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%I #16 Dec 05 2019 04:17:19
%S 1,2,15,3,20,4,25,5,1,3,20,4,25,5,1,4,25,5,1,5,1,6,40,8,50,10,2,15,3,
%T 20,4,25,5,1,7,45,9,55,11,70,14,85,17,105,21,130,26,160,32,195,39,235,
%U 47,285,57,345,69,415,83,500,100,20,4,25,5,1,8,50
%N Irregular triangle read as rows in which row n is the result of iterating the operation f(n) = n/5 if n == 0 (mod 5), otherwise f(n) = 5*(floor(n/5) + n + 1), terminating at the first occurrence of 1.
%C f(n) is the operation C(n,m) = n/m if n == 0 (mod m), m*(floor(n/m) + n + 1) otherwise where m = 5. C(n,2) is the operation in the Collatz problem (A070165) and C(n,8) is the operation in A329263.
%C Conjecture: For any initial value n >= 1, there is a number k such that f^{k}(n) = 1, where f^{0}(n) = n and f^{k}(n) = f(f^{k - 1}(n)).
%C For any number n, if n is a power of 5 multiplied by 1, 2, 3, or 4, then there is a number k such that f^{k}(n) = 1. If n is congruent to 10, 15, 20, or 25 (mod 30) and there is a k such that f^{k}(n) = 1, then f^{k + 1}(floor(n/6)) = 1.
%F T(n,0) = n, T(n,k + 1) = T(n,k)/5 if T(n,k) == 0 (mod 5), 5*(T(n,k) + floor(T(n,k)/5) + 1) otherwise, for n >= 1.
%e The irregular array T(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 15 3 20 4 25 5 1
%e 3: 3 20 4 25 5 1
%e 4: 4 25 5 1
%e 5: 5 1
%e 6: 6 40 8 50 10 2 15 3 20 4 25 5 1
%e 7: 7 45 9 55 11 70 14 85 17 105 21 130 26 160 ...
%e 8: 8 50 10 2 15 3 20 4 25 5 1
%e 9: 9 55 11 70 14 85 17 105 21 130 26 160 32 195 ...
%e 10: 10 2 15 3 20 4 25 5 1
%e T(7,31) = 1 and T(9,29) = 1.
%t Array[NestWhileList[If[Mod[#, 5] == 0, #/5, 5 (Floor[#/5] + # + 1)] &, #, # > 1 &] &, 8] // Flatten (* _Michael De Vlieger_, Dec 01 2019 *)
%o (PARI) row(n)=my(N=[n],m=5);while(n>1,N=concat(N,n=if(n%m,m*(n+floor(n/m)+1),n/m)));N
%Y Cf. A070165, A329263.
%K nonn,easy,tabf
%O 1,2
%A _Davis Smith_, Nov 30 2019
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