cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Aug 09 2015

Keywords

Comments

Maps between sequences A219648 and A261076.

Links

Crossrefs

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

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

Views

Author

Antti Karttunen, Aug 08 2015

Keywords

Crossrefs

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.

Programs

  • Scheme
    (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))))))

Formula

a(n) = A219642(A000071(n+2)) - A219642(A000071(n+1)). [By definition.]
a(n) = A219642(A000045(n+2)) - A219642(A000045(n+1)). [Equally.]

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

Views

Author

Antti Karttunen, Aug 08 2015

Keywords

Crossrefs

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

Formula

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

Views

Author

Antti Karttunen, Aug 09 2015

Keywords

Links

Crossrefs

Auxiliary sequence for constructing A261102 and A219648.
Cf. A261081.

Programs

  • Scheme
    (define (A261101 n) (let loop ((k 0)) (if (>= (A261081 k) n) k (loop (+ 1 k)))))
Showing 1-4 of 4 results.

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