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 1-7 of 7 results.

A042964 Numbers that are congruent to 2 or 3 mod 4.

Original entry on oeis.org

2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 67, 70, 71, 74, 75, 78, 79, 82, 83, 86, 87, 90, 91, 94, 95, 98, 99, 102, 103, 106, 107, 110, 111, 114, 115, 118, 119, 122, 123, 126, 127
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Also numbers m such that binomial(m+2, m) mod 2 = 0. - Hieronymus Fischer, Oct 20 2007
Also numbers m such that floor(1+(m/2)) mod 2 = 0. - Hieronymus Fischer, Oct 20 2007
Partial sums of the sequence 2, 1, 3, 1, 3, 1, 3, 1, 3, 1, ... which has period 2. - Hieronymus Fischer, Oct 20 2007
In groups of four add and divide by two the odd and even numbers. - George E. Antoniou, Dec 12 2001
From Jeremy Gardiner, Jan 22 2006: (Start)
Comments on the "mystery calculator". There are 6 cards.
Card 0: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, ... (A005408 sequence).
Card 1: 2, 3, 6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, ... (this sequence).
Card 2: 4, 5, 6, 7, 12, 13, 14, 15, 20, 21, 22, 23, 28, 29, 30, 31, 36, 37, 38, 39, ... (A047566).
Card 3: 8, 9, 10, 11, 12, 13, 14, 15, 24, 25, 26, 27, 28, 29, 30, 31, 40, 41, 42, ... (A115419).
Card 4: 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 48, 49, 50, ... (A115420).
Card 5: 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, ... (A115421).
The trick: You secretly select a number between 1 and 63 from one of the cards. You indicate to me the cards on which that number appears; I tell you the number you selected!
The solution: I add together the first term from each of the indicated cards. The total equals the selected number. The numbers in each sequence all have a "1" in the same position in their binary expansion. Example: You indicate cards 1, 3 and 5. Your selected number is 2 + 8 + 32 = 42.
Numbers having a 1 in position 1 of their binary expansion. One of the mystery calculator sequences: A005408, A042964, A047566, A115419, A115420, A115421. (End)
Complement of A042948. - Reinhard Zumkeller, Oct 03 2008
Also the 2nd Witt transform of A040000 [Moree]. - R. J. Mathar, Nov 08 2008
In general, sequences of numbers congruent to {a,a+i} mod k will have a closed form of (k-2*i)*(2*n-1+(-1)^n)/4+i*n+a, from offset 0. - Gary Detlefs, Oct 29 2013
Union of A004767 and A016825; Fixed points of A098180. - Wesley Ivan Hurt, Jan 14 2014, Oct 13 2015

Links

Crossrefs

Cf. A000040, A133620, A133621, A133622, A133630, A133635.
Cf. A133872, A133882, A133890, A133900, A133910.
Card trick: A005408, A047566, A115419, A115420, A115421.
Cf. A004767, A016825 A040000, A042948, A047406, A098180.

Programs

  • Magma
    [2*n+((-1)^(n-1)-1)/2 : n in [1..100]]; // Wesley Ivan Hurt, Oct 13 2015
  • Magma
    [n: n in [1..150] | n mod 4 in [2, 3]]; // Vincenzo Librandi, Oct 13 2015
  • Maple
    A042964:=n->2*n+((-1)^(n-1)-1)/2; seq(A042964(n), n=1..100); # Wesley Ivan Hurt, Jan 07 2014
  • Mathematica
    Flatten[Table[4n + {2, 3}, {n, 0, 31}]] (* Alonso del Arte, Feb 07 2013 *)
Select[Range[200],MemberQ[{2,3},Mod[#,4]]&] (* or *) LinearRecurrence[ {1,1,-1},{2,3,6},90] (* Harvey P. Dale, Nov 28 2018 *)
  • PARI
    a(n)=2*n+2-n%2
  • PARI
    Vec((2+x+x^2)/((1-x)*(1-x^2)) + O(x^100)) \\ Altug Alkan, Oct 13 2015

Formula

a(n) = A047406(n)/2.
From Michael Somos, Jan 12 2000: (Start)
G.f.: x*(2+x+x^2)/((1-x)*(1-x^2)).
a(n) = a(n-1) + 2 + (-1)^n. (End)
a(n) = 2n if n is odd, otherwise n = 2n - 1. - Amarnath Murthy, Oct 16 2003
a(n) = (3 + (-1)^(n-1))/2 + 2*(n-1) = 2n + 2 - (n mod 2). - Hieronymus Fischer, Oct 20 2007
A133872(a(n)) = 0. - Reinhard Zumkeller, Oct 03 2008
a(n) = 4*n - a(n-1) - 3 (with a(1) = 2). - Vincenzo Librandi, Nov 17 2010
a(n) = 2*n + ((-1)^(n-1) - 1)/2. - Gary Detlefs, Oct 29 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/8 - log(2)/4. - Amiram Eldar, Dec 05 2021
E.g.f.: 1 + ((4*x - 1)*exp(x) - exp(-x))/2. - David Lovler, Aug 08 2022

Extensions

Edited by N. J. A. Sloane, Jun 30 2008 at the suggestion of R. J. Mathar
Corrected by Jaroslav Krizek, Dec 18 2009

A047431 Numbers that are congruent to {1, 4, 5, 6} mod 8.

Original entry on oeis.org

1, 4, 5, 6, 9, 12, 13, 14, 17, 20, 21, 22, 25, 28, 29, 30, 33, 36, 37, 38, 41, 44, 45, 46, 49, 52, 53, 54, 57, 60, 61, 62, 65, 68, 69, 70, 73, 76, 77, 78, 81, 84, 85, 86, 89, 92, 93, 94, 97, 100, 101, 102, 105, 108, 109, 110, 113, 116, 117, 118, 121, 124
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A016813, A047404, A047406, A047546, A056594.

Programs

Formula

G.f.: x*(1+2*x-x^2+2*x^3)/((x^2+1)*(x-1)^2). - R. J. Mathar, Oct 08 2011
a(n) = (-2-(-i)^n-i^n+4n)/2 where i=sqrt(-1). - Colin Barker, Jun 06 2012
From Wesley Ivan Hurt, May 30 2016: (Start)
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4) for n>4.
a(2k) = A047406(k), a(2k-1) = A016813(k-1) k>0. (End)
E.g.f.: 2 - cos(x) - (1 - 2*x)*exp(x). - Ilya Gutkovskiy, May 30 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*Pi/16 + 3*log(2)/8. - Amiram Eldar, Dec 24 2021

A047425 Numbers that are congruent to {3, 4, 5, 6} mod 8.

Original entry on oeis.org

3, 4, 5, 6, 11, 12, 13, 14, 19, 20, 21, 22, 27, 28, 29, 30, 35, 36, 37, 38, 43, 44, 45, 46, 51, 52, 53, 54, 59, 60, 61, 62, 67, 68, 69, 70, 75, 76, 77, 78, 83, 84, 85, 86, 91, 92, 93, 94, 99, 100, 101, 102, 107, 108, 109, 110, 115, 116, 117, 118, 123, 124
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

Complement of numbers congruent to {0, 1, 2, 7} mod 8. - Jaroslav Krizek, Dec 19 2009
In general, sequences congruent to {a, a + i, a + 2i, ..., a + pi} mod k and a + p*i < k have a general form of (k - i*p)*floor(n/p) + i*n + a, from offset 0. - Gary Detlefs, Oct 20 2013

Links

Crossrefs

Cf. A047406, A047621.

Programs

Formula

G.f.: x*(3+x+x^2+x^3+2*x^4) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
a(n) = 8*floor((n-1)/4) + ((n-1) mod 4) + 3.
a(n) = OR(n-1, 1) + OR(n-1, 2). - Gary Detlefs, Oct 20 2013
From Wesley Ivan Hurt, May 31 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (4*n-1-i^(2*n)-(1-i)*i^(-n)-(1+i)*i^n)/2 where i=sqrt(-1).
a(2k) = A047406(k), a(2k-1) = A047621(k). (End)
E.g.f.: 2 + sin(x) - cos(x) + 2*x*sinh(x) + (2*x - 1)*cosh(x). - Ilya Gutkovskiy, May 31 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/16 + (3-sqrt(2))*log(2)/8 + sqrt(2)*log(2-sqrt(2))/4. - Amiram Eldar, Dec 26 2021

A047430 Numbers that are congruent to {0, 4, 5, 6} mod 8.

Original entry on oeis.org

0, 4, 5, 6, 8, 12, 13, 14, 16, 20, 21, 22, 24, 28, 29, 30, 32, 36, 37, 38, 40, 44, 45, 46, 48, 52, 53, 54, 56, 60, 61, 62, 64, 68, 69, 70, 72, 76, 77, 78, 80, 84, 85, 86, 88, 92, 93, 94, 96, 100, 101, 102, 104, 108, 109, 110, 112, 116, 117, 118, 120, 124
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A047406, A047615.

Programs

  • Magma
    [n : n in [0..150] | n mod 8 in [0, 4, 5, 6]]; // Wesley Ivan Hurt, May 25 2016
  • Maple
    A047430:=n->(8*n-5+I^(2*n)-(2+I)*I^(-n)-(2-I)*I^n)/4: seq(A047430(n), n=1..100); # Wesley Ivan Hurt, May 25 2016
  • Mathematica
    Table[(8n-5+I^(2n)-(2+I)*I^(-n)-(2-I)*I^n)/4, {n, 80}] (* Wesley Ivan Hurt, May 25 2016 *)
Select[Range[0, 124], MemberQ[{0, 4, 5, 6}, Mod[#, 8]] &] (* Michael De Vlieger, May 25 2016 *)
LinearRecurrence[{1,0,0,1,-1},{0,4,5,6,8},100] (* Harvey P. Dale, Aug 05 2023 *)

Formula

G.f.: x^2*(4+x+x^2+2*x^3) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Dec 07 2011
From Wesley Ivan Hurt, May 25 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (8*n-5+i^(2*n)-(2+i)*i^(-n)-(2-i)*i^n)/4 where i=sqrt(-1).
a(2k) = A047406(k), a(2k-1) = A047615(k). (End)
E.g.f.: (4 - sin(x) - 2*cos(x) + (4*x - 3)*sinh(x) + (4*x - 2)*cosh(x))/2. - Ilya Gutkovskiy, May 25 2016
Sum_{n>=2} (-1)^n/a(n) = sqrt(2)*log(2+sqrt(2))/8 - (2-sqrt(2))*(Pi-log(2))/16. - Amiram Eldar, Dec 23 2021

A047433 Numbers that are congruent to {2, 4, 5, 6} mod 8.

Original entry on oeis.org

2, 4, 5, 6, 10, 12, 13, 14, 18, 20, 21, 22, 26, 28, 29, 30, 34, 36, 37, 38, 42, 44, 45, 46, 50, 52, 53, 54, 58, 60, 61, 62, 66, 68, 69, 70, 74, 76, 77, 78, 82, 84, 85, 86, 90, 92, 93, 94, 98, 100, 101, 102, 106, 108, 109, 110, 114, 116, 117, 118, 122, 124
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A047406, A047617.

Programs

  • Magma
    [n : n in [0..150] | n mod 8 in [2, 4, 5, 6]]; // Wesley Ivan Hurt, May 26 2016
  • Maple
    A047433:=n->(8*n-3-I^(2*n)-(2-I)*I^(-n)-(2+I)*I^n)/4: seq(A047433(n), n=1..100); # Wesley Ivan Hurt, May 26 2016
  • Mathematica
    Select[Range[120], MemberQ[{2,4,5,6}, Mod[#,8]]&]  (* Harvey P. Dale, Mar 04 2011 *)

Formula

G.f.: x*(2+2*x+x^2+x^3+2*x^4) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Dec 07 2011
From Wesley Ivan Hurt, May 26 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (8*n-3-i^(2*n)-(2-i)*i^(-n)-(2+i)*i^n)/4 where i=sqrt(-1).
a(2k) = A047406(k), a(2k-1) = A047617(k). (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = (3-sqrt(2))*Pi/16 + log(2)/4 + sqrt(2)*log(sqrt(2)-1)/8. - Amiram Eldar, Dec 25 2021

A160108 Lodumo_8 of Fibonacci numbers.

Original entry on oeis.org

0, 1, 9, 2, 3, 5, 8, 13, 21, 10, 7, 17, 16, 25, 33, 18, 11, 29, 24, 37, 45, 26, 15, 41, 32, 49, 57, 34, 19, 53, 40, 61, 69, 42, 23, 65, 48, 73, 81, 50, 27, 77, 56, 85, 93, 58, 31, 89, 64, 97, 105, 66, 35, 101, 72, 109, 117, 74, 39, 113, 80, 121, 129, 82, 43, 125, 88, 133, 141
Offset: 0

Views

Author

Philippe Deléham, May 02 2009

Keywords

Comments

Some integers (see A047406) are missing.

Links

Crossrefs

Cf. A000045, A079344.

Formula

Lodumo_8 transform of A000045 (for definition see Transforms).
Conjecture: a(n) = 2*a(n-6)-a(n-12). - Colin Barker, Oct 04 2014
Empirical g.f.: x*(7*x^10 +x^9 +6*x^8 +3*x^7 +11*x^6 +8*x^5 +5*x^4 +3*x^3 +2*x^2 +9*x +1) / ((x -1)^2*(x +1)^2*(x^2 -x +1)^2*(x^2 +x +1)^2). - Colin Barker, Oct 04 2014

Extensions

Replaced 35 by 33 - R. J. Mathar, May 03 2009

A255842 a(n) = 2*n^2 + 12.

Original entry on oeis.org

12, 14, 20, 30, 44, 62, 84, 110, 140, 174, 212, 254, 300, 350, 404, 462, 524, 590, 660, 734, 812, 894, 980, 1070, 1164, 1262, 1364, 1470, 1580, 1694, 1812, 1934, 2060, 2190, 2324, 2462, 2604, 2750, 2900, 3054, 3212, 3374, 3540, 3710, 3884, 4062, 4244, 4430
Offset: 0

Views

Author

Avi Friedlich, Mar 08 2015

Keywords

Comments

This is the case k=6 of the form (n + sqrt(k))^2 + (n - sqrt(k))^2. Also, it is noted that a(n)*n = (n + sqrt(2))^3 + (n - sqrt(2))^3.
Equivalently, numbers m such that 2*m - 24 is a square.
For n = 0..10, a(n) - 1 is prime (see A092968).

Links

Crossrefs

Cf. A016825 (first differences), A092968, A114949.
Subsequence of A047238 and A047406.
Cf. similar sequences listed in A255843.

Programs

  • Magma
    [2*n^2+12: n in [0..50]];
  • Mathematica
    Table[2 n^2 + 12, {n, 0, 50}]
  • PARI
    vector(50, n, n--; 2*n^2+12)
  • Sage
    [2*n^2+12 for n in (0..50)]

Formula

G.f.: 2*(6 - 11*x + 7*x^2)/(1 - x)^3.
a(n) = a(-n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 2*A114949(n).
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = (1 + sqrt(6)*Pi*coth(sqrt(6)*Pi))/24.
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(6)*Pi*cosech(sqrt(6)*Pi))/24. (End)
E.g.f.: 2*exp(x)*(6 + x + x^2). - Elmo R. Oliveira, Jan 24 2025

Extensions

Edited by Bruno Berselli, Mar 11 2015
Showing 1-7 of 7 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.