%I #19 Jan 20 2023 01:31:40
%S 1,2,24,3,32,4,40,5,48,6,56,7,64,8,1,3,32,4,40,5,48,6,56,7,64,8,1,4,
%T 40,5,48,6,56,7,64,8,1,5,48,6,56,7,64,8,1,6,56,7,64,8,1,7,64,8,1,8,1,
%U 9,88,11,104,13,120,15,136,17,160,20,184,23,208
%N Irregular triangle read by rows in which row n is the result of iterating the operation f(n) = n/8 if n == 0 (mod 8), otherwise f(n) = 8*(floor(n/8) + n + 1), terminating at the first occurrence of 1.
%C The operation f(n) can be generalized to C(n,m) = n/m if n == 0 (mod m), m*(floor(n/m) + n + 1) otherwise. The operation for the 3x+1 (Collatz) problem is equivalent to C(n,2) and f(n) = C(n,8).
%C Conjecture: For any initial value of n >= 1 there is a number k such that f^{k}(n) = 1; in other words, every row of the triangle is finite.
%F a(n,0) = n, a(n,k + 1) = a(n,k)/8 if a(n,k) == 0 (mod 8), 8*(floor(a(n,k)/8) + a(n,k) + 1) otherwise, for n >= 1.
%e The irregular array a(n,k) starts:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ...
%e 1: 1
%e 2: 2 24 3 32 4 40 5 48 6 56 7 64 8 1
%e 3: 3 32 4 40 5 48 6 56 7 64 8 1
%e 4: 4 40 5 48 6 56 7 64 8 1
%e 5: 5 48 6 56 7 64 8 1
%e 6: 6 56 7 64 8 1
%e 7: 7 64 8 1
%e 8: 8 1
%e 9: 9 88 11 104 13 120 15 136 17 160 20 184 23 208 ...
%e 10: 10 96 12 112 14 128 16 2 24 3 32 4 40 5 ...
%e a(9,100) = 1 and a(10,20) = 1.
%o (PARI) collatz8(n)=N=[n];while(n>1,N=concat(N,n=if(n%8,8*(floor(n/8)+n+1),n/8)));N
%Y Cf. A070165.
%K nonn,easy,tabf
%O 1,2
%A _Davis Smith_, Nov 09 2019
|