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 1-10 of 11 results. Next

A219643 Least inverse of A219642; a(n) = minimal i such that A219642(i) = n.

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 8, 10, 13, 15, 18, 21, 23, 26, 29, 32, 34, 36, 39, 42, 45, 47, 50, 54, 55, 57, 60, 63, 66, 68, 71, 75, 78, 81, 84, 89, 91, 94, 97, 100, 102, 105, 109, 112, 115, 118, 123, 126, 130, 134, 138, 143, 144, 146, 149, 152, 155, 157, 160, 164, 167
Offset: 0

Views

Author

Antti Karttunen, Nov 24 2012

Keywords

Crossrefs

Cf. A219645 for the greatest inverse. A219644 gives the first differences.
This sequence is based on Fibonacci number system (Zeckendorf expansion): A014417. For analogous sequence in binary system, see: A213708, for factorial number system: A219653.

A219645 Greatest inverse of A219642; a(n) = maximal i such that A219642(i) = n.

Original entry on oeis.org

0, 1, 2, 4, 6, 7, 9, 12, 14, 17, 20, 22, 25, 28, 31, 33, 35, 38, 41, 44, 46, 49, 53, 54, 56, 59, 62, 65, 67, 70, 74, 77, 80, 83, 88, 90, 93, 96, 99, 101, 104, 108, 111, 114, 117, 122, 125, 129, 133, 137, 142, 143, 145, 148, 151, 154, 156, 159, 163, 166, 169
Offset: 0

Views

Author

Antti Karttunen, Nov 24 2012

Keywords

Crossrefs

Cf. A219643 for the least inverse. A219644 gives the first differences.
This sequence is based on Fibonacci number system (Zeckendorf expansion): A014417. Analogous sequence for binary system: A173601, for factorial number system: A219655.

Formula

a(n) = A219643(n)+A219644(n)-1.

A219644 Run lengths in A219642.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 3, 2, 3, 3, 2, 3, 3, 3, 2, 2, 3, 3, 3, 2, 3, 4, 1, 2, 3, 3, 3, 2, 3, 4, 3, 3, 3, 5, 2, 3, 3, 3, 2, 3, 4, 3, 3, 3, 5, 3, 4, 4, 4, 5, 1, 2, 3, 3, 3, 2, 3, 4, 3, 3, 3, 5, 3, 4, 4, 4, 5, 3, 3, 3, 5, 5, 3, 5, 5, 3, 2, 3, 3, 3, 2, 3, 4, 3, 3, 3
Offset: 0

Views

Author

Antti Karttunen, Nov 24 2012

Keywords

Comments

a(n) tells from how many starting values one can end to 0 in n steps, with the iterative process described in A219642 (if going around in 0->0 loop is disallowed).

Crossrefs

a(n) = 1+(A219645(n)-A219643(n)).
This sequence is based on Fibonacci number system (Zeckendorf expansion): A014417. Analogous sequence for binary system: A086876, for factorial number system: A219654.

Formula

a(n) = A219643(n+1)-A219643(n). (The first differences of A219643).

A219646 Partial sums of A219642.

Original entry on oeis.org

0, 1, 3, 6, 9, 13, 17, 22, 28, 34, 41, 48, 55, 63, 71, 80, 89, 98, 108, 118, 128, 139, 150, 162, 174, 186, 199, 212, 225, 239, 253, 267, 282, 297, 313, 329, 346, 363, 380, 398, 416, 434, 453, 472, 491, 511, 531, 552, 573, 594, 616, 638, 660, 682, 705, 729, 753
Offset: 0

Views

Author

Antti Karttunen, Nov 24 2012

Keywords

Crossrefs

Used to compute A219647. Analogous sequence for binary system: A213706, for factorial number system: A219656.

A219652 Number of steps to reach 0 starting with n and using the iterated process: x -> x - (sum of digits in factorial expansion of x).

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 18, 18, 19, 19, 19, 19, 19, 19
Offset: 0

Views

Author

Antti Karttunen, Nov 25 2012

Keywords

Comments

See A007623 for the factorial number system representation.

Crossrefs

Analogous sequence for binary system: A071542, for Zeckendorf expansion: A219642. Cf. A007623, A034968, A219650, A219651, A219653-A219655, A219659, A219661, A219666.

Programs

  • Mathematica
    nn = 72; m = 1; While[Factorial@ m < nn, m++]; m; Table[Length@ NestWhileList[# - Total@ IntegerDigits[#, MixedRadix[Reverse@ Range[2, m]]] &, n, # > 0 &] - 1, {n, 0, nn}] (* Michael De Vlieger, Jun 27 2016, Version 10.2 *)

Formula

a(0)=0; for n>0, a(n) = 1 + a(A219651(n)).

Extensions

Erroneous description corrected by Antti Karttunen, Dec 03 2012

A219641 a(n) = n minus (number of 1's in Zeckendorf expansion of n).

Original entry on oeis.org

0, 0, 1, 2, 2, 4, 4, 5, 7, 7, 8, 9, 9, 12, 12, 13, 14, 14, 16, 16, 17, 20, 20, 21, 22, 22, 24, 24, 25, 27, 27, 28, 29, 29, 33, 33, 34, 35, 35, 37, 37, 38, 40, 40, 41, 42, 42, 45, 45, 46, 47, 47, 49, 49, 50, 54, 54, 55, 56, 56, 58, 58, 59, 61, 61, 62, 63, 63, 66
Offset: 0

Views

Author

Antti Karttunen, Nov 24 2012

Keywords

Comments

See A014417 for the Fibonacci number system representation, also known as Zeckendorf expansion.

Crossrefs

Cf. A007895, A014417. A022342 gives the positions of records, resulting the same sequence with duplicates removed: A219640. A035336 gives the positions of values that occur only once: A219639. Cf. also A219637, A219642. Analogous sequence for binary system: A011371, for factorial number system: A219651.

Programs

  • Mathematica
    zeck = DigitCount[Select[Range[0, 500], BitAnd[#, 2*#] == 0&], 2, 1];
    Range[0, Length[zeck]-1] - zeck (* Jean-François Alcover, Jan 25 2018 *)
  • Python
    from sympy import fibonacci
    def a(n):
        k=0
        x=0
        while n>0:
            k=0
            while fibonacci(k)<=n: k+=1
            x+=10**(k - 3)
            n-=fibonacci(k - 1)
        return str(x).count("1")
    print([n - a(n) for n in range(101)]) # Indranil Ghosh, Jun 09 2017
  • Scheme
    (define (A219641 n) (- n (A007895 n)))
    

Formula

a(n) = n - A007895(n).

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).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Nov 24 2012

Keywords

Comments

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.

Crossrefs

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.

Programs

Formula

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

A255072 Number of steps to reach 0 starting with n and using the iterated process: x -> x - (number of runs in binary representation of x).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 4, 4, 4, 4, 5, 5, 5, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 11, 12, 12, 12, 12, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 20, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23, 23, 24, 24, 24, 25
Offset: 0

Views

Author

Antti Karttunen, Feb 14 2015

Keywords

Crossrefs

Cf. A255053 (least inverse), A255055 (greatest inverse), A255054 (run lengths).
Cf. A255061 & A255062 (values at points (2^n)-2 and (2^n)-1).
Analogous sequences: A071542, A219642, A219652

Formula

a(0) = 0; for n >= 1, a(n) = 1 + a(A236840(n)) = 1 + a(n - A005811(n)).
Other identities. For all n >= 0:
a(A255053(n)) = a(A255055(n)) = n.
a(A255056(n)) = n. [This sequence works also as an inverse function for number-of-runs beanstalk A255056.]

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

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.]

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

Views

Author

Antti Karttunen, Aug 08 2015

Keywords

Crossrefs

One less than A261082.
Partial sums of A261091.

Formula

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.
Showing 1-10 of 11 results. Next