A194281 Number of cycles under iteration of sum of cubes of digits in base b.
0, 1, 0, 1, 1, 8, 1, 4, 4, 6, 2, 12, 3, 7, 8, 7, 3, 16, 3, 6, 7, 7, 4, 14, 1, 8, 11, 7, 2, 20, 7, 5, 16, 9, 7, 18, 4, 7, 10, 6, 4, 24, 5, 5, 13, 6, 7, 25, 2, 10, 20, 6, 5, 23, 7, 7, 17, 9, 7, 29, 3, 10, 14, 14, 6, 21, 7, 10, 17, 18, 9, 30, 8, 10, 24, 12, 4, 28, 4, 19, 12, 11, 6, 36
Offset: 2
A226026 Maximum fixed points under iteration of sum of cubes of digits in base n.
1, 17, 62, 118, 251, 250, 433, 1052, 407, 1280, 2002, 1968, 793, 3052, 5614, 1456, 5337, 5939, 2413, 5615, 20217, 11648, 11080, 31024, 5425, 1737, 28027, 26846, 17451, 33535, 10261, 64019, 23552, 44937, 30086, 84870, 17353, 55243, 48824, 108936, 58618, 87977
Offset: 2
Comments
1 is considered a fixed point in all bases, 0 is not.
a(n)=1 iff A194025(n)=1.
In order for a number with d digits in base n to be a fixed point, it must satisfy the condition d*(n-1)^32. Because all binary numbers are "happy" (become 1 under iteration), there are no fixed points with more than 4 digits in any base.
Furthermore, 4-digit solutions of the form x0mm or xmmm (where m is n-1) represent extreme values of sum of cubed digits, and so 4-digit numbers can only be solutions if xn^3+n^2-1<=2n^3+x^3. For x=2 this reduces to n<=3, so any 4-digit solution must begin with 1 in bases above 3.
Examples
In base 5, the numbers 1, 28 and 118 are written as 1, 103, and 433. The sum of the cubes of their digits are 1, 1+0^3+3^3=28, and 4^3+3^3+3^3=118. There are no other solutions, so a(5)=118.
Links
- Christian N. K. Anderson, Table of n, a(n) for n = 2..1000
- Christian N. K. Anderson, Table of base, maximum fixed point, number of fixed points, and all fixed points for base 2 to 1000.
Crossrefs
Programs
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R
inbase=function(n,b) { x=c(); while(n>=b) { x=c(n%%b,x); n=floor(n/b) }; c(n,x) } yfp=vector("list",100) for(b in 2:100) { fp=c() for(w in 0:1) for(x in 1:b-1) for(y in 1:b-1) if((u1=w^3+x^3+y^3)<=(u2=w*b^3+x*b^2+y*b) & u1+b^3>u2+b-1) if(length((z=which((1:b-1)*((1:b-1)^2-1)==u2-u1)-1))) fp=c(fp,u2+z) yfp[[b]]=fp[-1] cat("Base",b,":",fp,"\n") }
A336783 Integers b where the number of cycles under iteration of sum of cubes of digits in base b is exactly four.
5, 90, 188
Offset: 1
Comments
Let b > 1 be an integer, and write the base b expansion of any nonnegative integer m as m = x_0 + x_1 b + ... +x_d b^d with x_d > 0 and 0 <= x_i < b for i=0,...,d.
Consider the map S_{x^3,b}: N to N, with S_{x^3,b}(m) := x_0^3+ ... + x_d^3.
It is known that the orbit set {m,S_{x^3,b}(m), S_{x^3,b}(S_{x^3,b}(m)), ...} is finite for all m>0. Each orbit contains a finite cycle, and for a given base b, the union of such cycles over all orbit sets is finite. Let us denote by L(x^3,i) the set of bases b such that the set of cycles associated to S_{x^3,b} consists of exactly i elements. In this notation, the sequence is the set of known elements of L(x^3,4).
Meanwhile, the known terms of the sequence L(x^3,1) is {2}, L(x^3,2) is empty, and L(x^3,3) is {3, 26}. It's undetermined whether the complete sequences are finite, if so, whether the above give all terms.
Examples
For instance, when b=5 the associated four cycles are (1),(28),(118) and (9,65,35).
Links
- H. Hasse and G. Prichett, A conjecture on digital cycles, J. reine angew. Math. 298 (1978), 8--15. Also on GDZ.
- D. Lorenzini, M. Melistas, A. Suresh, M. Suwama, and H. Wang, Integer Dynamics, preprint.
A375036 Perimeter of n when it is considered as a "histonumber" (see Comments for definition).
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 6, 6, 8, 10, 12, 14, 16, 18, 20, 22, 8, 8, 8, 10, 12, 14, 16, 18, 20, 22, 10, 10, 10, 10, 12, 14, 16, 18, 20, 22, 12, 12, 12, 12, 12, 14, 16, 18, 20, 22, 14, 14, 14, 14, 14, 14, 16, 18, 20, 22, 16, 16, 16, 16, 16, 16, 16, 18
Offset: 0
Comments
A histonumber is a geometric figure that appears when we transform each digit "d" of n into a column of unit squares of height "d". The number zero becomes a horizontal segment of unit length.
All histonumbers are even.
Examples
Histonumbers "1234", "2024" and "1000"
.
+---+ +---+
| | | |
+---+ + + +
| | | | |
+---+ + + +---+ +---+ +
| | | | | | | | |
+---+ + + + + + + + + +---+
| 1 2 | 3 | 4 | | 2 | 0 | 2 | 4 | | 1 | 0 0 0
+---+---+---+---+ +---+---+---+---+ +---+---+---+---+
.
The “1234” histonumber is a figure with a perimeter P of 16 units and a surface S of 10 square units (the surface of a histonumber is always the sum of its digits).
The “2024” histonumber is a figure with a perimeter P of 20 units and a surface S of 8 square units.
The “1000” histonumber is a figure with a perimeter P of 10 units and a surface S of 1 square unit.
By definition, a histonumber can always be drawn in one go on a sheet of paper, without ever lifting the pencil. This seems obvious for the 1234 histonumber above but is less so for 2024 and 1000. However, this is also the case here because the pencil will pass twice below the zeros. This justifies the value P = 20 of the perimeter of 2024 and the value P = 10 of the perimeter of 1000.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..23000
- Eric Angelini, Histonumbers
Programs
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Maple
a:= n-> (l-> (h-> 2*h+l[1]+l[-1]+add(abs(l[i]-l[i-1]) , i=2..h))(nops(l)))(convert(n, base, 10)): seq(a(n), n=0..67); # Alois P. Heinz, Jul 29 2024 -
Python
def a(n): d = [0] + list(map(int, str(n))) + [0] return 2*(len(d)-2) + sum(abs(d[i+1]-d[i]) for i in range(len(d)-1)) print([a(n) for n in range(68)]) # Michael S. Branicky, Jul 28 2024
Comments
Examples
Links
Crossrefs
Programs
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 Z,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: Z:={}: 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 Z:={op(Z),C}: fi: A:=A minus {op(B)}; od: return(nops(Z)); end: # Martin Renner, Aug 24 2011Sage