A301729 -id:A301729 - cp's OEIS frontend

A047273 Numbers that are congruent to {0, 1, 3, 5} mod 6.

0, 1, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19, 21, 23, 24, 25, 27, 29, 30, 31, 33, 35, 36, 37, 39, 41, 42, 43, 45, 47, 48, 49, 51, 53, 54, 55, 57, 59, 60, 61, 63, 65, 66, 67, 69, 71, 72, 73, 75, 77, 78, 79, 81, 83, 84, 85, 87, 89, 90, 91, 93, 95, 96, 97, 99, 101, 102, 103

Offset: 1

N. J. A. Sloane

Complement of A047235. - Reinhard Zumkeller, Oct 01 2008

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Index entries for two-way infinite sequences.
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Cf. A004524, A007310, A047228, A047235, A047241, A047261, A093719, A096981.
First differences of A281026.
See A301729 for an essentially identical sequence.

a047273 n = a047273_list !! (n-1)
a047273_list = 0 : 1 : 3 : 5 : map (+ 6) a047273_list
-- Reinhard Zumkeller, Feb 19 2013

[(6*n-6+(-1)^(n div 2)+(-1)^(-n div 2))/4: n in [1..100]]; // Wesley Ivan Hurt, May 20 2016

seq(2*(n-floor(n/4)) - (3-I^n-(-I)^n-(-1)^n)/4, n = 0..69); # Gary Detlefs, Mar 19 2010

LinearRecurrence[{2,-2,2,-1},{0,1,3,5},80] (* Harvey P. Dale, Jan 04 2015 *)

PARI
```
a(n)=n+(n+1)\4+(n+2)\4
```

[(lucas_number1(n+2, 0, 1)+3*n)/2 for n in range(0, 70)] # Zerinvary Lajos, Mar 09 2009

G.f.: x*(1+x+x^2)/((1-x)^2*(1+x^2)) = x*(1-x^2)*(1-x^3)/((1-x)^3*(1-x^4)).
a(n) = n + A004524(n+1) = -a(-n) for all n in Z.
Starting (1, 3, 5, ...) = partial sums of (1, 2, 2, 1, 1, 2, 2, 1, 1, ...). - Gary W. Adamson, Jun 19 2008
A093719(a(n)) = 1. - Reinhard Zumkeller, Oct 01 2008
a(n) = 2*(n-floor(n/4)) - (3-I^n-(-I)^n-(-1)^n)/4, with offset 0..a(0)=0. - Gary Detlefs, Mar 19 2010
a(n) = (3*n-3+cos(Pi*n/2))/2. - R. J. Mathar, Oct 08 2010
From Wesley Ivan Hurt, May 20 2016: (Start)
a(n) = 2*a(n-1)-2*a(n-2)+2*a(n-3)-a(n-4) for n>4.
a(n) = (6*n-6+(-1)^(n/2)+(-1)^(-n/2))/4. (End)
Euler transform of length 4 sequence [3, -1, -1, 1]. - Michael Somos, Jun 24 2017
Sum_{n>=2} (-1)^n/a(n) = log(2)/3 + log(3)/2. - Amiram Eldar, Dec 16 2021
E.g.f.: (2 + 3*exp(x)*(x - 1) + cos(x))/2. - Stefano Spezia, Jul 26 2024

A352747 Array read by ascending antidiagonals. A(n, k) = F(k, n) mod n for n >= 1 and k >= 0, where F(n, k) = A352744(n, k) are the Fibonacci numbers, A(0, k) = 1 for k >= 0.

Original entry on oeis.org

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

Offset: 0

Peter Luschny, Apr 08 2022

This array aims the study of the divisibility properties of the Fibonacci numbers A352744. The identity F(n, k) = (-1)^k*F(1 - n, -k) from A352744 shows that negative indices do not add to the divisibility properties of F(n, k).
All rows A(n, .) are pure periodic sequences. The length of the periods is given by (1, A270313). For n > 0 the length of the period of row A(n, .) is <= n.
The period length is 1 for n in (1, A023172) and n for n in (1, A074215), as observed by Robert Israel in A270313. In particular, if n is a power of 2 or a prime (A174090), then the period length is n.
The indices of the zero-free rows are in A353280. A zero-free row A(n, .) means that n will not divide F(k, n) whatever value k takes. For that it is sufficient to check that period(A(n, .)) is zero-free.
If period(A(n, .)) = [k | 0 <= k < n] we call n a 'Fibonacci friend'. In other words, in this case F(k, n) mod n = k for 0 <= k < n. A Fibonacci friend does not have to be prime (since 1 is a Fibonacci friend), but if it is  prime then it is congruent to {1, 4} mod 5 (A045468), and all such primes are Fibonacci friends.
To say that n is a Fibonacci friend is equivalent to saying that A(n, n) = 0 and that n divides F(n, n). Fibonacci friends are the indices of the zeros in A002752.
Integers n > 0 that divide Sum{k=0..n-1} (F(k, n) mod n) are congruent to {0, 1, 3, 5} mod 6 (A301729).

			Array starts (periods are indicated with () ):
[n\k] 0   1   2   3   4  5  6   7   8   9  10  11  12
----------------------------------------------------------
[ 0] (1), 1,  1,  1,  1, 1, 1,  1,  1,  1,  1,  1,  1, ...
[ 1] (0), 0,  0,  0,  0, 0, 0,  0,  0,  0,  0,  0,  0, ...
[ 2] (1,  0), 1,  0,  1, 0, 1,  0,  1,  0,  1,  0,  1, ...
[ 3] (1,  0,  2), 1,  0, 2, 1,  0,  2,  1,  0,  2,  1, ...
[ 4] (2,  1,  0,  3), 2, 1, 0,  3,  2,  1,  0,  3,  2, ...
[ 5] (3), 3,  3,  3,  3, 3, 3,  3,  3,  3,  3,  3,  3, ...
[ 6] (5,  1,  3), 5,  1, 3, 5,  1,  3,  5,  1,  3,  5, ...
[ 7] (1,  0,  6,  5,  4, 3, 2), 1,  0,  6,  5,  4,  3, ...
[ 8] (5,  2,  7,  4,  1, 6, 3,  0), 5,  2,  7,  4,  1, ...
[ 9] (3,  1,  8,  6,  4, 2, 0,  7,  5), 3,  1,  8,  6, ...
[10] (4,  9), 4,  9,  4, 9, 4,  9,  4,  9,  4,  9,  4, ...
[11] (0,  1,  2,  3,  4, 5, 6,  7,  8,  9, 10), 0,  1, ...
[12] (5), 5,  5,  5,  5, 5, 5,  5,  5,  5,  5,  5,  5, ...

Cf. A352744, A002752, A045468, A074215, A088209, A174090, A212804, A270313, A301729, A353280.

f := n -> combinat:-fibonacci(n + 1):
F := proc(n, k) option remember; (n-1)*f(k-1) + f(k) end:
A := (n, k) -> ifelse(n = 0, 1, modp(F(k, n), n)):
for n from 0 to 12 do seq(A(n, k), k = 0..10) od;

F[n_, k_] := (n - 1)*Fibonacci[k] + Fibonacci[k + 1];
A[n_, k_] := If[n == 0, 1, Mod[F[k, n], n]];
Table[A[n, k], {n, 0, 12}, {k, 0, 10}] // TableForm

def F(n, k): return (n - 1)*fibonacci(k) + fibonacci(k + 1)
def A(n,k): return mod(F(k, n), n)
for n in range(13): print([A(n,k) for k in range(13)])

A(n, 0) = A(n, n) = A002752(n).

Clearly 0 <= A(n, k) < n for all k and n > 0.

cp's OEIS Frontend

A047273 Numbers that are congruent to {0, 1, 3, 5} mod 6.

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