A215892 a(n) = 2^n - n^k, where k is the largest integer such that 2^n >= n^k.
0, 5, 0, 7, 28, 79, 192, 431, 24, 717, 2368, 5995, 13640, 29393, 0, 47551, 157168, 393967, 888576, 1902671, 3960048, 1952265, 8814592, 23788807, 55227488, 119868821, 251225088, 516359763, 344741824, 1259979967, 3221225472, 7298466623, 15635064768
Offset: 2
Keywords
Examples
a(2) = 2^2 - 2^2 = 0, a(3) = 2^3 - 3 = 5, a(4) = 2^4 - 4^2 = 0, a(5) = 2^5 - 5^2 = 7, a(6)..a(9) are 2^n - n^2, a(10)..a(15) are 2^n - n^3, a(16)..a(22) are 2^n - n^4, and so on.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 2..1000
Crossrefs
Programs
-
Magma
[2^n - n^Floor(n*Log(n, 2)): n in [2..40]]; // Vincenzo Librandi, Jan 14 2019
-
Mathematica
Table[2^n - n^Floor[n*Log[n, 2]], {n, 2, 35}] (* T. D. Noe, Aug 27 2012 *) -
Python
for n in range(2,100): a = 2**n k = 0 while n**(k+1) <= a: k += 1 print(a - n**k, end=',')
Formula
a(n) = 2^n - n^floor(n*log_n(2)), where log_n is the base-n logarithm.