A261091 -id:A261091 - cp's OEIS frontend

A261102 Simple self-inverse permutation of natural numbers: after zero, list each block of A261091(n) numbers in reverse order, from A261082(n) to A261081(n+1).

0, 1, 2, 3, 5, 4, 7, 6, 10, 9, 8, 15, 14, 13, 12, 11, 23, 22, 21, 20, 19, 18, 17, 16, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 76, 75, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 113, 112, 111

Offset: 0

Antti Karttunen, Aug 09 2015

nonn

Maps between sequences A219648 and A261076.

Antti Karttunen, Table of n, a(n) for n = 0..7624
Index entries for sequences that are permutations of the natural numbers

Cf. A261081, A261082, A261091, A261101.
Cf. A219648, A261076.
Cf. also A218602 (analogous sequence for base-2).

A261090 First differences of A261091: a(n) = A261091(n+1) - A261091(n).

Original entry on oeis.org

1, 0, 0, 1, 0, 1, 2, 3, 3, 6, 8, 12, 19, 29, 45, 69, 106, 164, 254, 395, 615, 960, 1500, 2346, 3672, 5751, 9012, 14129, 22159, 34768, 54578, 85730, 134763, 212017, 333860, 526226, 830230, 1311088, 2072264, 3277910, 5188460, 8217094

Offset: 0

Antti Karttunen, Aug 09 2015

nonn

Cf. A261091.

(define (A261090 n) (- (A261091 (+ n 1)) (A261091 n)))

a(n) = A261091(n+1) - A261091(n).

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)))))

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.

cp's OEIS Frontend

A261102 Simple self-inverse permutation of natural numbers: after zero, list each block of A261091(n) numbers in reverse order, from A261082(n) to A261081(n+1).

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A261090 First differences of A261091: a(n) = A261091(n+1) - A261091(n).

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

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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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