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A056527 Numbers where iterated sum of digits of square settles down to a cyclic pattern (in fact 13, 16, 13, 16, ...).

%I A056527 #12 Dec 19 2017 18:37:25
%S A056527 2,4,5,7,11,13,14,16,20,22,23,25,29,31,32,34,38,40,41,43,47,49,50,52,
%T A056527 56,58,59,61,65,67,68,70,74,76,77,79,83,85,86,88,92,94,95,97,101,103,
%U A056527 104,106,110,112,113,115,119,121,122,124,128,130,131,133,137,139,140
%N A056527 Numbers where iterated sum of digits of square settles down to a cyclic pattern (in fact 13, 16, 13, 16, ...).
%C A056527 Numbers == 2, 4, 5 or 7 mod 9, i.e. such that n^4 is not congruent to n^2 mod 9.
%C A056527 Numbers congruent to {2, 4, 5, 7} mod 9.
%H A056527 Colin Barker, <a href="/A056527/b056527.txt">Table of n, a(n) for n = 1..1000</a>
%H A056527 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1, 0, 0, 1, -1).
%F A056527 a(n) = a(n-1) + a(n-4) - a(n-5) for n>5. - _Harvey P. Dale_, Apr 05 2015
%F A056527 From _Colin Barker_, Dec 19 2017: (Start)
%F A056527 G.f.: x*(2 + 2*x + x^2 + 2*x^3 + 2*x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)).
%F A056527 a(n) = (-9 + (-1)^(1+n) - (3-3*i)*(-i)^n - (3+3*i)*i^n + 18*n) / 8 where i=sqrt(-1).
%F A056527 (End)
%e A056527 a(1)=2 because iteration starts 2, 4, 7, 13, 16, 13, 16, ....
%t A056527 Flatten[Table[9n+{2,4,5,7},{n,0,20}]] (* or *) LinearRecurrence[{1,0,0,1,-1},{2,4,5,7,11},100] (* _Harvey P. Dale_, Apr 05 2015 *)
%o A056527 (PARI) Vec(x*(2 + 2*x + x^2 + 2*x^3 + 2*x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)) + O(x^80)) \\ _Colin Barker_, Dec 19 2017
%Y A056527 Cf. A004159 for sum of digits of square, A056020 where iteration settles to 1, A056020 where iteration settles to 9, also A056528, A056529. Unhappy numbers A031177 deal with iteration of square of sum of digits not settling to a single result.
%K A056527 base,easy,nonn
%O A056527 1,1
%A A056527 _Henry Bottomley_, Jun 19 2000