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 1-6 of 6 results.

A062878 a(n) is the position of A050614(n) in A062877.

Original entry on oeis.org

1, 3, 6, 15, 24, 60, 102, 255, 384, 960, 1632, 4080, 6168, 15420, 26214, 65535, 98304, 245760, 417792, 1044480, 1579008, 3947520, 6710784, 16776960, 25166208, 62915520, 106956384, 267390960, 404232216, 1010580540, 1717986918, 4294967295, 6442450944
Offset: 0

Views

Author

Antti Karttunen, Jun 26 2001

Keywords

Comments

In binary this sequence looks like 1, 11, 110, 1111, 11000, 111100, 1100110, 11111111, 110000000, 1111000000, 11001100000, 111111110000, 1100000011000, 11110000111100, 110011001100110, ...
Sequence A282387 may be the same, but I cannot prove nor disprove this beyond a(22). - Robert Price, Feb 13 2017
Agrees with A282387 for at least 1000 terms. - Sean A. Irvine, Apr 14 2023

Programs

  • Mathematica
    A050614 = Table[k = Floor[Log[2, n + 1]]; Product[j = 2^(i + 1); l = Fibonacci[j + 1] + Fibonacci[j - 1]; If[BitAnd[2^i, n] == 0, b = 0, b = 1]; l^b, {i, 0, k}], {n, 0, 200}]; A062877 = Union[Total /@ Subsets[Fibonacci[Range[1, 46, 2]]]]; Flatten[Table[Position[ A062877, A050614[[i]] ] - 1, {i, 1, 25}]] (* Robert Price, Feb 13 2017 *)

Formula

a(2^n-1) = 2^(2^n) - 1. - Philippe Deléham, Apr 05 2007
a(n) = Sum_{k=0..n} A127872(n,k)*2^k. - Philippe Deléham, Oct 09 2007

Extensions

a(15)-a(22) from Robert Price, Feb 13 2017
More terms from Sean A. Irvine, Apr 14 2023

A022290 Replace 2^k in binary expansion of n with Fibonacci(k+2).

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 5, 6, 5, 6, 7, 8, 8, 9, 10, 11, 8, 9, 10, 11, 11, 12, 13, 14, 13, 14, 15, 16, 16, 17, 18, 19, 13, 14, 15, 16, 16, 17, 18, 19, 18, 19, 20, 21, 21, 22, 23, 24, 21, 22, 23, 24, 24, 25, 26, 27, 26, 27, 28, 29, 29, 30, 31, 32, 21, 22, 23, 24, 24, 25, 26
Offset: 0

Views

Author

Keywords

Examples

			n=4 = 2^2 is replaced by A000045(2+2) = 3. n=5 = 2^2 + 2^0 is replaced by A000045(2+2) + A000045(0+2) = 3+1 = 4. - _R. J. Mathar_, Jan 31 2015
From _Philippe Deléham_, Jun 05 2015: (Start)
This sequence regarded as a triangle with rows of lengths 1, 1, 2, 4, 8, 16, ...:
  0
  1
  2, 3
  3, 4, 5, 6
  5, 6, 7, 8, 8, 9, 10, 11
  8, 9, 10, 11, 11, 12, 13, 14, 13, 14, 15, 16, 16, 17, 18, 19
  ...
(End)
		

Crossrefs

Other sequences that are built by replacing 2^k in the binary representation with other numbers: A029931 (naturals), A054204 (even-indexed Fibonacci numbers), A062877 (odd-indexed Fibonacci numbers), A059590 (factorials), A089625 (primes).

Programs

  • Haskell
    a022290 0 = 0
    a022290 n = h n 0 $ drop 2 a000045_list where
       h 0 y _      = y
       h x y (f:fs) = h x' (y + f * r) fs where (x',r) = divMod x 2
    -- Reinhard Zumkeller, Oct 03 2012
    
  • Maple
    A022290 := proc(n)
        dgs := convert(n,base,2) ;
        add( op(i,dgs)*A000045(i+1),i=1..nops(dgs)) ;
    end proc: # R. J. Mathar, Jan 31 2015
    # second Maple program:
    b:= (n, i, j)-> `if`(n=0, 0, j*irem(n, 2, 'q')+b(q, j, i+j)):
    a:= n-> b(n, 1$2):
    seq(a(n), n=0..127);  # Alois P. Heinz, Jan 26 2022
  • Mathematica
    Table[Reverse[#].Fibonacci[1 + Range[Length[#]]] &@ IntegerDigits[n, 2], {n, 0, 54}] (* IWABUCHI Yu(u)ki, Aug 01 2012 *)
  • PARI
    my(m=Mod('x,'x^2-'x-1)); a(n) = subst(lift(subst(Pol(binary(n)), 'x,m)), 'x,2); \\ Kevin Ryde, Sep 22 2020
    
  • Python
    def A022290(n):
        a, b, s = 1,2,0
        for i in bin(n)[-1:1:-1]:
            s += int(i)*a
            a, b = b, a+b
        return s # Chai Wah Wu, Sep 10 2022

Formula

G.f.: (1/(1-x)) * Sum_{k>=0} F(k+2)*x^2^k/(1+x^2^k), where F = A000045.
a(n) = Sum_{k>=0} A030308(n,k)*A000045(k+2). - Philippe Deléham, Oct 15 2011
a(A003714(n)) = n. - R. J. Mathar, Jan 31 2015
a(A000225(n)) = A001911(n). - Philippe Deléham, Jun 05 2015
From Jeffrey Shallit, Jul 17 2018: (Start)
Can be computed from the recurrence:
a(4*k) = a(k) + a(2*k),
a(4*k+1) = a(k) + a(2*k+1),
a(4*k+2) = a(k) - a(2*k) + 2*a(2*k+1),
a(4*k+3) = a(k) - 2*a(2*k) + 3*a(2*k+1),
and the initial terms a(0) = 0, a(1) = 1. (End)
a(A003754(n)) = n-1. - Rémy Sigrist, Jan 28 2020
From Rémy Sigrist, Aug 04 2022: (Start)
Empirically:
- a(2*A003714(n)) = A022342(n+1),
- a(3*A003714(n)) = a(4*A003714(n)) = A026274(n) for n > 0.
(End)

A050614 Products of distinct terms of A001566: a(n) = Product_{i=0..floor(log_2(n+1))} L(2^(i+1))^bit(n,i).

Original entry on oeis.org

1, 3, 7, 21, 47, 141, 329, 987, 2207, 6621, 15449, 46347, 103729, 311187, 726103, 2178309, 4870847, 14612541, 34095929, 102287787, 228929809, 686789427, 1602508663, 4807525989, 10749959329, 32249877987, 75249715303
Offset: 0

Views

Author

Antti Karttunen, Dec 02 1999

Keywords

Comments

Each subset a(0..(2^k)-1) gives all the divisors of F(2^(k+1)) up to k=4 (F_32) and after that a subset of such divisors. E.g., the terms a(0)-a(7) are the divisors of F_16 = 987 (A018760).

Crossrefs

Bisection of A075149 and A050613 (see there for the other Maple procedures), subset of A062877. Cf. also A050615.

Programs

  • Maple
    [seq(A050614(n),n=0..30)]; A050614 := n -> product('luc(2^(i+1))^bit_i(n,i)','i'=0..floor_log_2(n+1));
  • Mathematica
    Table[k = Floor[Log[2, n + 1]]; Product[j = 2^(i + 1); l = Fibonacci[j + 1] + Fibonacci[j - 1]; If[BitAnd[2^i, n] == 0, b = 0, b = 1]; l^b, {i, 0, k}], {n, 0, 200}] (* Robert Price, Feb 13 2017 *)

Formula

a(n) = Sum_{k=0..n} A127872(n,k)*Fibonacci(2*k+1), see A000045 and A001519. - Philippe Deléham, Aug 30 2007

A054204 Integers expressible as sums of distinct even-subscripted Fibonacci numbers.

Original entry on oeis.org

1, 3, 4, 8, 9, 11, 12, 21, 22, 24, 25, 29, 30, 32, 33, 55, 56, 58, 59, 63, 64, 66, 67, 76, 77, 79, 80, 84, 85, 87, 88, 144, 145, 147, 148, 152, 153, 155, 156, 165, 166, 168, 169, 173, 174, 176, 177, 199, 200, 202, 203, 207, 208, 210, 211, 220, 221, 223, 224, 228, 229
Offset: 1

Views

Author

Marjorie Bicknell-Johnson (marjohnson(AT)earthlink.net), Apr 30 2000

Keywords

Comments

Number of partitions of a(n) into sums of distinct Fibonacci numbers is (n+1)st term of Stern's Diatomic series A002487. This sequence has A046815 as a subsequence.

Examples

			a(9)=22 since 9=2^3+2^0 and 22=F(2(3+1)) + F(2(0+1)) = F(8) + F(2).
		

Crossrefs

Cf. A022290, A062877 (odd-indexed Fibonaccis).
Distinct additive closure of A001906.

Programs

  • Mathematica
    fibEvenCount[n_] := Plus @@ (Reverse @ IntegerDigits[n, 2])[[2 ;; -1 ;; 2]]; evenIndexed = fibEvenCount /@ Select[Range[10000], BitAnd[#, 2 #] == 0 &]; Position[evenIndexed, ?(# == 0 &)] // Flatten (* _Amiram Eldar, Jan 20 2020*)
  • PARI
    my(m=Mod('x,'x^2-3*'x+1)); a(n) = subst(lift(subst(Pol(binary(n)), 'x,m)), 'x,3); \\ Kevin Ryde, Nov 25 2020

Formula

Subscripts in Zeckendorf representation of a(n) are 2(e+1) where e is exponent used to write n as sum of powers of 2.

A062879 Integers whose Zeckendorf expansion does not contain ones at even positions.

Original entry on oeis.org

0, 2, 5, 7, 13, 15, 18, 20, 34, 36, 39, 41, 47, 49, 52, 54, 89, 91, 94, 96, 102, 104, 107, 109, 123, 125, 128, 130, 136, 138, 141, 143, 233, 235, 238, 240, 246, 248, 251, 253, 267, 269, 272, 274, 280, 282, 285, 287, 322, 324, 327, 329, 335, 337, 340, 342, 356
Offset: 1

Views

Author

Antti Karttunen, Jun 26 2001

Keywords

Crossrefs

Bisection of A062877.
Subset of A022342.

Programs

  • Mathematica
    fibOddCount[n_] := Plus @@ (Reverse@IntegerDigits[n, 2])[[1 ;; -1 ;; 2]]; oddIndexed = fibOddCount /@ Select[Range[0, 10000], BitAnd[#, 2 #] == 0 &]; -1 + Position[oddIndexed, ?(# == 0 &)] // Flatten (* _Amiram Eldar, Jan 20 2020  *)

Formula

A062880(n) = A003714(a(n)).
A165276(a(n)) = 0. - Amiram Eldar, Jan 20 2020

Extensions

Offset corrected by Amiram Eldar, Jan 20 2020

A356326 The terms in the negaFibonacci representation of a(n) are the terms in common in the negaFibonacci representations of n and -n.

Original entry on oeis.org

0, 0, 0, 0, -1, 0, 0, 0, 0, -1, -3, -3, -1, 0, 0, 0, 0, 4, 0, 0, 0, 0, -1, -3, -3, -1, -8, -8, -8, -8, -1, -3, -3, -1, 0, 0, 0, 0, 4, 0, 0, 0, 0, 12, 10, 10, 12, 0, 0, 0, 0, 4, 0, 0, 0, 0, -1, -3, -3, -1, -8, -8, -8, -8, -1, -3, -3, -1, -21, -21, -21, -21, -17
Offset: 0

Views

Author

Rémy Sigrist, Aug 03 2022

Keywords

Examples

			For n = 11:
- using F(-k) = A039834(k):
- 11 = F(-1) + F(-4) + F(-7),
- -11 = F(-4) + F(-6),
- so a(11) = F(-4) = -3.
		

Crossrefs

Programs

  • PARI
    See Links section.

Formula

a(n) = 0 iff n belongs to A062877.
a(n) = A356327(A215024(n) AND A215025(n)) (where AND denotes the bitwise AND operator).
Empirically:
- a(A000045(k)+m) = a(A000045(k+1)-m) for k >= 0, m = 0..A000045(k+1)-A000045(k).
Showing 1-6 of 6 results.