%I #20 Nov 13 2016 16:53:24
%S 1,5,1,6,9,13,10,8,9,9,20,13,12,35,7,15,7,21,27,37,24,36,32,26,10,36,
%T 27,28,10,56,22,26,23,63,39,27,19,67,9,36,40,54,54,48,18,73,52,75,18,
%U 117,52,74,22,65,48,53,45,44,43,18,30,67,39,49,87,111,15
%N Number of attractors under iteration of sum of squares of digits in base n.
%C If b>=2 and a>=b^2 then S(a,2,b)<a. For each positive integer a, there is an positive integer m such that S^m(a,2,b)<b^2. (Grundman/Teeple, 2001, Lemma 6 and Corollary 7)
%H Martin Renner, <a href="/A193586/b193586.txt">Table of n, a(n) for n = 2..300</a>
%H H. G. Grundman, E. A. Teeple, <a href="http://www.fq.math.ca/Scanned/39-5/grundman.pdf">Generalized Happy Numbers</a>, Fibonacci Quarterly 39 (2001), nr. 5, p. 462-466.
%e In the decimal system all integers go to (1) or (4, 16, 37, 58, 89, 145, 42, 20) under the iteration of sum of squares of digits, hence there is one fixed point and one 8-cycle. Therefore a(10) = 1 + 8 = 9.
%p S:=proc(n,p,b) local Q,k,N,z; Q:=[convert(n,base,b)]; for k from 1 do N:=Q[k]; z:=convert(sum(N['i']^p,'i'=1..nops(N)),base,b); if not member(z,Q) then Q:=[op(Q),z]; else Q:=[op(Q),z]; break; fi; od; return Q; end:
%p NumberOfAttractors:=proc(b) local A,i,Q; A:=[]: for i from 1 to b^2 do Q:=S(i,2,b); A:=[op(A),Q[nops(Q)]]; od: return(nops({op(A)})); end:
%p seq(NumberOfAttractors(b),b=2..50);
%Y Cf. A193583, A193585.
%K nonn,base
%O 2,2
%A _Martin Renner_, Jul 31 2011