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.

Previous Showing 21-28 of 28 results.

A173598 Period 6: repeat [1, 8, 7, 2, 4, 5].

Original entry on oeis.org

1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8
Offset: 0

Views

Author

Paul Curtz, Nov 23 2010

Keywords

Comments

For A141425 = 1,2,4,5,7,8 permutations, see A153130. a(n) is based on A022998. Successive differences are linked to A070366.

Crossrefs

Programs

Formula

a(n) = A166138(n) mod 9.
a(2n+1) + a(2n+2) = 9.
G.f.: (1+8*x+7*x^2+2*x^3+4*x^4+5*x^5) / ((1-x)*(1+x)*(1+x+x^2)*(x^2-x+1)). - R. J. Mathar, Mar 08 2011
From Wesley Ivan Hurt, Jun 23 2016: (Start)
a(n) = a(n-6) for n>5.
a(n) = (9 - cos(n*Pi) - 6*cos(2*n*Pi/3) + 2*sqrt(3)*sin(n*Pi/3))/2. (End)

A177883 Period 6: repeat [4, 5, 7, 2, 1, 8].

Original entry on oeis.org

4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5
Offset: 0

Views

Author

Paul Curtz, Dec 14 2010

Keywords

Comments

Represents also the decimal expansion of 16934/37037 and the continued fractions of 0.23839... = (sqrt(496555)-667)/158 or of 4.194699... = (667+sqrt(496555))/327. - R. J. Mathar, Dec 20 2010

Crossrefs

Cf. A173598, A141425, A153130 (permutations).

Programs

Formula

a(n) = A166304(n) mod 9 = A022998(3n+2) mod 9.
a(2n) + a(2n+1) = 9.
G.f.: (4+5*x+7*x^2+2*x^3+x^4+8*x^5) / ( (1-x)*(1+x)*(1+x+x^2)*(x^2-x+1) ). - R. J. Mathar, Dec 20 2010
From Wesley Ivan Hurt, Jun 18 2016: (Start)
a(n) = a(n-6) for n>5.
a(n) = (9 -cos(n*Pi) + 3*cos(n*Pi/3) - 3*cos(2*n*Pi/3) + sqrt(3)*sin(n*Pi/3) - 3*sqrt(3)*sin(2*n*Pi/3))/2. (End)

A254076 a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n>2, a(0)=-1, a(1)=-2, a(2)=-4.

Original entry on oeis.org

-1, -2, -4, 1, 2, 13, 26, 61, 122, 253, 506, 1021, 2042, 4093, 8186, 16381, 32762, 65533, 131066, 262141, 524282, 1048573, 2097146, 4194301, 8388602, 16777213, 33554426, 67108861, 134217722, 268435453, 536870906, 1073741821, 2147483642, 4294967293
Offset: 0

Views

Author

Paul Curtz, Jan 29 2015

Keywords

Comments

The main diagonal of the difference table is -A000079(n) = -2^n.
a(n) mod 9 is of period 6: repeat 8, 7, 5, 1, 2, 4.
a(n) + a(n+1) = -3, -6, -3, 3, 15, ...; all are multiples of 3.

Crossrefs

Programs

  • Mathematica
    a[0] = -1; a[n_] := 2^(n-1) + 3*Mod[n, 2] - 6; Table[a[n], {n, 0, 33}] (* Jean-François Alcover, Feb 04 2015 *)
  • PARI
    Vec((9*x^3+x^2-1)/((x-1)*(x+1)*(2*x-1)) + O(x^100)) \\ Colin Barker, Jan 30 2015

Formula

a(2n+1) = A141725(n-1), a(2n+2) = 2*a(2n+1).
a(n+1) = 2*a(n) + (period 2: repeat 0, 9), n>0.
a(n) = -A157823(n) - (period 2: repeat 6, 3).
a(n+1) = a(n) - A156067(n).
a(n+2) = a(n) + 3*2^(n-1), n>0.
a(n+4) = a(n) + 15*2^(n-1), n>0.
a(n+6) = a(n) + 63*2^(n-1), n>0.
a(n) = (2^n - 3*(-1)^n - 9)/2 for n>0. - Colin Barker, Jan 30 2015
G.f.: (9*x^3+x^2-1) / ((x-1)*(x+1)*(2*x-1)). - Colin Barker, Jan 30 2015

A154529 A090040 mod 9.

Original entry on oeis.org

1, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8
Offset: 0

Views

Author

Paul Curtz, Jan 11 2009

Keywords

Comments

For n>2, equal to 2^(n-2) mod 9 [From Michael B. Porter, Feb 02 2010]
Apart from leading terms the same as A146501, A153130 and A029898. [From R. J. Mathar, Apr 13 2010]

Programs

  • Mathematica
    Join[{1},LinearRecurrence[{1,0,-1,1},{5,1,2,4},101]] (* Ray Chandler, Jul 15 2015 *)

Formula

a(n)=a(n-1)-a(n-3)+a(n-4), n>4. G.f.: (6*x^4+2*x^3+4*x+1-4*x^2)/((1-x)*(1+x)*(x^2-x+1)). [From R. J. Mathar, Feb 25 2009]

Extensions

Edited by N. J. A. Sloane, Jan 12 2009
Extended by Ray Chandler, Jul 15 2015

A158927 a(n) = -3a(n-1) - 3a(n-2) - 2a(n-3), n > 3.

Original entry on oeis.org

2, 2, 2, -7, 11, -16, 29, -61, 128, -259, 515, -1024, 2045, -4093, 8192, -16387, 32771, -65536, 131069, -262141, 524288, -1048579, 2097155, -4194304, 8388605, -16777213, 33554432, -67108867, 134217731, -268435456, 536870909, -1073741821, 2147483648
Offset: 0

Views

Author

Paul Curtz, Mar 31 2009

Keywords

Comments

The inverse binomial transform of A153130, after dropping A153130(0).
The inverse binomial transform of the full A153130 is A158916.
Dropping two initial terms of A153130 yields A158935, dropping three yields essentially a sign-reversed version of A158916, dropping 4 essentially the sequence here.

Crossrefs

Same recurrence as A131562, A158916, A158926.

Programs

  • GAP
    a := [2,2,2,-7];; for n in [5..10^3] do a[n] := -3*a[n-1] - 3*a[n-2] - 2*a[n-3]; od; a; # Muniru A Asiru, Jan 27 2018
  • Maple
    a := proc(n) option remember: if n=0 then 2 elif n=1 then 2 elif n=2 then 2 elif n=3 then -7 elif n>=4 then -3*procname(n-1) - 3*procname(n-2) - 2*procname(n-3) fi; end:
    seq(a(n), n=0..100); # Muniru A Asiru, Jan 27 2018

Formula

a(n) = -3a(n-1) - 3a(n-2) - 2a(n-3), with a(0)=a(1)=a(2)=2, a(3)=-7.
a(n) = (-1)^(n+1)*A157823(n) - A099838(n+3).
G.f.: (2+8*x+14*x^2+9*x^3)/((2*x+1)*(1+x+x^2)). - R. J. Mathar, Apr 09 2009
a(0)=2; a(n) = (1/2)*(-2)^n - 3*cos(2*Pi*n/3) + sqrt(3)*sin(2*Pi*n/3) for n >= 1. - Richard Choulet, Apr 23 2009

Extensions

Edited and extended by R. J. Mathar, Apr 09 2009

A158935 a(n)= -3a(n-1)-3a(n-2)-2a(n-3), n>3. a(0)=4, a(1)=4, a(2)=-5, a(3)=4.

Original entry on oeis.org

4, 4, -5, 4, -5, 13, -32, 67, -131, 256, -509, 1021, -2048, 4099, -8195, 16384, -32765, 65533, -131072, 262147, -524291, 1048576, -2097149, 4194301, -8388608, 16777219, -33554435, 67108864, -134217725, 268435453, -536870912, 1073741827, -2147483651
Offset: 0

Views

Author

Paul Curtz, Mar 31 2009

Keywords

Comments

The third column of the array of differences described in A153130. The first two columns are in A158916 and A158987. Taking differences like in A158926 keeps the recurrence.
Also the inverse binomial transform of A153130 if the first two items of A153130 are omitted.

Programs

  • Mathematica
    Join[{4},LinearRecurrence[{-3,-3,-2},{4,-5,4},50]] (* Harvey P. Dale, May 25 2011 *)

Formula

a(n)= A154589(n) + A099838(n+2).
G.f.: (4+16*x+19*x^2+9*x^3)/((2*x+1)*(1+x+x^2)). - R. J. Mathar, Apr 08 2009

Extensions

Partially edited and extended by R. J. Mathar, Apr 08 2009
Edited by N. J. A. Sloane, Apr 08 2009

A225539 Numbers n where 2^n and n have the same digital root.

Original entry on oeis.org

5, 16, 23, 34, 41, 52, 59, 70, 77, 88, 95, 106, 113, 124, 131, 142, 149, 160, 167, 178, 185, 196, 203, 214, 221, 232, 239, 250, 257, 268, 275, 286, 293, 304, 311, 322, 329, 340, 347, 358, 365, 376, 383, 394, 401, 412, 419, 430, 437, 448
Offset: 1

Views

Author

Marcus Hedbring, May 17 2013

Keywords

Comments

The digital roots of n have a cycle length of 9 (A010888) and the digital roots of 2^n have a cycle length of 6 (A153130). Therefore, if n is a term so is n+18.
The only values of the digital roots of a(n) are 5 and 7 (A010718).

Examples

			For n=23, the digital root of n is 5. 2^n equals 8388608 so the digital root of 2^n is 5 as well.
		

Crossrefs

Programs

  • Mathematica
    digitalRoot[n_] :=  Module[{r = n}, While[r > 9, r = Total[IntegerDigits[ r]]]; r]; Select[Range[448], digitalRoot[2^#] == digitalRoot[#] &] (* T. D. Noe, May 19 2013 *)
    LinearRecurrence[{1,1,-1},{5,16,23},60] (* Harvey P. Dale, Dec 29 2018 *)
  • PARI
    forstep(n=16,500,[7,11],print1(n", ")) \\ Charles R Greathouse IV, May 19 2013

Formula

a(n) = 9*n - 3 + (-1)^n.
a(n) = a(n-1) + 7 (odd n), a(n) = a(n-1) + 11 (even n) with a(1) = 5.
G.f. x*(5 + 11*x + 2*x^2) / ((1-x)^2 * (1+x)). - Joerg Arndt, May 17 2013

A294364 Linear recurrence with signature (1,1,-1,1,1), where the first terms are powers of 2 (1,2,4,8,16).

Original entry on oeis.org

1, 2, 4, 8, 16, 23, 37, 56, 94, 152, 250, 401, 649, 1046, 1696, 2744, 4444, 7187, 11629, 18812, 30442, 49256, 79702, 128957, 208657, 337610, 546268, 883880, 1430152, 2314031, 3744181, 6058208, 9802390, 15860600, 25662994, 41523593, 67186585, 108710174, 175896760, 284606936
Offset: 0

Views

Author

Keywords

Comments

The interest of this sequence mainly lies in the peculiarities of its array of successive differences, which begins:
1, 2, 4, 8, 16, 23, 37, 56, 94, ...
1, 2, 4, 8, 7, 14, 19, 38, 58, ...
1, 2, 4, -1, 7, 5, 19, 20, 40, ...
1, 2, -5, 8, -2, 14, 1, 20, 13, ...
1, -7, 13, -10, 16, -13, 19, -7, 31, ...
-8, 20, -23, 26, -29, 32, -26, 38, -23, ...
28, -43, 49, -55, 61, -58, 64, -61, 67, ...
The main diagonal is A000079 (powers of 2).
The first upper subdiagonal is A254076.
The second upper subdiagonal (4, 8, 7, 14, 19, 38, ...) is not in the OEIS.
The third upper subdiagonal is A185346 (2^n-9).
Every row, once computed mod 9, is 6-periodic, repeating (1, 2, 4, 8, 7, 5) (A153130).

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{1, 1, -1, 1, 1}, {1, 2, 4, 8, 16}, 40]

Formula

G.f.: (1+x+x^2+3*x^3+5*x^4) / (1-x-x^2+x^3-x^4-x^5).
a(n) = (9/2)*fibonacci(n) + (-1)^n - sqrt(3)*sin(n*Pi/3).
a(n) ~ (9/2)*fibonacci(n).
Previous Showing 21-28 of 28 results.