A268354 Highest power of 7 dividing n.
1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 49, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 49, 1, 1, 1, 1, 1, 1, 7
Offset: 1
Examples
Since 14 = 7 * 2, a(14) = 7. Likewise, since 7 does not divide 13, a(13) = 1.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..20790
- Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
- Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.
Crossrefs
Programs
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Magma
[7^Valuation(n,7): n in [1..150]]; // Vincenzo Librandi, Feb 03 2016
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Mathematica
7^Table[IntegerExponent[n, 7], {n, 150}] (* Vincenzo Librandi, Feb 03 2016 *) -
PARI
a(n) = 7^valuation(n, 7) \\ Michel Marcus, Feb 05 2016
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Sage
[7^valuation(i, 7) for i in [1..100]]
Formula
a(n) = 7^valuation(n,7).
a(n) = 7^A214411(n).
Completely multiplicative with a(7) = 7, a(p) = 1 for prime p and p <> 7. - Andrew Howroyd, Jul 20 2018
From Peter Bala, Feb 21 2019: (Start)
a(n) = gcd(n,7^n).
a(n) = n/A242603(n).
O.g.f.: x/(1 - x) + 6*Sum_{n >= 1} 7^(n-1)*x^(7^n)/ (1 - x^(7^n)). (End)
Sum_{k=1..n} a(k) ~ (6/(7*log(7)))*n*log(n) + (4/7 + 6*(gamma-1)/(7*log(7)))*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 15 2022
Dirichlet g.f.: zeta(s)*(7^s-1)/(7^s-7). - Amiram Eldar, Jan 03 2023
Extensions
More terms from Antti Karttunen, Dec 22 2017
Comments