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Digital root of A069778(n-1) = n^3 - n^2 + 1, n >= 1. Repeat(1, 6, 3, 7, 6, 6, 4, 6, 9).
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%I #22 Jul 25 2016 09:55:35

%S 1,6,3,7,6,6,4,6,9,1,6,3,7,6,6,4,6,9,1,6,3,7,6,6,4,6,9,1,6,3,7,6,6,4,

%T 6,9,1,6,3,7,6,6,4,6,9,1,6,3,7,6,6,4,6,9,1,6,3,7,6,6,4,6,9,1,6,3,7,6,

%U 6,4,6,9,1,6,3,7,6,6,4,6,9,1,6,3,7,6,6,4,6,9,1,6,3,7,6,6,4,6,9,1,6,3,7,6,6,4,6,9

%N Digital root of A069778(n-1) = n^3 - n^2 + 1, n >= 1. Repeat(1, 6, 3, 7, 6, 6, 4, 6, 9).

%C Periodic with cycle of 9: {1, 6, 3, 7, 6, 6, 4, 6, 9}.

%C The decimal expansion of 54588823/333333333 = 0.repeat(163766469).

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DigitalRoot.html">Digital Root</a>.

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 0, 0, 0, 0, 0, 1).

%F a(n) = digital root of n^3 - n^2 + n.

%e For a(3) = 3 because 3^3 - 3^2 + 3 = 27 - 9 + 3 = 21 with digit sum 3 which is also the digital root of 21.

%t LinearRecurrence[{0, 0, 0, 0, 0, 0, 0, 0, 1},{1, 6, 3, 7, 6, 6, 4, 6, 9},108] (* _Ray Chandler_, Jul 25 2016 *)

%o (PARI) DR(n)=s=sumdigits(n);while(s>9,s=sumdigits(s));s

%o for(n=1,100,print1(DR(abs(n^2-n-n^3)),", ")) \\ _Derek Orr_, Dec 30 2014

%Y Cf. A069778, A251754, A010888, A056992, A073636.

%K base,nonn,easy

%O 1,2

%A _Peter M. Chema_, Dec 08 2014

%E More terms from _Derek Orr_, Dec 30 2014

%E Edited: name changed; formula, comment and example rewritten; digital root link added. - _Wolfdieter Lang_, Jan 05 2015