A318528 Least number > 1 that equals the sum of the n-th powers of its first k divisors for some k.

6, 130, 36, 41860, 276, 1015690, 2316, 921951940, 20196, 10009766650, 179196

Offset

1,1

Comments

From Giovanni Resta, Aug 28 2018: (Start)

If n = p^k, with p an odd prime, then a(n) = 1 + 2^n + 3^n. We also have

a(12) <= 2387003305930334914 (with divisors 1, 2, 17, 34),

a(14) = 100006103532010 (1, 2, 5, 10),

a(15) = 14381676 (1, 2, 3),

a(16) <= 1880100018939820249188604888836, (1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78),

a(18) = 1000003814697527770 (1, 2, 5, 10),

a(20) <= 19105043663614041367780, (1, 2, 4, 5, 7, 10, 13),

a(21) = 10462450356 (1, 2, 3),

a(22) = 10000002384185795209930, (1, 2, 5, 10), and

a(24) <= 226500219158007133816826003223992308820431641700, (1, 2, 4, 5, 10, 20, 25, 47, 50, 94).

In general, if n = 4*k+2, then a(n) <= 1 + 2^n + 5^n + 10^n. (End)

Links

Table of n, a(n) for n=1..11.

Formula

a(n) = 1 + 2^n + 3^n for n = p^k, p > 2 prime. - Giovanni Resta, Aug 28 2018

From Charlie Neder, Jan 24 2019: (Start)

a(n) = 1 + 2^n + 3^n for n odd,

a(n) = 1 + 2^n + 5^n + 10^n for n congruent to 2 modulo 4,

a(n) = 1 + 2^n + 4^n + 5^n + 7^n + 10^n + 13^n for n congruent to 4 or 8 modulo 12 and not 16 modulo 20.

All other a(n) contain a term at least 24^n. (End)

Example

a(2) = 130 since 130 has the divisors 1, 2, 5, 10, ... and 1^2 + 2^2 + 5^2 + 10^2 = 130.

Mathematica

a[k_] := Module[{n = 2}, While[! MemberQ[Accumulate[Divisors[n]^k], n], n++]; n]; Do[Print[a[n]], {n, 1, 10}]

Prog

(PARI) a(n) = for(x=2, oo, my(div=divisors(x), s=0); for(k=1, #div, s=sum(i=1, k, div[i]^n); if(s==x, return(x)))) \\ Felix Fröhlich, Aug 28 2018

Crossrefs

Cf. A185584, A185960.

Sequence in context: A012842 A012638 A338709 * A095695 A156475 A000907

Adjacent sequences: A318525 A318526 A318527 * A318529 A318530 A318531

Keyword

nonn,more

Author

Amiram Eldar, Aug 28 2018

Status

approved