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Numbers k such that A000005(k) = A000005(A000217(k)).
3

%I #18 Jan 27 2017 13:04:26

%S 1,2,6,10,18,22,30,42,46,58,66,70,78,82,102,106,126,130,138,150,162,

%T 166,178,190,198,210,222,226,238,250,262,270,282,306,310,330,346,358,

%U 366,378,382,418,430,438,442,462,466,478,486,490,498,502,522,546,562

%N Numbers k such that A000005(k) = A000005(A000217(k)).

%C Numbers k such that the number of divisors of k is equal to the number of divisors of k*(k+1)/2.

%H Colin Barker, <a href="/A281664/b281664.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = A045326(n) - 1. - _Alois P. Heinz_, Jan 27 2017

%p a:= proc(n) option remember; local p; p:= a(n-1)+2;

%p while irem(p, 4)<>3 do p:= nextprime(p) od; p-1

%p end: a(1):=1:

%p seq(a(n), n=1..100); # _Alois P. Heinz_, Jan 27 2017

%t Select[Range@ 562, DivisorSigma[0, #] == DivisorSigma[0, PolygonalNumber@ #] &] (* _Michael De Vlieger_, Jan 27 2017, Version 10.4 *)

%o (PARI) select(n->numdiv(n)==numdiv(n*(n+1)/2), vector(1000, n, n))

%Y Cf. A000005, A000217, A006093, A045326.

%K nonn

%O 1,2

%A _Colin Barker_, Jan 26 2017