefs

Cf. A100555, A101777, A101778.

Sequence in context: A087842 A379258 A201206 * A189721 A202304 A047745

Adjacent sequences: A101773 A101774 A101775 * A101777 A101778 A101779

Keyword

nonn

Author

Ray Chandler, Jan 10 2005

Status

approved