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A330358
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a(n) = n mod 5 + n mod 25 + ... + n mod 5^k, where 5^k <= n < 5^(k+1).
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5
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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, 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, 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
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OFFSET
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1,7
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COMMENTS
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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.)
Here b = 5 and a(n) = s(n,5).
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).
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LINKS
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FORMULA
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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.
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).
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MAPLE
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a:= n-> add(irem(n, 5^j), j=1..ilog[5](n)):
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MATHEMATICA
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a[n_] := Sum[Mod[n, 5^j], {j, 1, Length[IntegerDigits[n, 5]] - 1}];
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PROG
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(PARI) a(n) = sum(k=1, logint(n, 5), n % 5^k);
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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