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q[n_] := Module[{f = FactorInteger[n]}, p = f[[;; , 1]]; e = f[[;; , 2]]; odde = Select[e, OddQ]; Length[e] > 1 && Length[odde] == 1 && Divisible[odde[[1]] - 1, 4] && Divisible[p[[Position[e, odde[[1]]][[1, 1]]]] - 1, 4]]; f[2, e_] := 1; f[p_, e_] := NextPrime[p, -1]^e; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[1, 10000, 2], q[#] && s[DivisorSigma[1, #]] > s[#] &] (* Amiram Eldar, Nov 04 2021 *)
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A064989(n) = { my(f = factor(n)); if((n>1 && f[1, 1]==2), f[1, 2] = 0); for (i=1, #f~, f[i, 1] = precprime(f[i, 1]-1)); factorback(f) };
Any Obviously, any hypothetical odd perfect number would be neither in this sequence nor in A348938.
<a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
<a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>
A003961A064989(n) = { my(f = factor(n); if((n>1 && f[1, 1]==2), f[1, 2] = 0); for (i=1, #f~, f[i, 1] = nextprimeprecprime(f[i, 1]+-1)); factorback(f) }; \\ From A003961
A064989(n) = {my(f); f = factor(n); if((n>1 && f[1, 1]==2), f[1, 2] = 0); for (i=1, #f~, f[i, 1] = precprime(f[i, 1]-1)); factorback(f)};
A326042(n) = A064989(sigma(A003961(n)));
isA348749(n) = ((n%2)&&(A326042(A064989(sigma(n)) > A064989(n)));
Any hypothetical odd perfect number would be neither in this sequence nor in A348938.
45, 117, 325, 333, 405, 549, 605, 657, 925, 1053, 1413, 1445, 1525, 1737, 1825, 2205, 2493, 2817, 2825, 2925, 2997, 3033, 3573, 3645, 3789, 3825, 3925, 4113, 4825, 4869, 4941, 5445, 5517, 5733, 5913, 5949, 6057, 6425, 6525, 6597, 6813, 6925, 7025, 7497, 7605, 7825, 7893, 8125, 8325, 8425, 8973, 9225, 9477, 9837, 9925
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A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ From A003961
A064989(n) = {my(f); f = factor(n); if((n>1 && f[1, 1]==2), f[1, 2] = 0); for (i=1, #f~, f[i, 1] = precprime(f[i, 1]-1)); factorback(f)};
A326042(n) = A064989(sigma(A003961(n)));
isA228058(n) = if(!(n%2)||(omega(n)<2), 0, my(f=factor(n), y=0); for(i=1, #f~, if(1==(f[i, 2]%4), if((1==y)||(1!=(f[i, 1]%4)), return(0), y=1), if(f[i, 2]%2, return(0)))); (y));
isA348749(n) = (n%2&&(A326042(A064989(n)) > A064989(n)));
isA348939(n) = (isA228058(n)&&isA348749(n));
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Antti Karttunen, Nov 04 2021
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