A000989 3-adic valuation of binomial(2*n, n): largest k such that 3^k divides binomial(2*n, n).

0, 0, 1, 0, 0, 2, 1, 1, 2, 0, 0, 1, 0, 0, 3, 2, 2, 3, 1, 1, 2, 1, 1, 3, 2, 2, 3, 0, 0, 1, 0, 0, 2, 1, 1, 2, 0, 0, 1, 0, 0, 4, 3, 3, 4, 2, 2, 3, 2, 2, 4, 3, 3, 4, 1, 1, 2, 1, 1, 3, 2, 2, 3, 1, 1, 2, 1, 1, 4, 3, 3, 4, 2, 2, 3, 2, 2, 4, 3, 3

Offset: 0

Author

N. J. A. Sloane, R. K. Guy

Keywords

• nonn
• easy

Comments

a(n) = 0 if and only if n is in A005836. - Charles R Greathouse IV, May 19 2013

sign(a(n+1) - a(n)) is repeat [0, 1, -1]. - Filip Zaludek, Oct 29 2016

By Kummer's theorem, number of carries when adding n + n in base 3. - Robert Israel, Oct 30 2016

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..2000
• Michael Gilleland, Some Self-Similar Integer Sequences.
• E. E. Kummer, Über die Ergänzungssätze zu den allgemeinen Reciprocitätsgesetzen, Journal für die reine und angewandte Mathematik, Vol. 44 (1852), pp. 93-146; alternative link.
• Dorel Miheţ, Legendre's and Kummer's theorems again, Resonance, Vol. 15, No. 12 (2010), pp. 1111-1121; alternative link.
• Armin Straub, Victor H. Moll and Tewodros Amdeberhan, The p-adic valuation of k-central binomial coefficients, Acta Arithmetica, Vol. 140, No. 1 (2009), pp. 31-42.
• Wikipedia, Kummer's theorem.

Crossrefs

Cf. A000984, A005836, A007949, A053735, A067397.

Programs

Haskell

a000989 = a007949 . a000984  -- Reinhard Zumkeller, Nov 19 2015

Maple

f:= proc(n) option remember; local k,m,d;
   k:= floor(log[3](n));
   d:= floor(n/3^k);
   m:= n-d*3^k;
   if d = 2 or 2*m > 3^k then procname(m)+1
   else procname(m)
   fi
end proc:
f(0):= 0:
map(f, [$0..100]); # Robert Israel, Oct 30 2016

Mathematica

p=3; Array[ If[ Mod[ bi=Binomial[ 2#, # ], p ]==0, Select[ FactorInteger[ bi ], Function[ q, q[ [ 1 ] ]==p ], 1 ][ [ 1, 2 ] ], 0 ]&, 27*3, 0 ]
Table[ IntegerExponent[ Binomial[2 n, n], 3], {n, 0, 100}] (* Jean-François Alcover, Feb 15 2016 *)

PARI

a(n) = valuation(binomial(2*n, n), 3)

PARI

a(n)=my(N=2*n,s);while(N\=3,s+=N);while(n\=3,s-=2*n);s \\ Charles R Greathouse IV, May 19 2013

Formula

a(n) = Sum_{k>=0} floor(2*n/3^k) - 2*Sum_{k>=0} floor(n/3^k). - Benoit Cloitre, Aug 26 2003

a(n) = A007949(A000984(n)). - Reinhard Zumkeller, Nov 19 2015

From Robert Israel, Oct 30 2016: (Start)

If 2*n < 3^k then a(3^k+n) = a(n).

If n < 3^k < 2*n then a(3^k+n) = a(n)+1.

If n < 3^k then a(2*3^k+n) = a(n)+1. (End)

a(n) = A053735(n) - A053735(2*n)/2. - Amiram Eldar, Feb 12 2021

From Miles Wilson, Jul 06 2025: (Start)

a(n) = A007949(n+1) + A067397(n).

G.f.: Sum_{k>=1} (x^(3^k/2+1/2)-x^(3^k))/((x-1)*(x^(3^k)-1)). (End)