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A263922 Highest exponent in prime factorization of n-th central binomial coefficient. 4
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 (list; graph; refs; listen; history; text; internal format)
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: A064122 A323424 A334098 * A057526 A033265 A096004

Adjacent sequences:  A263919 A263920 A263921 * A263923 A263924 A263925

KEYWORD

nonn,nice

AUTHOR

Robert Israel, Oct 29 2015

STATUS

approved

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Last modified June 5 14:33 EDT 2020. Contains 334840 sequences. (Running on oeis4.)