%I #14 Jul 25 2025 11:39:25

%S 1,1,4,4,5,4,8,4,8,6,10,10,9,14,13,13,14,18,13,20,17,14,17,20,23,21,

%T 25,20,22,22,24,20,20,28,23,25,24,33,25,29,34,38,37,37,38,41,37,42,38,

%U 42,32,45,38,44,44,39,46,45,42,45,49,54,41,55,48,54,49,55,60,54,59,60,61,50,59,51,65,66,59,72

%N a(n) = (1/2^n) * Sum_{k=0..2*n} ( binomial(2*n, k) (mod 2^n) ).

%C What is the limit of the average value of a(n)/n as n increases? It appears to be near Pi/4.

%H Chai Wah Wu, <a href="/A376178/b376178.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..6500 from Paul D. Hanna)

%e Term a(n) equals the sum of the residues of the binomial coefficients in (1 + x)^(2*n) modulo 2^n, divided by 2^n, as illustrated below.

%e a(1) = (1 + 0 + 1)/2 = 1;

%e a(2) = (1 + 0 + 2 + 0 + 1)/2^2 = 1;

%e a(3) = (1 + 6 + 7 + 4 + 7 + 6 + 1)/2^3 = 4;

%e a(4) = (1 + 8 + 12 + 8 + 6 + 8 + 12 + 8 + 1)/2^4 = 4;

%e a(5) = (1 + 10 + 13 + 24 + 18 + 28 + 18 + 24 + 13 + 10 + 1)/2^5 = 5;

%e a(6) = (1 + 12 + 2 + 28 + 47 + 24 + 28 + 24 + 47 + 28 + 2 + 12 + 1)/2^6 = 4;

%e a(7) = (1 + 14 + 91 + 108 + 105 + 82 + 59 + 104 + 59 + 82 + 105 + 108 + 91 + 14 + 1)/2^7 = 8;

%e a(8) = (1 + 16 + 120 + 48 + 28 + 16 + 72 + 176 + 70 + 176 + 72 + 16 + 28 + 48 + 120 + 16 + 1)/2^8 = 4;

%e a(9) = (1 + 18 + 153 + 304 + 500 + 376 + 132 + 80 + 238 + 492 + 238 + 80 + 132 + 376 + 500 + 304 + 153 + 18 + 1)/2^9 = 8;

%e a(10) = (1 + 20 + 190 + 116 + 749 + 144 + 872 + 720 + 18 + 24 + 436 + 24 + 18 + 720 + 872 + 144 + 749 + 116 + 190 + 20 + 1)/2^10 = 6;

%e ...

%e SPECIFIC VALUES.

%e Sum_{n>=1} a(n)*(4/5)^n = 17.30840243003270887738767198463123109202082549...

%e Sum_{n>=1} a(n)*(3/4)^n = 10.57892430627269874867780443159552606330691496...

%e Sum_{n>=1} a(n)*(2/3)^n = 5.417748705049056891272117551452109394174599147...

%e Sum_{n>=1} a(n)/2^n = 1.828467161197177620900317695715194643663356606030162443659...

%e Sum_{n>=1} a(n)/3^n = 0.6728993248236073006477750390537792815286492439817...

%e Sum_{n>=1} a(n)/4^n = 0.3970731138083449706711724063387465668154560090032...

%e Sum_{n>=1} a(n)/5^n = 0.2803736063617441421256259884859153810306621545786...

%o (PARI) {a(n) = sum(k=0, 2*n, binomial(2*n, k) % (2^n) )/2^n}

%o for(n=1, 80, print1(a(n),", "))

%Y Cf. A376179, A376531.

%K nonn

%O 1,3

%A _Paul D. Hanna_, Oct 08 2024