A306773 Triangle read by rows, 0 <= k < n, n >= 2: T(n,k) is the eventual period of the modified Fibonacci sequence x(j) (or 0 if x(j) never enters a cycle) defined as follows: x(0) = 0, x(1) = 1, and for j > 1 x(j) is obtained from x(j-1) + x(j-2) by deleting all occurrences of the digit k in base n.
1, 1, 3, 1, 3, 16, 1, 3, 24, 6, 1, 3, 13, 6, 100, 1, 3, 8, 252, 120, 24, 1, 3, 8, 42, 119, 96, 576, 1, 3, 60, 588, 378, 36, 624, 1932, 1, 3, 126, 600, 144, 381, 200, 936, 912, 1, 3, 480, 51, 9, 2760, 6220, 540, 1800, 5700, 1, 3, 170, 750, 2480, 14880, 10990, 300, 1440, 3660, 840, 1, 3, 13800, 5880, 432, 48096, 60528, 456, 17640, 8496, 10560
Offset: 2
Examples
Triangle begins: n\k 0 1 2 3 4 5 6 7 8 9 10 11 --------------------------------------------------------------------------- 2: 1 1 3: 3 1 3 4: 16 1 3 24 5: 6 1 3 13 6 6: 100 1 3 8 252 120 7: 24 1 3 8 42 119 96 8: 576 1 3 60 588 378 36 624 9: 1932 1 3 126 600 144 381 200 936 10: 912 1 3 480 51 9 2760 6220 540 1800 11: 5700 1 3 170 750 2480 14880 10990 300 1440 3660 12: 840 1 3 13800 5880 432 48096 60528 456 17640 8496 10560 T(10,0) = 912 because A243063 eventually enters a cycle of length 912.
Links
- Pontus von Brömssen, Rows n = 2..24, flattened
Programs
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PARI
isok(v) = {for (n=1, #v-2, if ((v[#v] == v[#v - n]) && (v[#v-1] == v[#v - n - 1]), return (n));); 0;} f(x, y, n, k) = {my(z=x+y, d = digits(z, n)); fromdigits(select(t->(t!=k), d), n);} T(n,k) = {my(v = [0, 1], len = 2); while (! (per = isok(v)), v = concat(v, f(v[len-1], v[len], n, k)); len++;); per;} \\ Michel Marcus, May 01 2019 -
Python
# Note: the function hangs if the sequence never enters a cycle. import functools,sympy def drop(x,n,k): return functools.reduce(lambda x,j:n*x+j if j!=k else x,sympy.ntheory.factor_.digits(x,n)[1:],0) # Drop all digits k from x in base n. def A306773(n,k): return next(sympy.cycle_length(lambda x:(x[1],drop(x[0]+x[1],n,k)),(0,1)))[0] # Pontus von Brömssen, May 09 2019
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