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 A318528 Least number > 1 that equals the sum of the n-th powers of its first k divisors for some k. 0
 6, 130, 36, 41860, 276, 1015690, 2316, 921951940, 20196, 10009766650, 179196 (list; graph; refs; listen; history; text; internal format)
 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 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 FrÃ¶hlich, 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

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Last modified June 18 10:54 EDT 2021. Contains 345098 sequences. (Running on oeis4.)