A261235 -id:A261235 - cp's OEIS frontend

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).

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

Antti Karttunen, Aug 13 2015

Hiroaki Yamanouchi, Table of n, a(n) for n = 0..100
Antti Karttunen, Naive C-program for computing the terms of A261234, A261236 and A261237 at the same time
Hiroaki Yamanouchi, Fast Python-program for computing terms of A213709, A261234 and analogous sequences in other bases

Cf. A000244, A054861, A261231, A261230.
First differences of A261232 and A261233.
Sum of A261236 and A261237.
Cf. A261235 (first differences of this sequence).
Cf. also A213709.

Table[Length@ NestWhileList[# - Total@ IntegerDigits[#, 3] &, 3^(n + 1) - 1, # > 3^n - 1 &] - 1, {n, 0, 16}] (* Michael De Vlieger, Jun 27 2016 *)

a(n) = A261236(n) + A261237(n).

a(23)-a(35) from Hiroaki Yamanouchi, Aug 16 2015

A261231 a(n) = number of steps to reach 0 when starting from k = n and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 20, 20, 20, 20, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 24, 24, 24

Offset: 0

Antti Karttunen, Aug 12 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..19683

Cf. A000244, A053735, A054861, A261232, A261233, A261234, A261235.

Cf. also A071542.

from sympy.ntheory.factor_ import digits
def a054861(n): return (n - sum(digits(n, 3)[1:]))/2
def a(n): return 0 if n==0 else 1 + a(2*a054861(n)) # Indranil Ghosh, May 22 2017

a(0) = 0; for n >= 1, a(n) = 1 + a(2*A054861(n)). [Note that A054861(n) = (n - A053735(n))/2, where A053735(n) = sum of digits of n, when written in base 3.]

cp's OEIS Frontend

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).

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