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Search: a069879 -id:a069879
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Let A_n be a square n X n matrix with entries A_n(i,j)=1 if i+j is prime, and A_n(i,j)=0 otherwise. Then a(n) counts the 1's in A_n.
+10
2
1, 3, 5, 9, 11, 15, 19, 23, 29, 37, 43, 51, 57, 63, 71, 81, 89, 97, 105, 113, 123, 135, 145, 157, 169, 181, 195, 209, 221, 235, 249, 263, 277, 293, 309, 327, 345, 363, 381, 401, 419, 439, 457, 475, 495, 515, 533, 551, 571, 591, 613, 637, 659, 683, 709, 735
OFFSET
1,2
COMMENTS
Bertrand's postulate guarantees for every integer n the existence of at least one prime q with n < q < 2n. Equivalently, A(n) has at least one skew diagonal below the main skew diagonal whose entries will be equal to 1.
LINKS
William Dowling and Nadia Lafreniere, Homomesy on permutations with toggling actions, arXiv:2312.02383 [math.CO], 2023. See page 10.
FORMULA
From Alois P. Heinz, Sep 29 2017: (Start)
a(n) = a(n-1) + 2 * (pi(2*n-1) - pi(n)) for n > 1, a(1) = 1.
a(n) = A069879(n) + 1 = 2*A071917(n) + 1. (End)
a(n) = Sum_{i=1..n} (pi(n+i) - pi(i)), where pi = A000720. - Ridouane Oudra, Aug 29 2019
a(n) = Sum_{p <= 2n+1, p prime} min(p-1, 2n+1-p). - Ridouane Oudra, Oct 30 2023
EXAMPLE
|1 1 0 1 0|
|1 0 1 0 1|
A_5 = |0 1 0 1 0| and so a(5) = 11.
|1 0 1 0 0|
|0 1 0 0 0|
MAPLE
with(numtheory):
a:= proc(n) option remember; `if`(n=1, 1,
a(n-1)+2*(pi(2*n-1)-pi(n)))
end:
seq(a(n), n=1..80); # Alois P. Heinz, Sep 29 2017
MATHEMATICA
A[n_] := Table[Boole[PrimeQ[i + j]], {i, 1, n}, {j, 1, n}]; a[n_] := Count[Flatten[A[n]], 1];
(* or, after Alois P. Heinz (200 times faster): *)
a[1] = 1; a[n_] := a[n] = a[n-1] + 2(PrimePi[2n-1] - PrimePi[n]);
Array[a, 80] (* Jean-François Alcover, Sep 29 2017 *)
PROG
(Python)
from sympy import primepi
from sympy.core.cache import cacheit
@cacheit
def a(n): return 1 if n==1 else a(n - 1) + 2*(primepi(2*n - 1) - primepi(n))
print([a(n) for n in range(1, 51)]) # Indranil Ghosh, Dec 13 2017, after Alois P. Heinz
(Magma) sol:=[]; for n in [1..56] do k:=0; for i, j in [1..n] do if IsPrime(i+j) then k:=k+1; end if; end for; Append(~sol, k); end for; sol; // Marius A. Burtea, Aug 29 2019
(PARI) first(n) = {my(res = vector(n), pn = 0, p2n1 = 1); res[1] = 1; for(i = 2, n,
if(isprime(i), pn++); if(isprime(2*i-1), p2n1++); res[i] = res[i-1] + 2*(p2n1 - pn)); res} \\ David A. Corneth, Aug 31 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Anthony Hernandez, Sep 26 2017
STATUS
approved

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