A261231 -id:A261231 - 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

A276623 The infinite trunk of ternary beanstalk: The only infinite sequence such that a(n-1) = a(n) - A053735(a(n)), where A053735(n) = base-3 digit sum of n.

Original entry on oeis.org

0, 2, 4, 8, 10, 12, 16, 20, 26, 28, 30, 34, 38, 42, 46, 52, 56, 62, 68, 72, 80, 82, 84, 88, 92, 96, 100, 106, 110, 116, 122, 126, 134, 140, 144, 152, 160, 164, 170, 176, 180, 188, 194, 198, 204, 212, 216, 224, 232, 242, 244, 246, 250, 254, 258, 262, 268, 272, 278, 284, 288, 296, 302, 306, 314, 322, 326, 332, 338, 342, 350, 356, 360

Offset: 0

Antti Karttunen, Sep 11 2016

Antti Karttunen, Table of n, a(n) for n = 0..6250

Cf. A004128, A024023, A053735, A054861, A261231 (left inverse), A261233, A276622, A276624, A276603 (terms divided by 2), A276604 (first differences).
Cf. A179016, A219648, A219666, A255056, A259934, A276573, A276583, A276613 for similar constructions.
Cf. also A263273.

(define (A276623 n) (A276624 (A276622 n)))

a(n) = A276624(A276622(n)).
Other identities. For all n >= 0:
A261231(a(n)) = n.
a(A261233(n)) = A024023(n) = 3^n - 1.

A261233 a(n) = number of steps to reach 0 when starting from k = (3^n)-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

0, 1, 3, 8, 20, 49, 123, 319, 849, 2294, 6250, 17112, 47013, 129605, 358838, 998805, 2796093, 7869800, 22251147, 63141639, 179701239, 512744599, 1466635089, 4205423895, 12087339723, 34816804310, 100469592521, 290336059740, 839932833290, 2432050970420, 7047731703137, 20440131344750, 59334695322383, 172409162871823, 501489513690423, 1460214792034791

Offset: 0

Antti Karttunen, Aug 13 2015

nonn

One less than A261232.
Cf. A000244, A261231.
Cf. A261234 (the first differences).
Cf. also A218600.

a(0) = 0; for n >= 1, a(n) = A261234(n-1) + a(n-1).
a(n) = A261231((3^n)-1).
a(n) = A261232(n)-1.

Terms from a(24) onward added from the output of Hiroaki Yamanouchi's program by Antti Karttunen, Aug 16 2015

A261232 a(n) = number of steps to reach 0 when starting from k = 3^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

1, 2, 4, 9, 21, 50, 124, 320, 850, 2295, 6251, 17113, 47014, 129606, 358839, 998806, 2796094, 7869801, 22251148, 63141640, 179701240, 512744600, 1466635090, 4205423896, 12087339724, 34816804311, 100469592522, 290336059741, 839932833291, 2432050970421, 7047731703138, 20440131344751, 59334695322384, 172409162871824, 501489513690424

Offset: 0

Antti Karttunen, Aug 13 2015

nonn

One more than A261233.
Cf. A000244, A261231.
Cf. also A213710.

a(0) = 1; for n >= 1, a(n) = A261234(n-1) + a(n-1).
a(n) = A261231(3^n).
a(n) = 1 + A261233(n).

Terms from a(24) onward added from the output of Hiroaki Yamanouchi's program by Antti Karttunen, Aug 16 2015

A356384 For any n >= 0, let x_n(1) = n, and for any b > 1, x_n(b) = x_n(b-1) minus the sum of digits of x_n(b-1) in base b; a(n) is the least b such that x_n(b) = 0.

Original entry on oeis.org

1, 2, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13

Offset: 0

Rémy Sigrist, Aug 05 2022

This sequence is well defined: for any n >= 0: if x_n(b) > 0, then x_n(b+1) < x_n(b), and we must eventually reach 0.
This sequence is weakly increasing; this is related to the fact that for any base b > 1, k -> (k minus the sum of digits of k in base b) is weakly increasing.
Note that some values (like 7) do not appear in this sequence (see also A356386).

			For n = 42:
- we have:
      b  x(b)
      -  ----
      1    42
      2    39
      3    36
      4    33
      5    28
      6    20
      7    12
      8     7
      9     0
- so a(42) = 9.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Rémy Sigrist, Colored scatterplot of (n, x_n(b)) for n <= 1000 and b = 1..a(n) (the color is function of b)
Rémy Sigrist, PARI program

Cf. A011371, A066568, A071542, A261231, A344853, A356386.

PARI
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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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A276623 The infinite trunk of ternary beanstalk: The only infinite sequence such that a(n-1) = a(n) - A053735(a(n)), where A053735(n) = base-3 digit sum of n.

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A261233 a(n) = number of steps to reach 0 when starting from k = (3^n)-1 and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

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A261232 a(n) = number of steps to reach 0 when starting from k = 3^n and repeatedly applying the map that replaces k with k - (sum of digits in base-3 representation of k).

Views

Author

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Formula

Extensions

A356384 For any n >= 0, let x_n(1) = n, and for any b > 1, x_n(b) = x_n(b-1) minus the sum of digits of x_n(b-1) in base b; a(n) is the least b such that x_n(b) = 0.

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PARI