A318528 - OEIS

%I #22 Jan 25 2019 10:36:44

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

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

%C From _Giovanni Resta_, Aug 28 2018: (Start)

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

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

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

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

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

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

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

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

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

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

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

%F a(n) = 1 + 2^n + 3^n for n = p^k, p > 2 prime. - _Giovanni Resta_, Aug 28 2018

%F From _Charlie Neder_, Jan 24 2019: (Start)

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

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

%F 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.

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

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

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

%o (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

%Y Cf. A185584, A185960.

%K nonn,more

%O 1,1

%A _Amiram Eldar_, Aug 28 2018