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A319927
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Numbers k such that the sum of the squares of the odd non-unitary divisors of k divides the sum of the squares of the non-unitary divisors of k.
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2
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9, 18, 25, 27, 45, 49, 50, 54, 63, 75, 81, 90, 98, 99, 117, 121, 125, 126, 135, 147, 150, 153, 162, 169, 171, 175, 189, 198, 207, 225, 234, 242, 243, 245, 250, 261, 270, 275, 279, 289, 294, 297, 306, 315, 325, 333, 338, 342, 343, 350, 351, 361, 363, 369, 375, 378, 387, 405
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OFFSET
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1,1
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COMMENTS
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Conjecture: For any nonnegative integer power p the sum of the p-th powers of the odd non-unitary divisors of a(n) divides the sum of the p-th powers of the non-unitary divisors of a(n).
The start of this sequence, including the 58 terms currently shown in the data section, is consistent with a definition "nonsquarefree numbers not divisible by 4", but some larger terms are divisible by 4: for example, a(1305) = 9216 = 2^10 * 3^2. - Peter Munn, Sep 21 2020
It seems that, for every n and odd prime p, p*a(n) is a term. - Ivan N. Ianakiev, Sep 11 2022
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LINKS
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EXAMPLE
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The non-unitary divisors of 18 are 3 and 6, of which only 3 is odd. 3^2=9 divides 3^3 + 6^2 = 45 and therefore 18 is in the sequence.
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MATHEMATICA
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sigmaNU[n_, p_]:=Total[Select[Divisors[n], GCD[#, n/#]>1&]^p];
sigmaNUOdd[n_, p_]:=Total[Select[Divisors[n], OddQ[#]&&GCD[#, n/#]>1&]^p];
p=2; (*if you want to check the conjecture for the power of 99, replace 2 with 99*)
Select[Range[1000], IntegerQ[sigmaNU[#, p]/sigmaNUOdd[#, p]]&]//Quiet
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PROG
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(PARI) su2(n) = sumdiv(n, d, if(gcd(d, n/d)!=1, d^2));
suo2(n) = sumdiv(n, d, if ((d%2) && (gcd(d, n/d)!=1), d^2));
isok(n) = my(suo = suo2(n)); if (suo, (su2(n) % suo) == 0); \\ Michel Marcus, Oct 28 2018
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CROSSREFS
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Cf. A048105 (number, i.e., sum of the 0th powers, of non-unitary divisors of n), A034444 (number of unitary divisors of n).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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