A278116 a(n) is the largest j such that A278115(n,k) strictly decreases for k=1..j.
1, 2, 3, 3, 4, 3, 2, 2, 5, 4, 4, 2, 2, 3, 3, 5, 3, 2, 2, 4, 3, 3, 2, 2, 3, 4, 6, 6, 2, 3, 4, 3, 3, 2, 2, 3, 5, 4, 4, 2, 4, 3, 4, 3, 2, 2, 3, 4, 3, 2, 2, 4, 3, 4, 3, 2, 2, 3, 4, 3, 2, 2, 3, 3, 5, 3, 2, 2, 4, 5, 4, 2, 2, 3, 3, 4, 3, 2, 3, 4, 7, 5, 2, 2, 3, 4, 2, 2, 2, 3, 5, 5, 5, 2, 2, 3, 4, 3, 2, 2, 4, 5, 3, 3, 2
Offset: 1
Links
- Jason Kimberley, Table of n, a(n) for n = 1..100000
Crossrefs
Programs
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Magma
A:=func
; A278116:=func A278115(n,P[j+1])} select j else #P) where P is PrimesUpTo(2*n^2)>; [A278116(n):n in[1..103]]; -
Mathematica
Map[1 + Length@ TakeWhile[Differences@ #, # < 0 &] &, #] &@ Table[# Floor[n Sqrt[2/#]]^2 &@ Prime@ k, {n, 105}, {k, PrimePi[2 n^2]}] (* Michael De Vlieger, Feb 17 2017 *)
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Python
def isqrt(n): if n < 0: raise ValueError('imaginary') if n == 0: return 0 a, b = divmod(n.bit_length(),2) x = 2**(a+b) while True: y = (x + n//x)//2 if y >= x: return x x = y; def next_prime(n): for p in range(n+1,2*n+1): for i in range(2,isqrt(n)+1): if p % i == 0: break else: return p return None def A278116(n): k = 0 p = 2 s2= (n**2)*p s = s2 while True: s_= s k+= 1 p = next_prime(p) s = (isqrt(s2//p)**2)*p if s > s_: break return k