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 1-3 of 3 results.

A109720 Periodic sequence {0,1,1,1,1,1,1} or n^6 mod 7.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1
Offset: 0

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Author

Bruce Corrigan (scentman(AT)myfamily.com), Aug 09 2005

Keywords

Comments

This sequence also represents n^12 mod 7; n^18 mod 7; (exponents are = 0 mod 6).
Characteristic sequence for numbers n>=1 to be relatively prime to 7. - Wolfdieter Lang, Oct 29 2008
a(n+4), n>=0, (periodic 1,1,1,0,1,1,1) is also the characteristic sequence for mod m reduced positive odd numbers (i.e., gcd(2*n+1,m)=1, n>=0) for each modulus m from 7*A003591 = [7,14,28,49,56,98,112,196,...]. [Wolfdieter Lang, Feb 04 2012]

Crossrefs

Cf. A010876 = n mod 7; A053879 = n^2 mod 7; A070472 = m^3 mod 7; A070512 = n^4 mod 7; A070593 = n^5 mod 7.

Programs

Formula

a(n) = 0 if n=0 mod 7; a(n)= 1 else.
G.f. = (x+x^2+x^3+x^4+x^5+x^6)/(1-x^7)= -x*(1+x)*(1+x+x^2)*(x^2-x+1) / ( (x-1)*(1+x+x^2+x^3+x^4+x^5+x^6) ).
a(n)=1-A082784(n); a(A047304(n))=1; a(A008589(n))=0; A033439(n) = SUM(a(k)*(n-k): 0<=k<=n). - Reinhard Zumkeller, Nov 30 2009
Multiplicative with a(p) = (if p=7 then 0 else 1), p prime. - Reinhard Zumkeller, Nov 30 2009
Dirichlet g.f. (1-7^(-s))*zeta(s). - R. J. Mathar, Mar 06 2011
For the general case: the characteristic function of numbers that are not multiples of m is a(n)=floor((n-1)/m)-floor(n/m)+1, m,n > 0. - Boris Putievskiy, May 08 2013

A282779 Period of cubes mod n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, 21, 22, 23, 24, 25, 26, 9, 28, 29, 30, 31, 32, 33, 34, 35, 12, 37, 38, 39, 40, 41, 42, 43, 44, 15, 46, 47, 48, 49, 50, 51, 52, 53, 18, 55, 56, 57, 58, 59, 60, 61, 62, 21, 64, 65, 66, 67, 68, 69, 70, 71, 24, 73, 74, 75, 76, 77, 78, 79, 80, 27
Offset: 1

Views

Author

Ilya Gutkovskiy, Feb 21 2017

Keywords

Comments

The length of the period of A000035 (n=2), A010872 (n=3), A109718 (n=4), A070471 (n=5), A010875 (n=6), A070472 (n=7), A109753 (n=8), A167176 (n=9), A008960 (n = 10), etc. (see also comment in A000578 from R. J. Mathar).
Conjecture: let a_p(n) be the length of the period of the sequence k^p mod n where p is a prime, then a_p(n) = n/p if n == 0 (mod p^2) else a_p(n) = n.
For example: sequence k^7 mod 98 gives 1, 30, 31, 18, 19, 48, 49, 50, 79, 80, 67, 68, 97, 0, 1, 30, 31, 18, 19, 48, 49, 50, 79, 80, 67, 68, 97, 0, ... (period 14), 7 is a prime, 98 == 0 (mod 7^2) and 98/7 = 14.

Examples

			a(9) = 3 because reading 1, 8, 27, 64, 125, 216, 343, 512, ... modulo 9 gives 1, 8, 0, 1, 8, 0, 1, 8, 0, ... with period length 3.
		

Crossrefs

Programs

  • Mathematica
    a[1] = 1; a[n_] := For[k = 1, True, k++, If[Mod[k^3, n] == 0 && Mod[(k + 1)^3 , n] == 1, Return[k]]]; Table[a[n], {n, 1, 81}]

Formula

Apparently: a(n) = 2*a(n-9) - a(n-18).
Empirical g.f.: x*(1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 7*x^6 + 8*x^7 + 3*x^8 + 8*x^9 + 7*x^10 + 6*x^11 + 5*x^12 + 4*x^13 + 3*x^14 + 2*x^15 + x^16) / ((1 - x)^2*(1 + x + x^2)^2*(1 + x^3 + x^6)^2). - Colin Barker, Feb 21 2017

A122606 n^(n+1) mod 7.

Original entry on oeis.org

0, 1, 1, 4, 2, 1, 6, 0, 1, 2, 5, 1, 5, 1, 0, 1, 4, 1, 4, 4, 6, 0, 1, 1, 3, 2, 6, 1, 0, 1, 2, 2, 1, 2, 6, 0, 1, 4, 6, 4, 3, 1, 0, 1, 1, 4, 2, 1, 6, 0, 1, 2, 5, 1, 5, 1, 0, 1, 4, 1, 4, 4, 6, 0, 1, 1, 3, 2, 6, 1, 0, 1, 2, 2, 1, 2, 6, 0, 1, 4, 6, 4, 3, 1, 0, 1, 1, 4, 2, 1, 6, 0, 1, 2, 5, 1, 5, 1, 0, 1, 4, 1, 4, 4, 6
Offset: 0

Views

Author

Zak Seidov, Sep 24 2006, Sep 25 2006

Keywords

Crossrefs

Programs

  • Mathematica
    Table[PowerMod[n,n+1,7],{n,0,120}] (* Harvey P. Dale, May 24 2012 *)

Formula

a(n) = n^(n+1) mod 7 = A007778(n) mod 7.
Periodic with cycle consisting of first 42 terms: {0,1,1,4,2,1,6,0,1,2,5,1,5,1,0,1,4,1,4,4,6,0,1,1,3,2,6,1,0,1,2,2,1,2,6,0,1,4,6,4,3,1}.
Showing 1-3 of 3 results.