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A229996
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For every positive integer m, let u(m) = (d(1),d(2),...,d(k)) be the unitary divisors of m. The sequence (a(n)) consists of successive numbers m which d(k)/d(1) + d(k-1)/d(2) + ... + d(k)/d(1) is an integer.
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9
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1, 10, 65, 130, 260, 340, 1105, 1972, 2210, 4420, 8840, 9860, 15650, 20737, 32045, 41474, 44200, 51272, 55250, 64090, 75140, 82948, 103685, 128180, 207370, 207553, 221000, 256360, 352529, 414740, 415106, 512720, 532100, 705058, 759025, 813800, 829480, 830212
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OFFSET
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1,2
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COMMENTS
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Also numbers m such that the sum of the squares of the unitary divisors of m is divisible by m (the unitary version of A046762) - Amiram Eldar, Jun 16 2018
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LINKS
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EXAMPLE
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The first 10 sums: 1, 5/2, 10/3, 17/4, 26/5, 25/3, 50/7, 65/8, 82/9, 13, so that a(1) = 1 and a(10) = 13.
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MATHEMATICA
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z = 1000; r[n_] := Select[Divisors[n], GCD[#, n/#] == 1 &]; k[n_] := Length[r[n]];
t[n_] := Table[r[n][[k[n] + 1 - i]]/r[n][[k[1] + i - 1]], {i, 1, k[n]}];
s = Table[Plus @@ t[n], {n, 1, z}]; a[n_] := If[IntegerQ[s[[n]]], 1, 0]; u = Table[a[n], {n, 1, z}]; Flatten[Position[u, 1]] (* A229996 *)
usigma2[n_] := If[n == 1, 1, Times @@ (1 + Power @@@ FactorInteger[n]^2)]; seqQ[n_] := Divisible[usigma2[n], n]; Select[Range[10^6], seqQ] (* Amiram Eldar, Jun 16 2018 *
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PROG
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(PARI) is(n) = {my(f = factor(n)); !(prod(i = 1, #f~, f[i, 1]^(2*f[i, 2]) + 1) % n); } \\ Amiram Eldar, Jun 16 2024
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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