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.

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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

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Author

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

Crossrefs

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

A133621 Numbers k such that binomial(k+p,k) mod k = 1, where p=10.

Original entry on oeis.org

3, 4, 11, 13, 17, 19, 23, 29, 31, 33, 37, 41, 43, 47, 53, 57, 59, 61, 67, 68, 71, 73, 79, 83, 85, 87, 89, 97, 101, 103, 107, 109, 111, 113, 121, 127, 131, 137, 139, 141, 143, 149, 151, 157, 163, 164, 167, 169, 173, 179, 181, 185, 187, 191, 193, 197, 199, 209, 211
Offset: 1

Views

Author

Hieronymus Fischer, Sep 30 2007

Keywords

Comments

All primes q > p are included, in that binomial(q+p,q) == (1+floor(p/q)) == 1 (mod q) holds for those primes.

Crossrefs

Programs

  • PARI
    isok(n) = ((binomial(n+10, n) % n) == 1) \\ Michel Marcus, Jul 15 2013

A133635 Nonprime numbers k such that binomial(k+p,k) mod k = 1, where p=5.

Original entry on oeis.org

26, 34, 49, 58, 74, 77, 82, 91, 98, 106, 119, 121, 122, 133, 143, 146, 154, 161, 169, 178, 187, 194, 202, 203, 209, 217, 218, 221, 226, 242, 247, 253, 259, 266, 274, 287, 289, 298, 299, 301, 314, 319, 322, 323, 329, 338, 341, 343, 346, 361, 362, 371, 377, 386
Offset: 1

Views

Author

Hieronymus Fischer, Sep 30 2007

Keywords

Crossrefs

Programs

  • Mathematica
    nn=400;With[{c=Complement[Range[nn],Prime[Range[PrimePi[nn]]]]}, Select[ c,Mod[Binomial[#+5,#],#]==1&]] (* Harvey P. Dale, Sep 24 2012 *)
  • Python
    from itertools import count, islice
    from math import comb
    from sympy import isprime
    def A133635_gen(startvalue=1): # generator of terms >= startvalue
        return filter(lambda k:comb(k+5,k)%k==1 and not isprime(k),count(max(startvalue,1)))
    A133635_list = list(islice(A133635_gen(),30)) # Chai Wah Wu, Feb 22 2023

A133623 Binomial(n+p, n) mod n where p=3.

Original entry on oeis.org

0, 0, 2, 3, 1, 0, 1, 5, 4, 6, 1, 11, 1, 8, 6, 9, 1, 16, 1, 11, 8, 12, 1, 21, 1, 14, 10, 15, 1, 26, 1, 17, 12, 18, 1, 31, 1, 20, 14, 21, 1, 36, 1, 23, 16, 24, 1, 41, 1, 26, 18, 27, 1, 46, 1, 29, 20, 30, 1, 51, 1, 32, 22, 33, 1, 56, 1, 35, 24, 36, 1, 61, 1, 38, 26, 39, 1, 66, 1, 41, 28, 42, 1
Offset: 1

Views

Author

Hieronymus Fischer, Sep 30 2007

Keywords

Comments

Let d(m)...d(2)d(1)d(0) be the base-n representation of n+p. The relation a(n)=d(1) holds, if n is a prime index. For this reason there are infinitely many terms which are equal to 1.
Appears to satisfy the recurrence: a(n) = -2*a(n-1) - a(n-2) + 2*a(n-3) + 4*a(n-4) + 2*a(n-5) - a(n-6) - 2*a(n-7) - a(n-8) for n > 14. - Chai Wah Wu, May 25 2016

Crossrefs

Programs

  • Mathematica
    Table[Mod[Binomial[n+3,n],n],{n,90}] (* Harvey P. Dale, Nov 22 2011 *)

Formula

a(n) = binomial(n+3,3) mod n.
a(n)=1 if n is a prime > 3, since binomial(n+3,n)==(1+floor(3/n))(mod n), provided n is a prime.
From Chai Wah Wu, May 26 2016: (Start)
a(n) = (n^3 + 5*n + 6)/6 mod n.
For n > 6:
if n mod 6 == 0, then a(n) = 5*n/6 + 1.
if n mod 6 is in {1, 5}, then a(n) = 1.
if n mod 6 is in {2, 4}, then a(n) = n/2 + 1.
if n mod 6 == 3, then a(n) = n/3 + 1.
(End)

A133622 a(n) = 1 if n is odd, a(n) = n/2+1 if n is even.

Original entry on oeis.org

1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 1, 10, 1, 11, 1, 12, 1, 13, 1, 14, 1, 15, 1, 16, 1, 17, 1, 18, 1, 19, 1, 20, 1, 21, 1, 22, 1, 23, 1, 24, 1, 25, 1, 26, 1, 27, 1, 28, 1, 29, 1, 30, 1, 31, 1, 32, 1, 33, 1, 34, 1, 35, 1, 36, 1, 37, 1, 38, 1, 39, 1, 40, 1, 41, 1, 42, 1, 43, 1, 44, 1
Offset: 1

Views

Author

Hieronymus Fischer, Sep 30 2007

Keywords

Comments

a(n) is the count of terms a(n+1) present so far in the sequence, with a(n+1) included in the count; example: a(1) = 1 "says" that there is 1 term "2" so far in the sequence; a(2) = 2 "says" that there are 2 terms "1" so far in the sequence... etc. This comment was inspired by A039617. - Eric Angelini, Mar 03 2020

Crossrefs

Programs

  • Haskell
    import Data.List (transpose)
    a133622 n = (1 - m) * n' + 1 where (n', m) = divMod n 2
    a133622_list = concat $ transpose [[1, 1 ..], [2 ..]]
    -- Reinhard Zumkeller, Feb 20 2015
    
  • Maple
    seq([1,n][],n=2..100); # Robert Israel, May 27 2016
  • Mathematica
    Riffle[Range[2,50],1,{1,-1,2}] (* Harvey P. Dale, Jan 19 2013 *)
  • PARI
    a(n)=if(n%2,1,n/2+1) \\ Charles R Greathouse IV, Sep 02 2015

Formula

a(n)=1+(binomial(n+1,2)mod n)=1+(binomial(n+1,n-1)mod n).
a(n)=binomial(n+2,2) mod n = binomial(n+2,n) mod n for n>2.
a(n)=1+(1+(-1)^n)*n/4.
a(n)=1+(A000217(n) mod n).
a(n)=a(n-2)+1, if n is even, a(n)=a(n-2) if n is odd.
a(n)=a(n-2)+1-(n mod 2)=a(n-2)+(1+(-1)^n)/2 for n>2.
a(n)=(a(n-3)+a(n-2))/a(n-1) for n>3.
G.f.: g(x)=x(1+2x-x^2-x^3)/(1-x^2)^2.
G.f.: (Q(0)-1-x)/x^2, where Q(k)= 1 + (k+1)*x/(1 - x/(x + (k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 23 2013
a(n) = 2*a(n-2)-a(n-4) for n > 4. - Chai Wah Wu, May 26 2016
E.g.f.: exp(x) - 1 + x*sinh(x)/2. - Robert Israel, May 27 2016

A133875 n modulo 5 repeated 5 times.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 5^2 = 25.

Crossrefs

Programs

  • Magma
    [(1 + Floor(n/5)) mod 5 : n in [0..50]]; // Wesley Ivan Hurt, Jun 06 2014
  • Maple
    A133875:=n->((1+floor(n/5)) mod 5); seq(A133875(n), n=0..100); # Wesley Ivan Hurt, Jun 06 2014
  • Mathematica
    Table[Mod[1 + Floor[n/5], 5], {n, 0, 100}] (* Wesley Ivan Hurt, Jun 06 2014 *)
    LinearRecurrence[{1,0,0,0,-1,1,0,0,0,-1,1,0,0,0,-1,1,0,0,0,-1,1},{1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,0},120] (* Harvey P. Dale, Dec 14 2017 *)

Formula

a(n) = (1 + floor(n/5)) mod 5.
a(n) = A010874(A002266(n+5)).
a(n) = 1 + floor(n/5) - 5*floor((n+5)/25).
a(n) = (((n+5) mod 25) - (n mod 5)) / 5.
a(n) = ((n + 5 - (n mod 5)) / 5) mod 5.
a(n) = A010874((n + 5 - A010874(n))/5).
a(n) = binomial(n+5, n) mod 5 = binomial(n+5, 5) mod 5.
a(n) = +a(n-1) -a(n-5) +a(n-6) -a(n-10) +a(n-11) -a(n-15) +a(n-16) -a(n-20) +a(n-21). - R. J. Mathar, Sep 03 2011
G.f.: ( 1+2*x^5+3*x^10+4*x^15 ) / ( (1-x)*(x^20+x^15+x^10+x^5+1) ). - R. J. Mathar, Sep 03 2011

A133885 Binomial(n+5,n) mod 5^2.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length 5^3=125.

Crossrefs

For the sequence regarding binomial(n+5, n) mod 5 see A133875.

Programs

  • Mathematica
    Table[Mod[Binomial[n+5,n],25],{n,0,90}] (* Harvey P. Dale, Jan 12 2023 *)

Formula

a(n)=binomial(n+5,5) mod 5^2.
G.f. g(x)=sum{0<=k<125, a(k)*x^k}/(1-x^125).

A133911 Number of prime factors (counted with multiplicity) of the period numbers defined by A133900.

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Examples

			a(6)=5, since A133900(6)=72=2*2*2*3*3.
a(12)=8, since A133900(12)=864=2*2*2*2*2*3*3*3.
		

Crossrefs

Formula

a(n)=A001222(A133900(n)).

A134331 Sum of prime factors (counted with multiplicity) of the period numbers defined by A133900.

Original entry on oeis.org

0, 4, 6, 8, 10, 12, 14, 12, 12, 18, 22, 19, 26, 22, 19, 16, 34, 22, 38, 22, 23, 32, 46, 23, 20, 36, 18, 26, 58, 37, 62, 20, 34, 46, 29, 29, 74, 50, 38, 31, 82, 38, 86, 36, 30, 58, 94, 30, 28, 32, 46, 40, 106, 30, 37, 37, 50, 70, 118, 41, 122, 74, 36, 24, 41, 48, 134, 50, 58, 50
Offset: 1

Views

Author

Hieronymus Fischer, Oct 20 2007

Keywords

Examples

			a(6)=12, since A133900(6)=72=2*2*2*3*3 and 2+2+2+3+3=12.
a(12)=19, since A133900(12)=864=2*2*2*2*2*3*3*3 and 2+2+2+2+2+3+3+3=19.
		

Crossrefs

A362686 Binomial(n+p, n) mod n where p=6.

Original entry on oeis.org

0, 0, 0, 2, 2, 0, 1, 3, 1, 8, 1, 0, 1, 8, 9, 5, 1, 10, 1, 10, 15, 12, 1, 15, 6, 14, 1, 8, 1, 12, 1, 9, 12, 18, 8, 10, 1, 20, 27, 19, 1, 36, 1, 12, 10, 24, 1, 45, 1, 36, 18, 14, 1, 28, 12, 15, 39, 30, 1, 48, 1, 32, 1, 17, 14, 12, 1, 18, 24, 50, 1, 19, 1, 38
Offset: 1

Views

Author

Ray Chandler, Apr 29 2023

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Mod[Binomial[n+6, n], n], {n, 90}]

Formula

a(n)=binomial(n+6,n) mod n.
For n > 1452, a(n) = 2*a(n-720) - a(n-1440).
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