A111869 -id:A111869 - cp's OEIS frontend

A073078 Least k such that n divides C(2k,k).

1, 1, 2, 3, 3, 2, 4, 7, 5, 3, 6, 5, 7, 4, 8, 15, 9, 5, 10, 3, 5, 6, 12, 7, 13, 7, 14, 5, 15, 8, 16, 31, 6, 9, 4, 5, 19, 10, 7, 13, 21, 5, 22, 6, 8, 12, 24, 15, 25, 13, 11, 7, 27, 14, 8, 11, 11, 15, 30, 14, 31, 16, 5, 63, 8, 6, 34, 9, 14, 4, 36, 14, 37, 19, 14, 10, 6, 7, 40, 15, 41, 21

Offset: 1

Benoit Cloitre, Aug 17 2002

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000984, A111869.

A073078 := proc(n)
    for k from 1 do
        if modp(binomial(2*k,k),n) = 0 then
            return k;
        end if;
    end do:
end proc: # R. J. Mathar, Aug 20 2014

lk[n_]:=Module[{k=1},While[!Divisible[Binomial[2k,k],n],k++];k]; Array[lk,90] (* Harvey P. Dale, Oct 09 2012 *)

a(n)=if(n<0,0,s=1; while(binomial(2*s,s)%n>0,s++); s)

a(2^k)=2^k-1, if n is an odd prime a(n)=(n+1)/2 (but there are also some composites satisfying this property, see A079290).

A118562 Least number k such that binomial(2k,k) is divisible by all squares to n squared but not (n+1) squared, or 0 if impossible.

Original entry on oeis.org

1, 3, 5, 15, 0, 23, 89, 95, 0, 123, 0, 215, 0, 0, 1117, 943, 0, 2003, 0, 0, 0, 3455, 0, 1439, 0, 7846, 0, 7916, 0, 14735, 13103, 0, 0, 0, 0, 23711, 0, 0, 0, 24049, 0, 44857, 0, 0, 0, 44711, 0, 47594, 0, 0, 0, 77021, 0, 0, 0, 0, 0, 195765, 0, 381398, 0, 0, 374435, 0, 0

Offset: 1

Robert G. Wilson v, Nov 23 2005

nonn

a(5)=0 because any number squared which would divide binomial(2k,k) would also be divided by 6^2 since 6=2*3.

			a(3)=5 because binomial(10,5)=252 which is divisible by the squares of 1, 2 & 3 but not 4 squared.
a(70)=385823.

Chai Wah Wu, Table of n, a(n) for n = 1..250

Cf. A059097, A111869, A080765, A024619.

f[n_] := Block[{k = 1, b = Binomial[2n, n]}, While[Mod[b, k^2] == 0, k++ ]; k - 1]; t = Table[0, {100}]; Do[ a = f[n]; If[a < 101 &t[[a]] == 0, t[[a]] = n; Print[{a, n}]], {n, 38000}] (* or *)
expoPF[k_, n_] := Module[{s = 0, x = n}, While[x > 0, x = Floor[x/k]; s += x]; s]; expoCF[k_, n_] := Min[expoPF[ #[[1]], n]/#[[2]] & /@ FactorInteger@k]; f[n_] := Module[{k = 2}, While[ expoCF[k, 2n] >= 2(1 + expoCF[k, n]), k++ ]; k-1]; t = Table[0, {100}]; Do[ a = f[n]; If[a < 101 &t[[a]] == 0, t[[a]] = n; Print[{a, n}]], {n, 400000}]; t

a(n)=0 iff n is a member of A080765: m such that m+1 divides lcm(1..m).

a(n-1)=0 iff n-1 is a member of A024619: Numbers that are not powers of primes.

cp's OEIS Frontend

A073078 Least k such that n divides C(2k,k).

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