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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).
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%I #39 Sep 15 2022 02:22:51

%S 1,3,3,5,15,21,7,9,45,55,33,39,91,105,15,17,153,171,95,105,231,253,69,

%T 75,325,351,189,203,435,465,31,33,561,595,315,333,703,741,195,205,861,

%U 903,473,495,1035,1081,141,147,1225,1275,663,689,1431,1485,385,399

%N 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).

%C The largest odd divisor of n-th triangular number.

%H T. D. Noe, <a href="/A069834/b069834.txt">Table of n, a(n) for n = 1..10000</a>

%F GCD(a(n),a(n+1)) = A000265(n+1). - _Ralf Stephan_, Apr 05 2003

%F a(n) = A000265(n) * A000265(n+1). - _Franklin T. Adams-Watters_, Nov 20 2009

%F From _Amiram Eldar_, Sep 15 2022: (Start)

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

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

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

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

%o (PARI) a(n)=local(t);t=n*(n+1)\2;t/2^valuation(t,2) \\ _Franklin T. Adams-Watters_, Nov 20 2009

%o (Python)

%o def A069834(n):

%o a, b = divmod(n*n+n, 2)

%o while b == 0:

%o a, b = divmod(a,2)

%o return 2*a+b # _Chai Wah Wu_, Dec 05 2021

%Y Cf. A000217, A000265, A050605.

%K nonn

%O 1,2

%A _Amarnath Murthy_, Apr 14 2002

%E More terms from _Ralf Stephan_, Apr 05 2003