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A385825 a(n) = Sum_{k=0..n} (binomial(n, k) mod 5).

%I A385825 #33 Jul 19 2025 18:07:18
%S A385825 1,2,4,8,11,2,4,8,16,22,4,8,16,22,34,8,16,22,44,48,11,22,34,48,61,2,4,
%T A385825 8,16,22,4,8,16,32,44,8,16,32,44,68,16,32,44,88,96,22,44,68,96,122,4,
%U A385825 8,16,22,34,8,16,32,44,68,16,32,59,68,106,22,44,68,116,142
%N A385825 a(n) = Sum_{k=0..n} (binomial(n, k) mod 5).
%H A385825 Chai Wah Wu, <a href="/A385825/b385825.txt">Table of n, a(n) for n = 0..10000</a>
%H A385825 Richard Garfield and Herbert S. Wilf, <a href="https://doi.org/10.1016/0022-314X(92)90078-4">The distribution of the binomial coefficients modulo p</a>, Journal of Number Theory, 41 (1) (1992), 1-5.
%F A385825 Row sums of A095140. - _R. J. Mathar_, Jul 19 2025
%t A385825 a[n_]:=Sum[Mod[Binomial[n,k],5],{k,0,n}];Array[a,60,0] (* _James C. McMahon_, Jul 10 2025 *)
%o A385825 (Python)
%o A385825 from gmpy2 import digits
%o A385825 from sympy.abc import x
%o A385825 from sympy import Poly, rem
%o A385825 def A385825(n):
%o A385825     k = (1,2,4,3)
%o A385825     s = digits(n,5)
%o A385825     t = [s.count(str(i)) for i in range(1,5)]
%o A385825     G = 2**t[0]*(x+2)**t[1]*(2*x**2+3)**t[3]*(2*x**3+2)**t[2]
%o A385825     c = Poly(rem(G,x**4-1),x).all_coeffs()[::-1]
%o A385825     return int(sum(k[i]*c[i] for i in range(len(c)) if c[i]))
%Y A385825 Cf. A001316, A051638, A384715, A385285, A385741.
%K A385825 nonn
%O A385825 0,2
%A A385825 _Chai Wah Wu_, Jul 09 2025