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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A204453 Period length 14: [0, 1, 2, 3, 4, 5, 6, 0, 6, 5, 4, 3, 2, 1] repeated.

Original entry on oeis.org

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

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Author

Wolfdieter Lang, Jan 17 2012

Keywords

Comments

This sequence can be continued periodically for negative values of n, and then a(-n) = a(n).
This is the seventh sequence of a k-family of sequences P_k, k>=1, which starts with A000007(n+1), n>=0, (the 0-sequence), A000035, A193680, A193682, A203572, A for k=1..6, respectively.
See a comment on A203571 for the general case of the P_k sequences. For a(n)=P_7(n) the nonnegative members of the equivalence classes [0], [1],...,[6], defined by p==q iff P_7(p)=P_7(q), are found in the array A113807 if there the last class [7], starting with 7, is replaced by 0,7,14,..., to become the first class [0] (nonnegative part).

Examples

			a(16) = 16(mod 7) = 2 because 16\7 = floor(16/7)=2 is even; the sign is +1.
a(9) = (7-9)(mod 7) = 5 because 9\7 = floor(9/7)=1 is odd; the sign is -1.
		

Crossrefs

Cf. A203572 (k=6), A113807, A010876.

Formula

a(n) = n(mod 7) if (-1)^floor(n/7)=+1 else (7-n)(mod 7), n>=0. (-1)^floor(n/7) is the sign corresponding to the parity of the quotient floor(n/7). This quotient is sometimes denoted by n\7.
O.g.f.: x*(1+2*x+3*x^2+4*x^3+5*x^4+6*x^5+6*x^7+5*x^8+4*x^9+ 3*x^10+2*x^11+x^12)/(1-x^14).
a(n) = (7*m*(m^4-21*m^3+175*m^2-735*m+1624)*((-1)^floor(n/7)-1)-10908*(-1)^floor(n/7)+12348)*m/1440 where m = n-7*floor(n/7). - Luce ETIENNE, Oct 13 2017

A337938 Irregular triangle read by rows: T(n, k) gives the primitive period of the sequence {k (Modd n)}_{k >= 0}, for n >= 1.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 2, 1, 0, 1, 2, 3, 0, 3, 2, 1, 0, 1, 2, 3, 4, 0, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 0, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 0, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 0, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 8, 7, 6, 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 9, 8, 7, 6, 5, 4, 3, 2, 1
Offset: 1

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Author

Wolfdieter Lang, Oct 25 2020

Keywords

Comments

The length of row n is 1 for n = 1, 2 for n = 2, and 2*n for n >= 3.
The modified modular equivalence relation Modd n is defined, for integer k and positive integer n, by k (Modd n) = k (mod n) if floor(k/n) is even, and -k (mod n) if floor(k/n) is odd. The smallest nonnegative complete residue system modulo n, namely RS(n) = {0, 1, ..., n-1}, is used. See the W. Lang link, Definition 4, eq. (69), p. 25 - 26.
In order to have row length 2*n for all n >= 1 one could use for n = 1 and 2 the imprimitive periods 0, 0 and 0, 1, 0, 1, respectively.
The name Modd n derives from the fact that the multiplicative (but not additive ) group Modd n has the smallest positive reduced residue system with only odd numbers, named RRSodd(n), as elements (for n = 0 RRS(n) = {0}, but here it is taken as {1}). This group is isomorphic to the Galois group G(rho(n)) = Gal(Q(rho(n))/Q), with rho(n) = 2*cos(pi/n). See the W. Lang link.

Examples

			The irregular triangle begins:
n \ k 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ..
1:    0
2:    0 1
3:    0 1 2 0 2 1
4:    0 1 2 3 0 3 2 1
5:    0 1 2 3 4 0 4 3 2 1
6:    0 1 2 3 4 5 0 5 4 3  2  1
7:    0 1 2 3 4 5 6 0 6 5  4  3  2  1
8:    0 1 2 3 4 5 6 7 0 7  6  5  4  3  2  1
9:    0 1 2 3 4 5 6 7 8 0  8  7  6  5  4  3  2  1
10   :0 1 2 3 4 5 6 7 8 9  0  9  8  7  6  5  4  3  2  1
...
T(1, 0) = 0 because {k (Modd 1)}_{k >= 0} is the 0 sequence A000007:  0 (Modd 1) =  0 (mod 1) = 0, 1 (Modd 1) = -1 (mod 1) = 0,  2 (Modd 1) = 2 (mod 1) = 0, ... .
T(7, 6) = 6 because floor(6/7) = 0, which is even, hence 6 (Modd 7) = 6 (mod 7) = 6.
T(7, 8) = 6 because  floor(8/7) = 1, which is odd, hence  8 (Modd 7) = -8 (mod 7) = 6.
		

Crossrefs

Cf. Periodic sequences for n = 1, 2, ..., 7: A000007, A000035, A193680, A193682, A203571, A203572.
Cf. A002262 (for mod n), A053616 (as a triangle, for mod* n).

Formula

T(n,k) = k (Modd n), for n >= 1, and k = 0 for n = 1, k = 0, 1 for n = 2, and k = 0, 1, ..., 2*n - 1, for n >= 3. For k (Modd n) see the comment above.
Showing 1-2 of 2 results.