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 A191218 Odd numbers n such that sigma(n) is congruent to 2 modulo 4. 8
 5, 13, 17, 29, 37, 41, 45, 53, 61, 73, 89, 97, 101, 109, 113, 117, 137, 149, 153, 157, 173, 181, 193, 197, 229, 233, 241, 245, 257, 261, 269, 277, 281, 293, 313, 317, 325, 333, 337, 349, 353, 369, 373, 389, 397, 401, 405, 409, 421, 425, 433, 449, 457, 461, 477 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Exactly the numbers of the form p^{4k+1}*m^2 with p a prime congruent to 1 modulo 4 and m a positive integer coprime with p. The odd perfect numbers are all of this form. See A228058 for the terms where m > 1. - Antti Karttunen, Apr 22 2019 LINKS Antti Karttunen, Table of n, a(n) for n = 1..20000 EXAMPLE For n=3 one has a(3)=17 since sigma(17) = 18 = 4*4 +2 is congruent to 2 modulo 4 MAPLE with(numtheory): genodd := proc(b) local n, s, d; for n from 1 to b by 2 do s := sigma(n); if modp(s, 4)=2 then print(n); fi; od; end; MATHEMATICA Select[Range[1, 501, 2], Mod[DivisorSigma[1, #], 4]==2&] (* Harvey P. Dale, Nov 12 2017 *) PROG (PARI) forstep(n=1, 10^3, 2, if(2==(sigma(n)%4), print1(n, ", "))) \\ Joerg Arndt, May 27 2011 (PARI) list(lim)=my(field=vectorsmall(lim\=1), v=List(), x2, t); for(x=2, sqrtint(lim-1), x2=x^2; t=sqrtint(lim-x2); forstep(y=if(t

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Last modified October 18 22:26 EDT 2019. Contains 328211 sequences. (Running on oeis4.)