A082490 Exponent of highest power of 3 dividing sum(0<=k

0, 1, 2, 0, 2, 3, 1, 2, 4, 0, 1, 2, 0, 3, 4, 2, 3, 5, 1, 2, 3, 1, 3, 4, 2, 3, 6, 0, 1, 2, 0, 2, 3, 1, 2, 4, 0, 1, 2, 0, 4, 5, 3, 4, 6, 2, 3, 4, 2, 4, 5, 3, 4, 7, 1, 2, 3, 1, 3, 4, 2, 3, 5, 1, 2, 3, 1, 4, 5, 3, 4, 6, 2, 3, 4, 2, 4, 5, 3, 4, 8, 0, 1, 2, 0, 2, 3, 1, 2, 4, 0, 1, 2, 0, 3, 4

Offset: 1

Author

Ralf Stephan, Apr 28 2003

Keywords

• nonn
• hear

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
• J. Shallit, k-regular Sequences
• J. Shallit, Number theory and formal languages, in D. A. Hejhal, J. Friedman, M. C. Gutzwiller and A. M. Odlyzko, eds., Emerging Applications of Number Theory, IMA Volumes in Mathematics and Its Applications, V. 109, Springer-Verlag, 1999, pp. 547-570. (Example 1.)

Crossrefs

Cf. A000989, A006134, A007949.

Programs

Maple

map(t -> padic:-ordp(t,3), ListTools:-PartialSums([seq(binomial(2*n,n),n=0..100)])); # Robert Israel, Mar 27 2018

Mathematica

IntegerExponent[#,3]&/@Accumulate[Table[Binomial[2n,n],{n,0,100}]]

PARI

s=0; for(n=1, 150, s=s+binomial(2*n-2, n-1); print1(valuation(s, 3)", "))

PARI

a(n) = valuation(n^2 * binomial(2*n, n), 3); \\ Michel Marcus, Mar 27 2018

Formula

a(n) = A007949(A006134(n)) = A007949 (n^2 * C(2n, n)) (Allouche, Shallit; Zagier) = 2*A007949(n) + A000989(n).