A261233 -id:A261233 - cp's OEIS frontend

A276622 Simple self-inverse permutation of natural numbers: after a(0)=0, list each block of A261234(n) numbers in reverse order, from A261232(n) to A261233(1+n).

0, 1, 3, 2, 8, 7, 6, 5, 4, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 123, 122, 121, 120, 119, 118, 117, 116, 115, 114, 113, 112, 111, 110, 109, 108, 107, 106, 105, 104, 103, 102, 101, 100, 99, 98, 97, 96

Offset: 0

Antti Karttunen, Sep 11 2016

Maps between A276623 and A276624.

Antti Karttunen, Table of n, a(n) for n = 0..6250
Index entries for sequences that are permutations of the natural numbers

Cf. A261232, A276621, A276623, A276624.

(define (A276622 n) (if (zero? n) n (+ (A261232 (+ -1 (A276621 n))) (- (A261232 (A276621 n)) n 1))))

a(0) = 0; for n >= 1, a(n) = A261232(A276621(n)-1) + A261232(A276621(n)) - n - 1.

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

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.

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.]

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

A276621 After a(0)=0, each n occurs A261234(n-1) times.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 0

Antti Karttunen, Sep 11 2016

Auxiliary function for computing A276622 & A276623.

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

Cf. A261233, A276622, A276623.

After a(0), a(n) differs from A111393(n+1) for the first time at n=46, where a(46)=5, while A111393(47)=6.

(define (A276621 n) (let loop ((k 0)) (if (>= (A261233 k) n) k (loop (+ 1 k)))))

cp's OEIS Frontend

A276622 Simple self-inverse permutation of natural numbers: after a(0)=0, list each block of A261234(n) numbers in reverse order, from A261232(n) to A261233(1+n).

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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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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).

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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).

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A276621 After a(0)=0, each n occurs A261234(n-1) times.

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