A318528 a(n) = 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, 2387003305930334914, 1602516, 100006103532010, 14381676, 1880100018939820249188604888836, 129271236, 1000003814697527770, 1162785756, 19105043663614041367780, 10462450356, 10000002384185795209930, 94151567436, 226500219158007133816826003223992308820431641700

Offset: 1

Author

Amiram Eldar, Aug 28 2018

Keywords

• nonn

Comments

a(48) > 10^90. - Max Alekseyev, Jan 17 2025

Examples

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

Links

• Max Alekseyev, Table of n, a(n) for n = 1..47

Crossrefs

Cf. A027750, A185584, A185960, A190940, A307137, A378313, A378314.

Programs

Mathematica

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

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

Formula

a(n) = 1 + 2^n + 3^n for n = p^k with prime p > 2. - 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)

Extensions

a(12)-a(24) from Giovanni Resta confirmed by Max Alekseyev, Jan 04 2025