A261102 -id:A261102 - cp's OEIS frontend

A219648 The infinite trunk of Zeckendorf beanstalk. The only infinite sequence such that a(n-1) = a(n) - number of 1's in Zeckendorf representation of a(n).

0, 1, 2, 4, 5, 7, 9, 12, 14, 17, 20, 22, 24, 27, 29, 33, 35, 37, 40, 42, 45, 47, 50, 54, 56, 58, 61, 63, 67, 70, 74, 76, 79, 83, 88, 90, 92, 95, 97, 101, 104, 108, 110, 113, 117, 121, 123, 126, 130, 134, 138, 143, 145, 147, 150, 152, 156, 159, 163, 165, 168

Offset: 0

Antti Karttunen, Nov 24 2012

nonn

a(n) tells in what number we end in n steps, when we start climbing up the infinite trunk of the "Zeckendorf beanstalk" from its root (zero).
There are many finite sequences such as 0,1,2; 0,1,2,4,5; etc. (see A219649) and as the length increases, so (necessarily) does the similarity to this infinite sequence.
There can be only one infinite trunk in "Zeckendorf beanstalk" as all paths downwards from numbers >= A000045(i) must pass through A000045(i)-1 (i.e. A000071(i)). This provides also a well-defined method to compute the sequence, for example, via a partially reversed version A261076.
See A014417 for the Fibonacci number system representation, also known as Zeckendorf expansion.

Antti Karttunen, Table of n, a(n) for n = 0..11817

Cf. A000045, A000071, A007895, A014417, A219641, A219649, A261076, A261102. For all n, A219642(a(n)) = n and A219643(n) <= a(n) <= A219645(n). Cf. also A261083 & A261084.

Other similarly constructed sequences: A179016, A219666, A255056.

(define (A219648 n) (A261076 (A261102 n)))

a(n) = A261076(A261102(n)).

A261091 a(n) = number of steps required to reach F(n+1)-1 from F(n+2)-1 by repeatedly subtracting from a natural number the number of ones in its Zeckendorf representation. Here F(n) = the n-th Fibonacci number, F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, ...

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 3, 5, 8, 11, 17, 25, 37, 56, 85, 130, 199, 305, 469, 723, 1118, 1733, 2693, 4193, 6539, 10211, 15962, 24974, 39103, 61262, 96030, 150608, 236338, 371101, 583118, 916978, 1443204, 2273434, 3584522, 5656786, 8934696, 14123156, 22340250

Offset: 0

Antti Karttunen, Aug 08 2015

nonn

Cf. A000045, A000071, A219641, A219642.
From a(1) onward the first differences of both A261081 and A261082.
Cf. A261090 (first differences of this sequence).
Cf. also A261102, A261076.

(define (A261091 n) (let ((end (- (A000045 (+ 1 n)) 1))) (let loop ((k (- (A000045 (+ 2 n)) 1)) (s 0)) (if (= k end) s (loop (A219641 k) (+ 1 s))))))

a(n) = A219642(A000071(n+2)) - A219642(A000071(n+1)). [By definition.]

a(n) = A219642(A000045(n+2)) - A219642(A000045(n+1)). [Equally.]

A261081 a(n) = number of steps required to reach 0 from F(n+2)-1 by repeatedly subtracting from a natural number the number of ones in its Zeckendorf representation. Here F(n) = the n-th Fibonacci number, F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, ...

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 10, 15, 23, 34, 51, 76, 113, 169, 254, 384, 583, 888, 1357, 2080, 3198, 4931, 7624, 11817, 18356, 28567, 44529, 69503, 108606, 169868, 265898, 416506, 652844, 1023945, 1607063, 2524041, 3967245, 6240679, 9825201, 15481987, 24416683, 38539839, 60880089

Offset: 0

Antti Karttunen, Aug 08 2015

nonn

One less than A261082.
Partial sums of A261091.
Cf. A000045, A219642, A261102.

a(n) = A219642(A000045(n+2)-1).
a(0) = 0; for n >= 1, a(n) = A261091(n) + a(n-1).
Other identities. For all n >= 0:
a(n) = A261082(n)-1.

A261076 The infinite trunk of Zeckendorf (Fibonacci) beanstalk, with reversed subsections.

Original entry on oeis.org

0, 1, 2, 4, 7, 5, 12, 9, 20, 17, 14, 33, 29, 27, 24, 22, 54, 50, 47, 45, 42, 40, 37, 35, 88, 83, 79, 76, 74, 70, 67, 63, 61, 58, 56, 143, 138, 134, 130, 126, 123, 121, 117, 113, 110, 108, 104, 101, 97, 95, 92, 90, 232, 226, 221, 217, 213, 209, 205, 201, 198, 193, 189, 185, 181, 178, 176, 172, 168, 165, 163, 159, 156, 152, 150, 147, 145

Offset: 0

Antti Karttunen, Aug 09 2015

This can be viewed as an irregular table: after the initial zero on row 0, start each row n with k = F(n+2)-1 and subtract repeatedly the number of "1-fibits" (number of terms in Zeckendorf expansion of k) from k to get successive terms, until the number that has already been listed (which is always F(n+1)-1) is encountered, which is not listed second time, but instead, the current row is finished and the next row starts with (F(n+3))-1, with the same process repeated. Here F(n) = the n-th Fibonacci number, A000045(n).

			As an irregular table, the sequence looks like:
  0;
  1;
  2;
  4;
  7, 5;
  12, 9;
  20, 17, 14;
  33, 29, 27, 24, 22;
  54, 50, 47, 45, 42, 40, 37, 35;
  ...
After zero, each row n is A261091(n) elements long.

Antti Karttunen, Table of n, a(n) for n = 0..11817; rows 0 .. 23 of the irregular table.
Indranil Ghosh, Python program to generate the sequence

Cf. A000045, A000071, A007895, A072649, A219641, A219648, A261091, A261102.

Cf. A218616 (analogous sequence for base-2).

For n <= 2, a(n) = n; for n >= 3, if A219641(a(n-1)) = F(k)-1 [i.e., one less than some Fibonacci number F(k)] then a(n) = F(k+2)-1, otherwise a(n) = A219641(a(n-1)).
As a composition:
a(n) = A219648(A261102(n)).

A261082 a(n) = number of steps required to reach 0 from F(n+2) by repeatedly subtracting from a natural number the number of ones in its Zeckendorf representation. Here F(n) = the n-th Fibonacci number, F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, ...

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 11, 16, 24, 35, 52, 77, 114, 170, 255, 385, 584, 889, 1358, 2081, 3199, 4932, 7625, 11818, 18357, 28568, 44530, 69504, 108607, 169869, 265899, 416507, 652845, 1023946, 1607064, 2524042, 3967246, 6240680, 9825202, 15481988, 24416684, 38539840, 60880090

Offset: 0

Antti Karttunen, Aug 08 2015

nonn

One more than A261081.
First differences give A261091.
Cf. A000045, A219642, A261102.

a(n) = A219642(A000045(n+2)).
a(0) = 1; for n >= 1, a(n) = A261091(n) + a(n-1).
Other identities. For all n >= 0:
a(n) = A261081(n)+1.

A261101 After zero, each n occurs A261091(n) times.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12

Offset: 0

Antti Karttunen, Aug 09 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..7624

Auxiliary sequence for constructing A261102 and A219648.

Cf. A261081.

(define (A261101 n) (let loop ((k 0)) (if (>= (A261081 k) n) k (loop (+ 1 k)))))

cp's OEIS Frontend

A219648 The infinite trunk of Zeckendorf beanstalk. The only infinite sequence such that a(n-1) = a(n) - number of 1's in Zeckendorf representation of a(n).

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A261091 a(n) = number of steps required to reach F(n+1)-1 from F(n+2)-1 by repeatedly subtracting from a natural number the number of ones in its Zeckendorf representation. Here F(n) = the n-th Fibonacci number, F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, ...

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A261081 a(n) = number of steps required to reach 0 from F(n+2)-1 by repeatedly subtracting from a natural number the number of ones in its Zeckendorf representation. Here F(n) = the n-th Fibonacci number, F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, ...

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A261076 The infinite trunk of Zeckendorf (Fibonacci) beanstalk, with reversed subsections.

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A261082 a(n) = number of steps required to reach 0 from F(n+2) by repeatedly subtracting from a natural number the number of ones in its Zeckendorf representation. Here F(n) = the n-th Fibonacci number, F(0) = 0, F(1) = 1, F(2) = 1, F(3) = 2, ...

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A261101 After zero, each n occurs A261091(n) times.

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