A238454 Difference between 2^(2*n-1) and the next larger square.
2, 1, 4, 16, 17, 68, 89, 356, 697, 1337, 2449, 4001, 4417, 17668, 24329, 4633, 18532, 74128, 296512, 1186048, 1778369, 1181833, 4727332, 18909328, 28184177, 17830441, 71321764, 285287056, 381898097, 9092137, 36368548, 145474192, 581896768, 2327587072, 9310348288
Offset: 1
Examples
a(1) = 4 - 2^1 = 2. a(2) = 9 - 2^3 = 1. a(3) = 36 - 2^5 = 4.
Programs
-
Mathematica
(Floor[Sqrt[#]]+1)^2-#&/@Table[2^(2n-1),{n,40}] (* Harvey P. Dale, Jul 05 2019 *) -
PARI
a(n) = my(r,s=sqrtint(1<<(2*n-1),&r)); 2*s+1-r; \\ Kevin Ryde, Oct 12 2022
-
Python
def isqrt(a): sr = 1 << (int.bit_length(int(a)) >> 1) while a < sr*sr: sr>>=1 b = sr>>1 while b: s = sr + b if a >= s*s: sr = s b>>=1 return sr def a(n): nn = 2**(2*n+1) s = isqrt(nn) return (s+1)**2-nn for n in range(77): print(str(a(n)), end=',') -
Sage
def a(n): return ceil(2^n/sqrt(2))^2 - 2^(2*n-1) # Ralf Stephan, Mar 08 2014
Formula
From Antti Karttunen, Feb 27 2014: (Start)
a(n) = ceiling(sqrt(2^(2*n-1)))^2 - 2^(2*n-1).
For all n, A000035(abs(A201125(n) - A238454(n))) = 1, because if the nearest square at the other side of 2^(2*n-1) is even, then the nearest square at the other side is odd.
(End)
Comments