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 1-5 of 5 results.

A203571 Period length 10: [0, 1, 2, 3, 4, 0, 4, 3, 2, 1] repeated.

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Jan 11 2012

Keywords

Comments

This sequence can be continued periodically for negative values of n.
This is the fifth 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, for k = 1, ..., 4, respectively.
In general, the sequence P_k, k >= 1 (periodically continued for negative values of n), is used to define the k equivalence classes [0], [1], ..., [k-1], with [j] := {n integer| P_k(n) = j}. Two integers are equivalent if and only if they are mapped by P_k to the same value. For P_5, P_6 and P_7 see the arrays (not the triangles) A090298, A092260 and A113807, respectively. In each of these cases the class [k] should be replaced by the class [0], and also negative n-values are allowed. Multiplication can be done class-wise. E.g., k = 5: P_5(n) = a(n), 7*12 == 3*2 = 6 == 4; a(7*12) = a(a(7)*a(12)) = a(3*2) = 4. This kind of multiplication could be called multiplication Modd n, in order to distinguish it from multiplication mod n. Addition cannot be done class-wise. E.g., k = 5: 7 + 12 = 19 == 1 is not equivalent to 3 + 2 = 5 == 0; a(7+12) = 1 is not equal to a(a(7) + a(12)) = a(3+2) = 0.
Periodic sequences of this type can be also calculated by a(n) = c + floor(q/(p^m-1)*p^n) mod p, where c is a constant, q is the number representing the periodic digit pattern and m is the period length. c, p and q can be calculated as follows: Let D be the array representing the number pattern to be repeated, m = size of D, max = maximum value of elements in D, min = minimum value of elements in D. Then c := min, p := max - min + 1 and q := p^m * Sum_{i=0..(m-1)} (D(i) - min)/p^i. Example: D = (0, 1, 2, 3, 4, 0, 4, 3, 2, 1), c = 0, m = 10, p = 5 and q = 3034180 for this sequence. - Hieronymus Fischer, Jan 04 2013 [Corrected by Rémi Guillaume, Aug 28 2024]
For periodic sequences with terms < 10 one can use the well-known fact that ab..z/99..9 = 0.ab..zab..zab..z... (infinite periodic decimal fraction), this leads to one of the given formulas. For the general case it is sufficient to shift the terms to nonnegative values and to switch to a sufficiently large basis instead of 10 (there are infinitely many choices). - M. F. Hasler, Jan 13 2013

Examples

			a(12) = 12 mod 5 = 2 since 12\5 = floor(12/5) = 2 is even; the sign is +1.
a(7) = -7 mod 5 = 3 since 7\5 = floor(7/5) = 1 is odd; the sign is -1.
		

Crossrefs

Programs

Formula

a(n) = n mod 5 if (-1)^floor(n/5) = +1 else -n mod 5, n >= 0. (-1)^floor(n/5) is the sign corresponding to the parity of the quotient floor(n/5). This quotient is sometimes denoted by n\5.
O.g.f.: x*(1+2*x+3*x^2+4*x^3+4*x^5+3*x^6+2*x^7+x^8)/(1-x^10) = -x*(1 +2*x +3*x^2 +4*x^3 +4*x^5 +3*x^6 +2*x^7 +x^8) / ( (x-1) *(1+x) *(x^4+x^3+x^2+x+1) *(x^4-x^3+x^2-x+1) ).
a(n) = (2/5)*cos(Pi*n) - cos(4*Pi*n/5) - (1/5)*cos(3*Pi*n/5) + (2/5)*5^(1/2)*cos(3*Pi*n/5) - cos(2*Pi*n/5) - (1/5)*cos(Pi*n/5) - (2/5)*5^(1/2)*cos(Pi*n/5) + 2. - Leonid Bedratyuk, May 13 2012
a(n) = floor(123404321/9999999999*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 04 2013
a(n) = floor(151709/2441406*5^(n+1)) mod 5. - Hieronymus Fischer, Jan 04 2013
a(n) = (5-abs(n-(10*ceiling(n/10)-5)))*(ceiling((n+5)/10)-floor((n+5)/10)). - Wesley Ivan Hurt, Mar 26 2014 [corrected by Jason Yuen, Feb 17 2025]
a(n+10) = a(n) for n in Z; a(-n) = a(n) for n in Z. - Rémi Guillaume, Aug 28 2024

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

Views

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

A203572 Period length 12: 0, 1, 2, 3, 4, 5, 0, 5, 4, 3, 2, 1 repeated.

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Jan 12 2012

Keywords

Comments

This sequence can be continued periodically for negative values of n.
This is the sixth 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, A203571 for k=1,...,5, respectively.
See a comment on A203571 for the general case of the P_k sequences. For a(n)=P_6(n) the nonnegative members of the equivalence classes [0], [1],...,[5], defined by p==q iff P_6(p)=P_6(q), are found in the array A092260 if there class [6], starting with 6, is replaced by 0,6,12,..., which is class [0] (nonnegative part).

Examples

			a(14) = 14(mod 6) = 2 because 14\6 = floor(14/6)=2 is even; the sign is +1.
a(8) = (6-8)(mod 6) = 4 because 8\6 = floor(8/6)=1 is odd; the sign is -1.
		

Crossrefs

Cf. A203571.

Programs

  • Mathematica
    PadRight[{},120,{0,1,2,3,4,5,0,5,4,3,2,1}] (* Harvey P. Dale, Nov 28 2015 *)

Formula

a(n) = n(mod 6) if (-1)^floor(n/6)=+1 else (6-n)(mod 6), n>=0. (-1)^floor(n/6) is the sign corresponding to the parity of the quotient floor(n/6). This quotient is sometimes denoted by n\6.
O.g.f.: x*(1+2*x+3*x^2+4*x^3+5*x^4+5*x^6+4*x^7+3*x^8+2*x^9+ x^10)/(1-x^12).

A203575 Array of certain four complete residue classes (nonnegative members), read by SW-NE antidiagonals.

Original entry on oeis.org

0, 1, 4, 2, 7, 8, 3, 6, 9, 12, 5, 10, 15, 16, 11, 14, 17, 20, 13, 18, 23, 24, 19, 22, 25, 28, 21, 26, 31, 32, 27, 30, 33, 36, 29, 34, 39, 40, 35, 38, 41, 44, 37, 42, 47, 48, 43, 46
Offset: 1

Views

Author

Wolfdieter Lang, Jan 12 2012

Keywords

Comments

See A193682 for the sequence called P_4, with period length 8, which defines the four complete residue classes [m], m = 0,1,2,3, via the equivalence relation p==q iff P_4(p) = P_4(q).
See a comment on A203571 for the general P_k sequences, and the multiplicative (but not additive) structure of these residue classes.
The row length sequence of this tabf array is [1,2,3,4,4,4,...].
This array defines a certain permutation of the nonnegative integers.

Examples

			The array starts
n\m  1   2   3   4
1:   0
2:   1   4
3:   2   7   8
4:   3   6   9  12
5:   5  10  15  16
6:  11  14  17  20
7:  13  18  23  24
8:  19  22  25  28
9:  21  26  31  32
10: 27  30  33  36
...
The sequence P_4(n)=A193682(n), n>=0, is repeated 0, 1, 2, 3, 0, 3, 2, 1, with period length 8. P_4(6)=2, hence 6 belongs to class [2].
Multiplicative structure: 11*23 == 3*1 = 3. Indeed: P_4(11*23) = P_4(253) = P_(5), because 253==5(mod 8), and P_(5)= 3, hence 11*23 belongs to class 3. In general, P_4(p*q) = P_4(P_4(p)*P_4(q)).
		

Crossrefs

Cf.A193682, A088520 (k=3), A090298 (k=5), A092260 (k=6), A113807 (k=7).

Formula

The nonnegative members of the four complete residue classes are (see a comment above for their definition):
[0]: 0, 4, 8, 12, 16, 20, 24, 28, 32, 36,... (A008586)
[1]: 1, 7, 9, 15, 17, 23, 25, 31, 33, 39,... (A047522)
[2]: 2, 6, 10, 14, 18, 22, 26, 30, 34, 38,... (A016825)
[3]: 3, 5, 11, 13, 19, 21, 27, 29, 35, 37,... (A047621)
In each class the corresponding negative numbers should be included.

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

Views

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 1-5 of 5 results.