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(Sum of digits of (2^n - 11)) - n.
3

%I #10 Oct 13 2024 12:27:30

%S 1,-2,2,2,3,-3,-5,1,5,5,6,9,7,-5,-1,8,9,3,10,25,11,2,12,6,13,10,5,14,

%T 24,27,25,22,26,17,27,30,19,7,11,20,12,15,22,19,23,14,24,27,25,22,26,

%U 35,27,21,10,16,20,11,30,33,22,19,41,41,33,18,-2,13,35,35,54,48,28,25,20,29,30

%N (Sum of digits of (2^n - 11)) - n.

%C Is this ever zero? If not, this would prove that A329492(11) = -1, and that A328882 is never -11. (-11 is the first negative open case.)

%H Robert Israel, <a href="/A329493/b329493.txt">Table of n, a(n) for n = 4..10000</a>

%p f:= proc(n) convert(convert(2^n-11,base,10),`+`)-n end proc:

%p map(f, [$4..100]); # _Robert Israel_, Nov 17 2019

%t Table[Total[IntegerDigits[2^n-11]]-n,{n,4,90}] (* _Harvey P. Dale_, Oct 13 2024 *)

%Y Cf. A328882, A329002, A329492.

%K sign,base

%O 4,2

%A _N. J. A. Sloane_, Nov 16 2019