A244415 - OEIS

%I #29 Jul 09 2023 20:05:43

%S 0,1,0,1,0,1,0,2,0,1,0,1,0,1,0,2,0,1,0,1,0,1,0,2,0,1,0,1,0,1,0,3,0,1,

%T 0,1,0,1,0,2,0,1,0,1,0,1,0,2,0,1,0,1,0,1,0,2,0,1,0,1,0,1,0,3,0,1,0,1,

%U 0,1,0,2,0,1,0,1,0,1,0,2,0,1,0,1,0,1,0,2,0,1,0,1,0,1,0,3,0,1,0,1

%N Exponent of 4 appearing in the 4-adic value of 1/n, n >= 1, given in A240226(n).

%C See the comment under A240226 for g-adic value of x and the Mahler reference, p. 7, where this exponent is called f.

%C Note that the exponent used in the g-adic value of 1/n is also called g-adic valuation of n if g is prime. See e.g., A007814 (g=2) and A007949 (g=3) and the corresponding A006519 and A038500 for the 2-adic and 3-adic value of 1/n, respectively.

%D Kurt Mahler, p-adic numbers and their functions, second ed., Cambridge University Press, 1981.

%H Antti Karttunen, <a href="/A244415/b244415.txt">Table of n, a(n) for n = 1..65537</a>

%F a(n) = 0 if n is odd, and if n is even a(n) = f(1/n) with f(1/n) the smallest positive integer such that the highest power of 2 in n (that is A006519(n)) divides 4^f(1/n).

%F a(n) = valuation(2*n, 4). - _Andrew Howroyd_, Jul 31 2018

%F Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2/3. - _Amiram Eldar_, Jun 30 2023

%e n = 2: A006519(2) = 1, 2 divides 4^1, hence f(1/2) = 1 = a(2).

%e n = 4: A006519(4) = 2^2, 4 divides 4^1, hence f(1/4) = 1 = a(4).

%e n = 8: A006519(8) = 2^3, 8 does not divide 4^1 but 4^2, hence f(1/8) = 2 = a(8).

%t Array[IntegerExponent[2 #, 4] &, 105] (* _Michael De Vlieger_, Nov 06 2018 *)

%o (PARI) a(n) = valuation(2*n, 4); \\ _Andrew Howroyd_, Jul 31 2018

%o (Python)

%o def A244415(n): return (~n&n-1).bit_length()+1>>1 # _Chai Wah Wu_, Jul 09 2023

%Y Cf. A240226, A007814 (case g=2), A007949 (case g=3).

%K nonn,easy

%O 1,8

%A _Wolfdieter Lang_, Jun 28 2014