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A194025
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Number of fixed points under iteration of sum of cubes of digits in base b.
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4
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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, 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, 24, 5, 4, 17, 5, 5, 9, 10, 1, 15, 4, 3, 13, 3, 5, 19, 4, 13, 7
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OFFSET
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2,2
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COMMENTS
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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.)
1 is considered a fixed point in all bases, 0 is not.
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.
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)
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LINKS
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EXAMPLE
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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.
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MAPLE
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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:
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:
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PROG
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(Sage)
# inefficient but straightforward
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
(R) #See A226026 for an optimized version
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)
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
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CROSSREFS
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Largest of the a(n) fixed points: A226026.
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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