A069834 a(n) = n-th reduced triangular number: n*(n+1)/{2^k} where 2^k is the largest power of 2 that divides product n*(n+1).

1, 3, 3, 5, 15, 21, 7, 9, 45, 55, 33, 39, 91, 105, 15, 17, 153, 171, 95, 105, 231, 253, 69, 75, 325, 351, 189, 203, 435, 465, 31, 33, 561, 595, 315, 333, 703, 741, 195, 205, 861, 903, 473, 495, 1035, 1081, 141, 147, 1225, 1275, 663, 689, 1431, 1485, 385, 399

Offset: 1

Author

Amarnath Murthy, Apr 14 2002

Keywords

• nonn

Comments

The largest odd divisor of n-th triangular number.

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000217, A000265, A050605.

Programs

Mathematica

Table[tri = n*(n + 1)/2; tri/2^IntegerExponent[tri, 2], {n, 100}] (* T. D. Noe, Oct 28 2013 *)

PARI

for(n=1,100,t=n*n+n;while(t%2==0,t=t/2);print1(t","))

PARI

a(n)=local(t);t=n*(n+1)\2;t/2^valuation(t,2) \\ Franklin T. Adams-Watters, Nov 20 2009

Python

def A069834(n):
    a, b = divmod(n*n+n, 2)
    while b == 0:
        a, b = divmod(a,2)
    return 2*a+b # Chai Wah Wu, Dec 05 2021

Formula

GCD(a(n),a(n+1)) = A000265(n+1). - Ralf Stephan, Apr 05 2003

a(n) = A000265(n) * A000265(n+1). - Franklin T. Adams-Watters, Nov 20 2009

From Amiram Eldar, Sep 15 2022: (Start)

a(n) = A000265(A000217(n)).

Sum_{n>=1} 1/a(n) = Sum_{i,j>=1} 2^(i+1)/(4^i*(2*j-1)^2 - 1) = 2.84288562849221553965... . (End)

Extensions

More terms from Ralf Stephan, Apr 05 2003