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 1-8 of 8 results.

A037098 Sequence A037093 shown in octal.

Original entry on oeis.org

0, 1, 3, 16, 71, 345, 1624, 17121, 71577, 345342, 1624171, 17121011, 71577060, 345342131, 1624173733, 17331156576, 71473314601, 345325637555, 3624144315044, 16333160460471, 71272765314407
Offset: 0

Views

Author

Antti Karttunen, Jan 29 1999

Keywords

Crossrefs

Compare to A037099.

Formula

a(n) = A007094(A037093(n)).

Extensions

Entry revised Dec 29 2007

A036284 Periodic vertical binary vectors of Fibonacci numbers.

Original entry on oeis.org

6, 24, 1440, 5728448, 92568198012160, 26494530374406845814111659520, 2095920895719545919920115988669687683503034097906010941440, 13128614603426246034591796912897206548807135027496968025827278400248602613784037111736380004928525614173642247188480
Offset: 0

Views

Author

Antti Karttunen, Nov 01 1998

Keywords

Comments

The sequence can be also computed with a recurrence that does not explicitly refer to Fibonacci numbers. See the given Maple and C programs.
Conjecture: For n>=1, each term a(n), when considered as a GF(2)[X]-polynomial, is divisible by GF(2)[X] -polynomial (x^3 + 1) ^ A000225(n-1). If this holds, then for n>=1, a(n) = A048720bi(A136380(n),A048723bi(9,A000225(n-1))). Conjecture 2: there is also one extra (x^1 + 1) factor present, see A136384.

Examples

			When Fibonacci numbers are written in binary (see A004685), under each other as:
0000000 (0)
0000001 (1)
0000001 (1)
0000010 (2)
0000011 (3)
0000101 (5)
0001000 (8)
0001101 (13)
0010101 (21)
0100010 (34)
0110111 (55)
1011001 (89)
it can be seen that the bits in the n-th column from right repeat after a period of A007283(n): 3, 6, 12, 24, ... (See also A001175). This sequence is formed from those bits: 011, reversed is 110, is binary for 6, thus a(0) = 6. 000110, reversed is 11000, is binary for 24, thus a(1) = 24, 000001011010, reversed is 10110100000, is binary for 1440, thus a(2) = 1440.

Links

Crossrefs

Same sequence in octal base: A036285. Bits reversed: A036286. See also A136378, A136379, A136380, A136382, A136384, A037096, A037093, A000045.

Programs

  • Maple
    A036284:=proc(n) option remember; local a, b, c, i, j, k, l, s, x, y, z; if (0 = n) then (6) else a := 0; b := 0; s := 0; x := 0; y := 0; k := 3*(2^(n-1)); l := 3*(2^n); j := 0; for i from 0 to l do z := bit_i(A036284(n-1),(j)); c := (a + b + (`if`((x = y),x,(z+1))) mod 2); if(c <> 0) then s := s + (2^i); fi; a := b; b := c; x := y; y := z; j := j + 1; if(j = k) then j := 0; fi; od; RETURN(s); fi; end:
bit_i := (x,i) -> `mod`(floor(x/(2^i)),2);
  • Mathematica
    a[n_] := Sum[Mod[Fibonacci[k]/2^n // Floor, 2]* 2^k, {k, 0, 3*2^n - 1}]; Table[a[n], {n, 0, 7}] (* Jean-François Alcover, Mar 04 2016 *)

Formula

a(n) = Sum_{k=0..A007283(n)-1} ([A000045(k)/(2^n)] mod 2) * 2^k, where [] stands for floor function, i.e. Sum (bit n of Fibonacci(k))*(2^k), k = 0 ... (3*(2^n))-1.

Extensions

Entry revised Dec 29 2007

A037095 "Sloping binary representation" of powers of 3 (A000244), slope = -1.

Original entry on oeis.org

1, 1, 3, 1, 3, 9, 11, 17, 19, 25, 123, 65, 195, 169, 171, 753, 435, 249, 2267, 4065, 8163, 841, 843, 31313, 29651, 39769, 38331, 30081, 160643, 49769, 53867, 563377, 700659, 1611961, 760731, 1207073, 5668771, 5566345, 11844619, 8699025, 10386067, 55868313
Offset: 0

Views

Author

Antti Karttunen, Jan 28 1999

Keywords

Examples

			When powers of 3 are written in binary (see A004656), under each other as:
  000000000001 (1)
  000000000011 (3)
  000000001001 (9)
  000000011011 (27)
  000001010001 (81)
  000011110011 (243)
  001011011001 (729)
  100010001011 (2187)
and one collects their bits from the column-0 to NW-direction (from the least to the most significant end), one gets 1 (1), 01 (1), 011 (3), 0001 (1), 00011 (3), 001001 (9), etc. (See A105033 for similar transformation done on nonnegative integers, A001477).

Links

Crossrefs

Cf. A105033, A000244, A037093, A037094, A037096, A037097, A339601.

Programs

  • Maple
    A037095:= n-> add(bit_n(3^(n-i), i)*(2^i), i=0..n):
bit_n := (x, n) -> `mod`(floor(x/(2^n)), 2):
seq(A037095(n), n=0..41);
# second Maple program:
b:= proc(n) option remember; `if`(n=0, 1, (p->
       expand((p-(p mod 2))*x/2)+3^n)(b(n-1)))
    end:
a:= n-> subs(x=2, b(n) mod 2):
seq(a(n), n=0..42);  # Alois P. Heinz, Dec 10 2020
  • PARI
    A339601(n) = { my(m=1, s=0); while(n>=m, s += bitand(m,n); m <<= 1; n \= 3); (s); };
A037095(n) = A339601(3^n); \\ Antti Karttunen, Dec 09 2020
  • PARI
    BINSLOPE(f) = n -> sum(i=0,n,bitand(2^(n-i),f(i))); \\ General transformation for these kinds of sequences.
A037095 = BINSLOPE(n -> 3^n); \\ And its application to A000244. - Antti Karttunen, Dec 09 2020

Formula

a(n) = A339601(A000244(n)). - Antti Karttunen, Dec 09 2020

Extensions

Entry revised Dec 29 2007
More terms from Sean A. Irvine, Dec 08 2020

A052005 Number of Fibonacci numbers (A000045) with length n in base 2.

Original entry on oeis.org

2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1
Offset: 1

Views

Author

Antti Karttunen and Patrick De Geest, Nov 15 1999

Keywords

Comments

There are no double 2's except at the very start because multiplying by phi^3 adds at least 2 to Fn's binary length. For a similar reason there aren't any 3's because multiplying by phi^2 increments at least by one F(n)'s binary length.
Also a(n) is the number of Fibonacci numbers F(k) between powers of 2 such that 2^n <= F(k) < 2^(n+1). - Frank M Jackson, Apr 14 2013

Examples

			F(17) = 1597{10} = 11000111101{2} the only one of length 11 and F(18) = 2584{10} = 101000011000{2} the only one of length 12 so both a(11) and a(12) equal 1.

Links

Crossrefs

Cf. A000079, A001622, A052006, A000045, A050815, A036284, A037093, A022927, A022934, A104287.

Programs

  • Mathematica
    nmax = 105; kmax = Floor[ k /. FindRoot[ Log[2, Fibonacci[k]] == nmax, {k, nmax, 2*nmax}]]; A052005 = Tally[ Length /@ IntegerDigits[ Fibonacci[ Range[kmax]], 2]][[All, 2]] (* Jean-François Alcover, May 07 2012 *)
termcount[n1_] := (m1=0; While[Fibonacci[m1]<2^n1, m1++]; m1); Table[termcount[n+1]-termcount[n], {n, 0, 200}] (* Frank M Jackson, Apr 14 2013 *)
Most[Transpose[Tally[Table[Length[IntegerDigits[Fibonacci[n], 2]], {n, 140}]]][[2]]] (* T. D. Noe, Apr 16 2013 *)

Formula

Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log(2)/log(phi) = A104287. - Amiram Eldar, Nov 21 2021

A037094 "Sloping binary representation" of Lucas numbers (A000032), slope = +1.

Original entry on oeis.org

0, 7, 29, 114, 971, 3695, 14684, 58639, 496705, 1892294, 7518347, 30023387, 258775984, 966632223, 3848859285, 32551146626, 123937019667, 492763242871, 1967451434524, 16666715013959, 63494909959113
Offset: 0

Views

Author

Antti Karttunen, Jan 28 1999

Keywords

Examples

			When Lucas numbers (A000032) are written in binary, under each other as:
0000010 (2)
0000001 (1)
0000011 (3)
0000100 (4)
0000111 (7)
0001011 (11)
0010010 (18)
0011101 (29)
0101111 (47)
1001100 (76)
and one starts collecting their bits from column-0 to SW-direction (from the least to the most significant end), one gets 000... (0), ...00111 (7), ...011101 (29), ...001110010 (114), etc. (See A102370 for similar transformation done on nonnegative integers).

Crossrefs

Cf. A000032, A037093, A037095, A037099 (same sequence in octal).

Formula

a(n) := Sum(bit_n(A000032(n+i), i)*(2^i), i=0..inf) [ bit_n := (x, n) -> `mod`(floor(x/(2^n)), 2); ]
In practice, 3n (2n?) can be used as an upper limit instead of infinity.

Extensions

Entry revised Dec 29 2007

A052006 Numbers k for which Fibonacci(k) is the first member of a 1,1 pair (A052005).

Original entry on oeis.org

17, 30, 43, 53, 66, 79, 89, 102, 115, 125, 138, 151, 161, 174, 187, 200, 210, 223, 236, 246, 259, 272, 282, 295, 308, 321, 331, 344, 357, 367, 380, 393, 403, 416, 429, 442, 452, 465, 478, 488, 501, 514, 524, 537, 550, 560, 573, 586, 599, 609, 622, 635, 645
Offset: 0

Views

Author

Antti Karttunen and Patrick De Geest, Nov 15 1999

Keywords

Comments

Keep adding the terms of sequence A052005 up to the first member of the next 1,1 pair to yield the terms of this sequence. - Patrick De Geest
Those k for which F(k-1) < 2^(floor(log_2(F(k)))) and F(k+1) >= 2^(floor(log_2(F(k)))+1) and F(k+2) >= 2^(floor(log_2(F(k)))+2).

Links

Crossrefs

The first differences are A051392.
Cf. A052005, A000045, A050815, A036284, A037093.

Programs

  • Mathematica
    With[{F = Fibonacci}, Reap[For[n=0, n<1000, n++, If[F[n-1] < 2^Floor[Log[2, F[n]]] && F[n+1] >= 2^(Floor[Log[2, F[n]]]+1) && F[n+2] >= 2^(Floor[Log[ 2, F[n]]]+2), Print[n]; Sow[n]]]][[2, 1]]] (* Jean-François Alcover, Feb 27 2016 *)

A037099 Sequence A037094 shown in octal.

Original entry on oeis.org

0, 7, 35, 162, 1713, 7157, 34534, 162417, 1712101, 7157706, 34534213, 162417333, 1733115660, 7147321437, 34532167225, 362414612202, 1633316153423, 7127276460567, 34502517467034, 362420550141507
Offset: 0

Views

Author

Antti Karttunen, Jan 29 1999

Keywords

Crossrefs

Cf. A037098, A037093.

Formula

a(n) = A007094(A037094(n)).

Extensions

Entry revised Dec 29 2007

A303655 Bit column sums in the binary expansions of Fibonacci(n)/2^n for n >= 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 5, 7, 12, 9, 10, 9, 14, 13, 18, 21, 17, 23, 16, 20, 24, 23, 23, 26, 26, 30, 29, 29, 32, 34, 32, 37, 34, 33, 43, 30, 37, 41, 46, 43, 44, 42, 52, 45, 51, 50, 53, 50, 51, 49, 55, 64, 48, 60, 53, 65, 73, 67, 58, 69, 62, 75, 65, 74, 71, 69, 68, 88, 89, 85, 67, 76, 82, 83, 76, 81, 89, 91, 98, 93, 92, 83, 104, 87, 95, 90, 85, 101, 91, 101, 105, 105, 114, 84, 104, 108, 116, 121, 104, 126, 104, 110, 131, 107, 111, 137, 109, 126, 124, 119, 127, 136, 127, 120, 122, 145, 132, 132, 127, 131, 122, 129, 130, 136, 144, 146
Offset: 1

Views

Author

Paul D. Hanna, Apr 27 2018

Keywords

Examples

			The binary expansions of Fibonacci(n)/2^n for n >= 1 begin:
.1
.01
.010
.0011
.00101
.001000
.0001101
.00010101
.000100010
.0000110111
.00001011001
.000010010000
.0000011101001
.00000101111001
.000001001100010
.0000001111011011
.00000011000111101
.000000101000011000
.0000001000001010101
.00000001101001101101
.000000010101011000010
.0000000100010100101111
.00000000110111111110001
.000000001011010100100000
.0000000010010010100010001
.00000000011101101000110001
.000000000101111111101000010
.0000000001001101100101110011
.00000000001111101100010110101
.000000000011001011001000101000
.0000000000101001000101011011101
.00000000001000010011110100000101
.000000000001101011100011111100010
.0000000000010101110000010011100111
.00000000000100011001100110011001001
.000000000000111000111101000110110000
.0000000000001011100001001111001111001
.00000000000010010101000111000000101001
.000000000000011110001010000111010100010
.0000000000000110000110010111111011001011
.00000000000001001110111101000110101101101
.000000000000001111111110000000110000111000
.0000000000000011001110101101001100110100101
.00000000000000101001110011101010010111011101
.000000000000001000011101001010011111110000010
.0000000000000001101101011100111110010101011111
.00000000000000010110001000110010010010011100001
.000000000000000100011110100011010000101001000000
.0000000000000000111001111101001100010111100100001
.00000000000000001011101110001100110011100101100001
...
the column sums of which form this sequence.
Thus, a(n) equals the number of 1-bits in column n in the binary expansions of Fibonacci(n)/2^n for n >= 1.

Links

Crossrefs

Cf. A000045, A037093.

Formula

Sum_{n>=1} a(n) / 2^n = 2.
Showing 1-8 of 8 results.

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