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A278116
a(n) is the largest j such that A278115(n,k) strictly decreases for k=1..j.
4
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
1,2
LINKS
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 *)
PROG
(Magma)
A:=func<n, k|Isqrt(2*n^2 div k)^2*k>;
A278116:=func<n|(exists(j){j:j in[1..#P-1]|A(n, P[j])le A278115(n, P[j+1])}
select j else #P) where P is PrimesUpTo(2*n^2)>;
[A278116(n):n in[1..103]];
(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
CROSSREFS
Cf. A278102.
This is the row length sequence for triangles A278117 and A278118.
A278119 lists first occurrences in this sequence.
Sequence in context: A214738 A353360 A098007 * A352420 A215469 A007554
KEYWORD
nonn,easy
AUTHOR
Jason Kimberley, Feb 12 2017
STATUS
approved