A073078 -id:A073078 - cp's OEIS frontend

A079290 Composite numbers satisfying A073078(n)=(n+1)/2.

9, 15, 25, 27, 49, 81, 121, 125, 169, 243, 289, 343, 361, 529, 625, 729, 841, 961, 1331, 1369, 1681, 1849, 2187, 2197, 2209, 2401, 2809, 3125, 3481, 3721, 4489, 4913, 5041, 5329, 6241, 6561, 6859, 6889, 7921, 9409, 10201, 10609, 11449

Offset: 1

Benoit Cloitre, Apr 09 2003

nonn

Charles R Greathouse IV, Table of n, a(n) for n = 1..94

Cf. A073078, A244623.

A073078 := proc(n)
    local bink,k ;
    bink := 1 ;
    for k from 1 do
        bink := 2*bink*(2-1/k) ;
        if modp(bink,n) = 0 then
            return k;
        end if;
    end do:
end proc:
A079290 := proc(n)
    option remember;
    local a;
    if n = 1 then
        9;
    else
        for a from procname(n-1)+1 do
            if not isprime(a) and 2*A073078(a) = a+1 then
                return a;
            end if;
        end do:
    end if;
end proc: # R. J. Mathar, Aug 20 2014

b[n_] := For[k=1, True, k++, If[Divisible[Binomial[2k, k], n], Return[k]]];
Select[Select[Range[12000], CompositeQ], b[#] == (# + 1)/2&] (* Jean-François Alcover, Oct 31 2019 *)

p=5;forprime(q=7,1e4,forstep(n=p+2,q-2,2, for(s=2,n\2, if(binomial(2*s,s)%n==0,next(2)));print1(n", ")); p=q) \\ Charles R Greathouse IV, May 24 2013

a(21)-a(43) from Charles R Greathouse IV, May 24 2013

A111869 Least number k such that C(2k,k) is divisible by n^2.

Original entry on oeis.org

1, 3, 5, 15, 13, 5, 25, 63, 41, 13, 61, 15, 85, 25, 14, 255, 145, 41, 181, 23, 25, 61, 265, 95, 313, 85, 365, 27, 421, 14, 481, 1023, 61, 145, 39, 53, 685, 181, 86, 63, 841, 25, 925, 61, 44, 265, 1105, 383, 1201, 313, 145, 85, 1405, 365, 63, 95, 181, 421, 1741, 23, 1861

Offset: 1

Robert G. Wilson v, Nov 23 2005

nonn

From David A. Corneth, Aug 15 2025: (Start)
Conjecture 1: a(n) = (n^2 - 1)/2 + 1 for odd prime n.
Conjecture 2: Let q be the largest prime factor of n. Let e be the multiplicity of q in the factorization of n. Then a(n) >= (q^(2*e) - 1)/2 + 1. for n != 2.
These conjectures hold for n = 1..4002.
Conjecture 3: a(2^k) = 4^k - 1 for k >= 1.
This conjecture holds for k = 1..11. (End)

			From _David A. Corneth_, Aug 15 2025: (Start)
a(4) = 15 as 4^2 = 16 | binomial(2*15, 15) = binomial(30, 15) and for any k < 15 we have 16 does not divide binomial(2*k, k). We don't really need to compute binomial(30, 15) and not the previous binomial(2*k, k) but just find how many factors 2 they have. binomial(30, 15) = 30! / (15!)^2.
We find the 2-adic valuation of 30! as follows: let b(0) = 30 and let b(n + 1) = floor(b(n) / 2). Then the 2-adic valuation of 30! is Sum{k = 1..floor(log(30)/log(2))} b(k) = b(1) + b(2) + b(3) + b(4) = 15 + 7 + 3 + 1 = 26.
Similar for 15! it is 7 + 3 + 1 = 11. 26 - 2*11 = 4 >= 4 so a(4) <= 15 and checking the others gives a(4) = 15. (End)

David A. Corneth, Table of n, a(n) for n = 1..4002 (first 500 terms from Seiichi Manyama, terms n = 501..840 from Chai Wah Wu)
David A. Corneth, PARI program
David A. Corneth, Upper bounds on a(n) for n = 1..10000

Cf. A000984, A059097, A073078, A370980.

f[n_] := Block[{k = 1, m = n^2}, While[ Mod[ Binomial[2k, k], m] != 0, k++ ]; k]; Array[f, 61]

PARI
```
See Corneth link
```

a(n) = A073078(n^2).

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A079290 Composite numbers satisfying A073078(n)=(n+1)/2.

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