A263922 Highest exponent in prime factorization of n-th central binomial coefficient.

1, 1, 2, 1, 2, 2, 3, 2, 2, 2, 3, 2, 3, 3, 4, 2, 3, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 2, 2, 3, 2, 3, 3, 4, 2, 4, 3, 4, 4, 4, 4, 5, 2, 3, 4, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 2, 2, 3, 4, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 4, 3, 3, 4, 3, 4, 4, 5

Offset

1,3

Comments

a(n) >= 2 for n > 4.

a(n) is the maximum number of carries in base-p addition of n+n for primes p <= 2n.

A000120(n) <= a(n) <= A070939(n).

It appears that a(n) >= 3 for n > 1056. Any further n must be greater than 10^1000. Similarly it appears that a(n) >= 4 for n > 557056 and a(n) >= 5 for n > 1090519552. - Charles R Greathouse IV, Oct 31 2015

Links

Robert Israel, Table of n, a(n) for n = 1..10000

MathOverflow, Divisibility of a binomial coefficient by p^2 — current status

A. Granville and O. Ramaré, Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients, Mathematika 43 (1996), 73-107, [DOI].

Formula

a(n) = A051903(A000984(n)).

Example

For n=3, C(6,3) = 20 = 2^2 * 5^1 so a(3) = 2.

Maple

f:= t -> max(seq(s[2], s=ifactors(t)[2])):
seq(f(binomial(2*n, n)), n=1..100); # Robert Israel, Oct 29 2015

Mathematica

a[n_] := FactorInteger[Binomial[2 n, n]][[All, 2]] // Max; Array[a, 100] (* Jean-François Alcover, Nov 27 2015 *)

Prog

(PARI) f(n, p)=my(d=Vecrev(digits(n, p)), c); sum(i=1, #d, c=(2*d[i]+c>=p))
a(n)=my(r=hammingweight(n), L=sqrtnint(n, r+1), t); forprime(p=3, L, t=f(n, p); if(t>r, L=sqrtnint(n, 1+r=t)); if(p>=L, return(r))); r \\ Charles R Greathouse IV, Oct 29 2015
(PARI) vector(200, n, vecmax(factor(binomial(2*n, n))[, 2])) \\ Altug Alkan, Oct 30 2015
(Sage)
max_exp = lambda n: max([p[1] for p in list(n.factor())])
print([max_exp(binomial(2*n, n)) for n in (1..87)]) # Peter Luschny, Oct 30 2015
def a(n):
    N = 2*n
    r = sum(N.digits(2))
    b = 1+ZZ(N).nth_root(r, truncate_mode=1)[0]
    for p in primes(3, b):
        t, q = 0, N
        while True:
            q //= p
            if q == 0: break
            if (q & 1) == 1: t += 1
        if t > r : r = t
    return r
print([a(n) for n in (1..87)]) # Peter Luschny, Nov 02 2015
(Haskell)
a263922 = a051903 . a000984  -- Reinhard Zumkeller, Nov 19 2015

Crossrefs

Cf. A000120, A000984, A051903, A070939, A263924.

Sequence in context: A323424 A334098 A358371 * A057526 A033265 A096004

Adjacent sequences: A263919 A263920 A263921 * A263923 A263924 A263925

Keyword

nonn,nice

Author

Robert Israel, Oct 29 2015

Status

approved