A194025 - OEIS

A194025

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

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

2,2

COMMENTS

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.)

From Christian N. K. Anderson, May 23 2013: (Start)

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)

LINKS

Christian N. K. Anderson, Table of n, a(n) for n = 2..1000

Christian N. K. Anderson, Histogram of a(n)

H. G. Grundman and E. A. Teeple, Generalized Happy Numbers, Fibonacci Quarterly 39 (2001), nr. 5, p. 462-466.

EXAMPLE

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.

MAPLE

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:

# Martin Renner, Aug 24 2011

PROG

(Sage)

def A194025(n):

# 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

CROSSREFS

Cf. A193594, A194281.

Solutions for a(10): A046197.