A108866 Numerator of Sum_{k=1..n} 2^k/k.

0, 2, 4, 20, 32, 256, 416, 4832, 8192, 42496, 74752, 1467392, 2650112, 62836736, 115552256, 42790912, 79691776, 2535587840, 4766040064, 170851041280, 1617069867008, 3070050172928, 5843921666048, 256460544016384, 490390373269504, 4697678227177472, 9016382767235072

Offset

0,2

Comments

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

References

A. M. Robert, A Course in p-adic Analysis, Springer, 2000; see p. 278.

Links

Harvey P. Dale, Table of n, a(n) for n = 0..1000 [a(0) = 0 adapted by Georg Fischer, Mar 07 2020]

Formula

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

Example

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, ...

Mathematica

Join[{0}, Accumulate[Table[2^n/n, {n, 30}]]//Numerator] (* Harvey P. Dale, Oct 28 2018 *)

Prog

(PARI) a(n) = numerator(sum(k=1, n, 2^k/k)); \\ Michel Marcus, Mar 07 2020

Crossrefs

Cf. A087910. The denominators are A229726 (repeated).

Sequence in context: A295826 A201502 A069535 * A259122 A059206 A294230

Adjacent sequences: A108863 A108864 A108865 * A108867 A108868 A108869

Keyword

nonn,frac

Author

N. J. A. Sloane, Jul 12 2005

Extensions

a(0) corrected by A.H.M. Smeets, Mar 06 2020

Status

approved