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%I #20 Dec 15 2019 21:59:43
%S 2,5,7,11,19,20,23,28,29,32,34,38,39,41,46,53,57,59,62,70,73,77,83,89,
%T 90,94,103,104,113,118,119,124,131,160,173,177,179,188,190,191,208,
%U 227,229,233,239,242,248,251,263,280,281,290,293,297,298,311,316,327,335
%N For two consecutive numbers, the sum of the divisors of the sum of the two numbers divides the sum of the divisors of the product of the numbers. That is, numbers n such that sigma(2n+1) divides sigma(n^2 + n).
%H Hugo Pfoertner, <a href="/A152216/b152216.txt">Table of n, a(n) for n = 1..10000</a>
%e For n=11, 11+12 = 23, sigma(23) = 24; sigma(11*12) = sigma(132) = 336 and 24|336.
%p for n from 1 to 500 do if numtheory[sigma](n*(n+1)) mod numtheory[sigma](2*n+1) = 0 then printf("%d,",n); fi; od: # _R. J. Mathar_, Dec 04 2008
%p with(numtheory): a := proc (n) if type(sigma(n^2+n)/sigma(2*n+1), integer) = true then n else end if end proc: seq(a(n), n = 1 .. 400); # _Emeric Deutsch_, Dec 03 2008
%t Select[Range[335], Mod @@ DivisorSigma[1, {#^2 + #, 2 # + 1}] == 0 &] (* _Michael De Vlieger_, Dec 14 2019 *)
%o (PARI) for(k=1,335,if(!(sigma(k^2+k)%sigma(2*k+1)),print1(k,", "))) \\ _Hugo Pfoertner_, Dec 10 2019
%Y Cf. A000203.
%K nonn
%O 1,1
%A _J. M. Bergot_, Nov 29 2008
%E Corrected and extended by _Emeric Deutsch_ and _R. J. Mathar_, Dec 03 2008