Original entry on oeis.org
-1, 0, 1, 2, 3, 6, 12, 28, 71, 186, 494, 1317, 3478, 9045, 23265, 59856, 155779, 412413, 1108874, 3009228, 8188150, 22257484, 60462422, 164715758, 452011067, 1253176571
Offset: 0
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Table[Length@ # - 2 First@ FirstPosition[#, k_ /; k != 2] + 1 &@ Map[First@ IntegerDigits[#, 3] &, #] &@ NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &], {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016, Version 10 *)
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(define (A261230 n) (- (A261236 n) (A261237 n)))
A261234
a(n) = number of steps to reach (3^n)-1 when starting from k = (3^(n+1))-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).
Original entry on oeis.org
1, 2, 5, 12, 29, 74, 196, 530, 1445, 3956, 10862, 29901, 82592, 229233, 639967, 1797288, 5073707, 14381347, 40890492, 116559600, 333043360, 953890490, 2738788806, 7881915828, 22729464587, 65652788211, 189866467219, 549596773550, 1592118137130, 4615680732717, 13392399641613, 38894563977633, 113074467549440, 329080350818600, 958725278344368, 2795854777347489
Offset: 0
Cf.
A261235 (first differences of this sequence).
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Table[Length@ NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &] - 1, {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016 *)
A261237
Number of steps needed when starting from (3^(n+1))-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k) to encounter the first number whose base-3 representation begins with a digit other than 2.
Original entry on oeis.org
1, 1, 2, 5, 13, 34, 92, 251, 687, 1885, 5184, 14292, 39557, 110094, 308351, 868716, 2458964, 6984467, 19890809, 56775186, 162427605, 465816503, 1339163192, 3858600035, 11138726760, 32199805820
Offset: 0
For n=0, we start from 3^(0+1) - 1 = 2 (also "2" in base-3), and subtract 2 to get 0, which doesn't begin with 2, thus a(0) = 1.
For n=1, we start from 3^(1+1) - 1 = 8 ("22" in base-3), and subtract 2*2 = 4 to get 4 ("11" in base-3) which doesn't begin with 2, thus a(1) = 1.
For n=2, we start from 3^(2+1) - 1 = 26 ("222" in base-3), and subtract first 6 to get 20 ("202" in base-3), from which we subtract 4, to get 16 ("121" in base-3), so in two steps we have reached the first such number that does not begin with "2" in base-3, thus a(2) = 2.
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/* Use the C-program given in A261234. */
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Flatten@ Table[FirstPosition[#, k_ /; k != 2] &@ Map[First@ IntegerDigits[#, 3] &, NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &]] - 1, {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016, Version 10 *)
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a(n)=my(k=3^(n+1)-1,t=2*3^n,s); while(k>=t, k-=sumdigits(k,3); s++); s \\ Charles R Greathouse IV, Aug 21 2015
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(definec (A261237 n) (let loop ((k (- (A000244 (+ 1 n)) 1)) (s 0)) (if (< (A122586 k) 2) s (loop (* 2 (A054861 k)) (+ 1 s)))))
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