|
%I #37 Mar 07 2020 08:54:17
%S 0,2,4,20,32,256,416,4832,8192,42496,74752,1467392,2650112,62836736,
%T 115552256,42790912,79691776,2535587840,4766040064,170851041280,
%U 1617069867008,3070050172928,5843921666048,256460544016384,490390373269504,4697678227177472,9016382767235072
%N Numerator of Sum_{k=1..n} 2^k/k.
%C Conjecture: for n > 3, numerator(-2/n + Sum_{k=1..n} 2^k/k) == 0 (mod n^2) if and only if n is prime. See my formula below. Cf. A332786. - _Thomas Ordowski_, Mar 02 2020
%D A. M. Robert, A Course in p-adic Analysis, Springer, 2000; see p. 278.
%H Harvey P. Dale, <a href="/A108866/b108866.txt">Table of n, a(n) for n = 0..1000</a> [a(0) = 0 adapted by _Georg Fischer_, Mar 07 2020]
%F a(n) = numerator(Sum_{k=1..n} (2^k-2)/k + Sum_{k=1..n} 2/k). This formula is a heuristic of my conjecture in the comments section. Cf. A330718. - _Thomas Ordowski_, Mar 02 2020
%e The initial values of the sum are 2, 4, 20/3, 32/3, 256/15, 416/15, 4832/105, 8192/105, 42496/315, 74752/315, 1467392/3465, 2650112/3465, 62836736/45045, 115552256/45045, 42790912/9009, 79691776/9009, 2535587840/153153, 4766040064/153153, 170851041280/2909907, ...
%t Join[{0},Accumulate[Table[2^n/n,{n,30}]]//Numerator] (* _Harvey P. Dale_, Oct 28 2018 *)
%o (PARI) a(n) = numerator(sum(k=1, n, 2^k/k)); \\ _Michel Marcus_, Mar 07 2020
%Y Cf. A087910. The denominators are A229726 (repeated).
%K nonn,frac
%O 0,2
%A _N. J. A. Sloane_, Jul 12 2005
%E a(0) corrected by _A.H.M. Smeets_, Mar 06 2020
|
