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Numbers k such that sum of the divisors d of k divides 1 + 2 + ... + k = k(k+1)/2.
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%I #35 Feb 21 2024 10:53:27

%S 1,2,15,20,24,95,104,207,224,287,464,1023,1199,1952,4095,4607,8036,

%T 12095,15872,16895,19359,22932,23519,28799,45440,45695,54144,77375,

%U 101567,102024,130304,159599,163295,223199,296207,317184,352799,522752,524160,635904

%N Numbers k such that sum of the divisors d of k divides 1 + 2 + ... + k = k(k+1)/2.

%C Alternately, numbers k such that sum of the divisors d of k divides the sum of the non-divisors d' of k, where 1 <= d, d' <= k.

%C Numbers k such that A232324(k) = antisigma(k) mod sigma(k) = A024816(n) mod A000203(n) = 0. - _Jaroslav Krizek_, Jan 24 2014

%H Donovan Johnson, <a href="/A076617/b076617.txt">Table of n, a(n) for n = 1..200</a>

%F a(n+2) = A066860(n) - _Alex Ratushnyak_, Jul 02 2013

%e The sum of the divisors of 15 is sigma(15) = 24; the sum of the non-divisors of 15 that are between 1 and 15 is 2 + 4 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 = 96. Since 24 divides 96, 15 is a term of the sequence.

%t a = {}; Do[ s = DivisorSigma[1, i]; n = (i (i + 1) / 2) - s; If[Mod[n, s] == 0, a = Append[a, i]], {i, 1, 10^5}]; a

%t Select[Range[640000],Divisible[(#(#+1))/2,DivisorSigma[1,#]]&] (* _Harvey P. Dale_, Aug 01 2019 *)

%o (PARI) is(n)=n*(n+1)/2%sigma(n)==0 \\ _Charles R Greathouse IV_, May 02 2013

%Y Cf. A000203, A024816, A066860.

%K nonn

%O 1,2

%A _Joseph L. Pe_, Oct 22 2002

%E New name from _J. M. Bergot_, May 02 2013

%E More terms from _T. D. Noe_, May 02 2013