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A330073
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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.
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2
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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, 20, 4, 25, 5, 1, 7, 45, 9, 55, 11, 70, 14, 85, 17, 105, 21, 130, 26, 160, 32, 195, 39, 235, 47, 285, 57, 345, 69, 415, 83, 500, 100, 20, 4, 25, 5, 1, 8, 50
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OFFSET
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1,2
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COMMENTS
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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.
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)).
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.
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LINKS
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FORMULA
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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.
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EXAMPLE
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The irregular array T(n,k) starts:
n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ...
1: 1
2: 2 15 3 20 4 25 5 1
3: 3 20 4 25 5 1
4: 4 25 5 1
5: 5 1
6: 6 40 8 50 10 2 15 3 20 4 25 5 1
7: 7 45 9 55 11 70 14 85 17 105 21 130 26 160 ...
8: 8 50 10 2 15 3 20 4 25 5 1
9: 9 55 11 70 14 85 17 105 21 130 26 160 32 195 ...
10: 10 2 15 3 20 4 25 5 1
T(7,31) = 1 and T(9,29) = 1.
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MATHEMATICA
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Array[NestWhileList[If[Mod[#, 5] == 0, #/5, 5 (Floor[#/5] + # + 1)] &, #, # > 1 &] &, 8] // Flatten (* Michael De Vlieger, Dec 01 2019 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy,tabf
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AUTHOR
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STATUS
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approved
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