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.
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
A. Granville and O. Ramaré, Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients, Mathematika 43 (1996), 73-107, [DOI].
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
KEYWORD
nonn,nice
AUTHOR
Robert Israel, Oct 29 2015
STATUS
approved