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A132739
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Largest divisor of n not divisible by 5.
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19
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1, 2, 3, 4, 1, 6, 7, 8, 9, 2, 11, 12, 13, 14, 3, 16, 17, 18, 19, 4, 21, 22, 23, 24, 1, 26, 27, 28, 29, 6, 31, 32, 33, 34, 7, 36, 37, 38, 39, 8, 41, 42, 43, 44, 9, 46, 47, 48, 49, 2, 51, 52, 53, 54, 11, 56, 57, 58, 59, 12, 61, 62, 63, 64, 13, 66, 67, 68, 69, 14, 71, 72, 73, 74, 3, 76, 77
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OFFSET
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1,2
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COMMENTS
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a(n) = A060791(n) when n is not divisible by 5. When n is divisible by 5 a(n) divides A060791(n). Tom Edgar, Feb 08 2014
As well as being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 21 2019
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LINKS
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FORMULA
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a(n) = n/gcd(n,5^n).
O.g.f.: F(x) - 4*F(x^5) - 4*F(x^25) - 4*F(x^125) - ..., where F(x) = x/(1 - x)^2 is the generating function for the positive integers. More generally, for m >= 1,
Sum_{n >= 0} a(n)^m*x^n = F(m,x) - (5^m - 1)(F(m,x^5) + F(m,x^25) + F(m,x^125) + ...), where F(m,x) = A(m,x)/(1 - x)^(m+1) with A(m,x) the m_th Eulerian polynomial: A(1,x) = x, A(2,x) = x*(1 + x), A(3,x) = x*(1 + 4*x + x^2) - see A008292.
Repeatedly applying the Euler operator x*d/dx or its inverse operator to the o.g.f. for the sequence produces generating functions for the sequences n^m*a(n), m in Z. Some examples are given below. (End)
Sum_{k=1..n} a(k) ~ (5/12) * n^2. - Amiram Eldar, Nov 28 2022
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EXAMPLE
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Sum_{n >= 1} n*a(n)*x^n = G(x) - (4*5)*G(x^5) - (4*25)*G(x^25) - (4*125)*G(x^125) - ..., where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} (1/n)*a(n)*x^n = H(x) - (4/5)*H(x^5) - (4/25)*H(x^25) - (4/125)*H(x^125) - ..., where H(x) = x/(1 - x).
Sum_{n >= 1} (1/n^2)*a(n)*x^n = L(x) - (4/5^2)*L(x^5) - (4/25^2)*L(x^25) - (4/125^2)*L(x^125) - ..., where L(x) = Log(1/(1 - x)).
Also, Sum_{n >= 1} 1/a(n)*x^n = L(x) + (4/5)*L(x^5) + (4/5)*L(x^25) + (4/5)*L(x^125) + ....
(End)
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MATHEMATICA
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Table[n/5^IntegerExponent[n, 5], {n, 100}] (* Amiram Eldar, Sep 15 2020 *)
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PROG
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(Haskell)
a132739 n | r > 0 = n
| otherwise = a132739 n' where (n', r) = divMod n 5
(Ruby) p (1..50).map { |n| n /= 5 while (n % 5) == 0; n } # Simon Strandgaard, Nov 01 2021
(Python)
a, b = divmod(n, 5)
while b == 0:
a, b = divmod(a, 5)
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CROSSREFS
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KEYWORD
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nonn,mult
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AUTHOR
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STATUS
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approved
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