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A194025 Number of fixed points under iteration of sum of cubes of digits in base b.

%I A194025 #56 Jan 05 2025 19:51:39
%S A194025 1,2,9,3,4,7,6,8,5,8,5,5,3,3,24,3,2,9,2,3,16,5,2,20,2,2,7,9,3,14,2,6,
%T A194025 8,4,10,12,2,8,8,7,2,12,4,5,17,5,4,27,6,5,10,4,2,11,9,5,9,6,3,25,5,6,
%U A194025 24,5,4,17,5,5,9,10,1,15,4,3,13,3,5,19,4,13,7
%N A194025 Number of fixed points under iteration of sum of cubes of digits in base b.
%C A194025 If b >= 2 and n >= 2*b^3, then S(n,3,b) < n. For each positive integer n, there is a positive integer m such that S^m(n,3,b) < 2*b^3. (Grundman/Teeple, 2001, Lemma 8 and Corollary 9.)
%C A194025 From _Christian N. K. Anderson_, May 23 2013: (Start)
%C A194025 1 is considered a fixed point in all bases, 0 is not.
%C A194025 In order for a number with d digits in base n to be a fixed point, it must satisfy the condition d*(n-1)^3 < n^d, which can only occur when d < 4 for n > 2. Because all binary numbers are "happy" (become 1 under iteration), there are no fixed points with more than 4 digits in any base. It can further be demonstrated that all 4-digit solutions begin with 1 in base n.
%C A194025 Unlike the number of fixed points under iteration of sum of squares of digits (A193583), this sequence contains many even numbers, and its histogram converges to a smooth distribution (approximately gamma(2.64,2.8); see "histogram" in links). (End)
%H A194025 Christian N. K. Anderson, <a href="/A194025/b194025.txt">Table of n, a(n) for n = 2..1000</a>
%H A194025 Christian N. K. Anderson, <a href="/A194025/a194025.gif">Histogram of a(n)</a>
%H A194025 H. G. Grundman and E. A. Teeple, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/39-5/grundman.pdf">Generalized Happy Numbers</a>, Fibonacci Quarterly 39 (2001), nr. 5, p. 462-466.
%e A194025 In the decimal system all integers go to (1); (153); (370); (371); (407) or (55, 250,133); (136, 244); (160, 217, 352); (919, 1459) under the iteration of sum of cubes of digits, hence there are five fixed points, two 2-cycles and two 3-cycles. Therefore a(10) = 5.
%p A194025 S:=proc(n,p,b) local Q,k,N,z; Q:=[n]; 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 A194025 a:=proc(b) local F,i,A,Q,B,C; A:=[]: for i from 1 to 2*b^3 do Q:=S(convert(i,base,b),3,b); A:={op(A),Q[nops(Q)]}; od: F:={}: for i from 1 while nops(A)>0 do B:=S(A[1],3,b); C:=[seq(B[i],i=1..nops(B)-1)]: if nops(C)=1 then F:={op(F),op(C)}: fi: A:=A minus {op(B)}; od: return(nops(F)); end:
%p A194025 # _Martin Renner_, Aug 24 2011
%o A194025 (Sage)
%o A194025 def A194025(n):
%o A194025     # inefficient but straightforward
%o A194025     return len([i for i in (1..2*n**3) if i==sum(d**3 for d in i.digits(base=n))]) # _D. S. McNeil_, Aug 23 2011
%o A194025 (R) #See A226026 for an optimized version
%o A194025 inbase=function(n, b) { x=c(); while(n>=b) { x=c(n%%b, x); n=floor(n/b) }; c(n, x) }; yn=rep(NA, 30)
%o A194025 for(b in 2:30) yn[b]=sum(sapply(1:(2*b^3), function(x) sum(inbase(x, b)^3))==1:(2*b^3)); yn # _Christian N. K. Anderson_, Jun 08 2013
%Y A194025 Cf. A193594, A194281.
%Y A194025 Solutions for a(10): A046197.
%Y A194025 Largest of the a(n) fixed points: A226026.
%Y A194025 Related sequences for sum of squared digits: A193583, A209242.
%K A194025 nonn,base
%O A194025 2,2
%A A194025 _Martin Renner_, Aug 22 2011