|
%I #57 Sep 22 2020 02:45:25
%S 1,1,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,3,1,1,1,1,2,1,1,1,1,
%T 2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,3,1,1,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,
%U 1,2,1,1,1,1,3,1,1,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,2,1,1,1,1,3,1,1,1,1,2
%N 5^a(n) exactly divides 5n. Or, 5-adic valuation of 5n.
%C More generally, consider the sequence defined by p^a(n) exactly divides p*n. For p = 3 we have A051064 and for p = 2 we have A001511.
%H T. D. Noe, <a href="/A055457/b055457.txt">Table of n, a(n) for n=1..1000</a>
%H Joseph Rosenbaum, <a href="https://doi.org/10.2307/2302451">Elementary Problem E319</a>, American Mathematical Monthly, volume 45, number 10, December 1938, pages 694-696. (The A indices in P at equations 1' and 2' for p=5.)
%F G.f.: Sum_{k>=0} x^(5^k)/(1-x^5^k). - _Ralf Stephan_, Apr 12 2002
%F Multiplicative with a(p^e) = e+1 if p = 5, 1 otherwise.
%F a(n) = -Sum_{d|n} mu(5d)*tau(n/d). - _Benoit Cloitre_, Jun 21 2007
%F Dirichlet g.f.: zeta(s)/(1-1/5^s). - _R. J. Mathar_, Feb 09 2011
%F a(n) = A112765(5n). - _R. J. Mathar_, Jul 17 2012
%F a(5n) = 1 + a(n). a(5n+k) = 1 for k = 1..4. - _Robert Israel_, Dec 07 2015
%F G.f. satisfies A(x^5) = A(x) - x/(1-x). - _Robert Israel_, Dec 08 2015
%F a(n) = A112765(n) + 1. - _Amiram Eldar_, Sep 21 2020
%F Sum_{k=1..n} a(k) ~ 5*n/4. - _Vaclav Kotesovec_, Sep 21 2020
%e a(5) = 2 since 5^2 exactly divides 5 times 5;
%e a(25) = 3 since 5^3 exactly divides 5 times 25;
%e a(125) = 4 since 5^4 exactly divides 5 times 125.
%p seq(padic:-ordp(5*n,5), n=1..1000); # _Robert Israel_, Dec 07 2015
%t max = 1000; s = (1/x)*Sum[x^(5^k)/(1-x^5^k), {k, 0, Log[5, max] // Ceiling }] + O[x]^max; CoefficientList[s, x] (* _Jean-François Alcover_, Dec 04 2015 *)
%t Table[IntegerExponent[n, 5] + 1, {n, 1, 100}] (* _Amiram Eldar_, Sep 21 2020 *)
%o (PARI) a(n)=-sumdiv(n,d,moebius(5*d)*numdiv(n/d)) \\ _Benoit Cloitre_, Jun 21 2007
%o (PARI) a(n)=valuation(5*n,5) \\ _Anders Hellström_, Dec 04 2015
%Y Cf. A001511, A007949, A051064, A112765, A191610 (partial sums).
%K nonn,mult,easy
%O 1,5
%A _Alford Arnold_, Jun 25 2000
|
