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Positive integers n such that n^3 divided by the digital root of n is a cube.
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%I #12 Sep 15 2019 16:54:59

%S 1,8,10,19,26,28,37,44,46,55,62,64,73,80,82,91,98,100,109,116,118,127,

%T 134,136,145,152,154,163,170,172,181,188,190,199,206,208,217,224,226,

%U 235,242,244,253,260,262,271,278,280,289,296,298,307,314,316,325

%N Positive integers n such that n^3 divided by the digital root of n is a cube.

%H Colin Barker, <a href="/A236653/b236653.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,-1).

%F a(n) = a(n-1)+a(n-3)-a(n-4).

%F G.f.: x*(8*x^3+2*x^2+7*x+1) / ((x-1)^2*(x^2+x+1)).

%e 26 is in the sequence because the digital root of 26 is 8, and 26^3/8 = 2197 = 13^2.

%t LinearRecurrence[{1,0,1,-1},{1,8,10,19},60] (* _Harvey P. Dale_, Sep 15 2019 *)

%o (PARI) s=[]; for(n=1, 400, d=(n-1)%9+1; if(n^3%d==0 && ispower(n^3\d, 3), s=concat(s, n))); s

%o (PARI) Vec(x*(8*x^3+2*x^2+7*x+1)/((x-1)^2*(x^2+x+1)) + O(x^100))

%Y Cf. A010888, A236652.

%K nonn,base,easy

%O 1,2

%A _Colin Barker_, Jan 29 2014