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a(n) = n-th even superperfect number divided by 2^n, minus 1.
1

%I #10 Mar 15 2020 03:40:30

%S 0,0,1,3,127,1023,2047,4194303,2251799813685247,

%T 302231454903657293676543,39614081257132168796771975167,

%U 20769187434139310514121985316880383

%N a(n) = n-th even superperfect number divided by 2^n, minus 1.

%H Jinyuan Wang, <a href="/A134711/b134711.txt">Table of n, a(n) for n = 1..18</a>

%F a(n) = (A061652(n)/(2^n)) - 1.

%e a(5) = 127 because the 5th even superperfect number is 4096 and 2^5 = 32 and (4096/32) - 1 = 127.

%Y Cf. A061652 (even superperfect numbers).

%K nonn

%O 1,4

%A _Omar E. Pol_, Nov 07 2007

%E Offset changed to 1 by and a(11)-a(12) from _Jinyuan Wang_, Mar 15 2020