A330358 - OEIS

%I #22 Dec 31 2021 09:29:12

%S 0,0,0,0,0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,1,2,3,4,0,2,4,6,8,5,7,9,11,

%T 13,10,12,14,16,18,15,17,19,21,23,20,22,24,26,28,0,2,4,6,8,5,7,9,11,

%U 13,10,12,14,16,18,15,17,19,21,23,20,22,24,26,28,0,2,4,6,8,5,7,9,11,13,10

%N a(n) = n mod 5 + n mod 25 + ... + n mod 5^k, where 5^k <= n < 5^(k+1).

%C Conjecture: For b >= 2, consider the function s(n,b) = Sum_{1 <= b^j <= n} (n mod b^j) from p. 8 in Dearden et al. (2011). Then s(b*n + r, b) = b*s(n,b) + r*N(n,b) for 0 <= r <= b-1, where N(n,b) = floor(log_b(n)) + 1 is the number of digits in the base-b representation of n. As initial conditions, we have s(n,b) = 0 for 1 <= n <= b. (This is a generalization of a result by _Robert Israel_ in A049802.)

%C Here b = 5 and a(n) = s(n,5).

%C We have N(n,2) = A070939(n), N(n,3) = A081604(n), N(n,4) = A110591(n), and N(n,5) = A110592(n).

%C If A_b(x) = Sum_{n >= 1} s(n,b)*x^n is the g.f. of the sequence (s(n,b): n >= 1) and the above conjecture is correct, then it can be proved that A_b(x) = b * A_b(x^b) * (1-x^b)/(1-x) + x * ((b-1)*x^b - b*x^(b-1) + 1)/((1-x)^2 * (1-x^b)) * Sum_{k >= 1} x^(b^k).

%H Robert Israel, <a href="/A330358/b330358.txt">Table of n, a(n) for n = 1..10000</a>

%F Conjecture: a(5*n+r) = 5*a(n) + r*A110592(n) = 5*a(n) + r*(floor(log_5(n)) + 1) for n >= 1 and r = 0, 1, 2, 3, 4.

%F If the conjecture above is true, the g.f. A(x) satisfies A(x) = 5*(1 + x + x^2 + x^3 + x^4)*A(x^5) + x*(1 + 2*x + 3*x^2 + 4*x^3)/(1 - x^5) * Sum_{k >= 1} x^(5^k).

%p a:= n-> add(irem(n, 5^j), j=1..ilog[5](n)):

%p seq(a(n), n=1..105);  # _Alois P. Heinz_, Dec 13 2019

%t a[n_] := Sum[Mod[n, 5^j], {j, 1, Length[IntegerDigits[n, 5]] - 1}];

%t Array[a, 105] (* _Jean-François Alcover_, Dec 31 2021 *)

%o (PARI) a(n) = sum(k=1, logint(n, 5), n % 5^k);

%o for(n=1, 100, print1(a(n), ", ")); \\ (after _Michel Marcus_'s program in A049804)

%Y Cf. A049802, A049803, A049804, A070939, A081604, A110591, A110592.

%K nonn,look

%O 1,7

%A _Petros Hadjicostas_, Dec 12 2019