A101777 -id:A101777 - cp's OEIS frontend

A101778 Last term in each row of triangle referenced in A101777.

3, 5, 3, 2, 7, 5, 3, 7, 7, 5, 3, 2, 11, 7, 7, 5, 3, 13, 11, 7, 7, 5, 3, 2, 13, 13, 11, 7, 7, 5, 3, 17, 13, 13, 11, 7, 7, 5, 3, 2, 19, 17, 13, 13, 11, 7, 7, 5, 3, 19, 19, 17, 13, 13, 11, 7, 7, 5, 3, 2, 23, 19, 19, 17, 13, 13, 11, 7, 7, 5, 3, 23, 23, 19, 19, 17, 13, 13, 11, 7, 7, 5, 3, 2, 23

Offset: 1

Ray Chandler, Jan 10 2005

nonn

Jinyuan Wang, Table of n, a(n) for n = 1..10000

Cf. A100555, A101776, A101777.

A020482(k) = forprime(q=2, k, if(isprime(2*k-q), return(2*k-q)));
a(n) = {my(r=(ceil(sqrt(2*n+1)))^2-2*n+3); if(r%2==0, r=A020482(r/2), if(isprime(r-2), r-=2, r=A020482(r\2))); r; } \\ Jinyuan Wang, Jan 29 2020

a(n) = A101777(A000217(n)).

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

Original entry on oeis.org

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

Ray Chandler, Jan 10 2005

nonn

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

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

Cf. A100555, A101777, A101778.

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 *)

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

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]

cp's OEIS Frontend

A101778 Last term in each row of triangle referenced in A101777.

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