A082115 -id:A082115 - cp's OEIS frontend

A131295 a(n)=ds_4(a(n-1))+ds_4(a(n-2)), a(0)=0, a(1)=1; where ds_4=digital sum base 4.

0, 1, 1, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3, 5, 5, 4, 3, 4, 4, 2, 3

Offset: 0

Hieronymus Fischer, Jun 27 2007

The digital sum analog (in base 4) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(3)=8.
Also a(n)==A004090(n) modulo 3 (A004090(n)=digital sum of Fib(n)).
For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(4) for the base p=4.
a(n) and Fib(n)=A000045(n) are congruent modulo 3 which implies that (a(n) mod 3) is equal to (Fib(n) mod 3)=A082115(n-1) (for n>0). Thus (a(n) mod 3) is periodic with the Pisano period = A001175(3)=8 too. - Hieronymus Fischer

			a(8)=3, since a(6)=5=11(base 4), ds_4(5)=2,
a(7)=4=10(base 4), ds_4(4)=1 and so a(8)=2+1.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for Colombian or self numbers and related sequences

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131296, A131297, A131318, A131319, A131320.

nxt[{a_,b_}]:={b,Total[IntegerDigits[a,4]]+Total[IntegerDigits[b,4]]}; NestList[ nxt,{0,1},110][[All,1]] (* Harvey P. Dale, Jul 30 2018 *)

a(n)=a(n-1)+a(n-2)-3*(floor(a(n-1)/4)+floor(a(n-2)/4)).
a(n)=floor(a(n-1)/4)+floor(a(n-2)/4)+(a(n-1)mod 4)+(a(n-2)mod 4).
a(n)=A002265(a(n-1))+A002265(a(n-2))+A010873(a(n-1))+A010873(a(n-2)).
a(n)=Fib(n)-3*sum{1A000045(n).

A082116 Fibonacci sequence (mod 5).

Original entry on oeis.org

0, 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0, 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0, 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0, 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0, 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0, 1, 1

Offset: 0

Eric W. Weisstein, Apr 03 2003

This sequence contains the complete set of residues modulo 5. See A079002. - Michel Marcus, Jan 31 2020

S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989. See p. 88. - N. J. A. Sloane, Feb 20 2013

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Brandon Avila and Yongyi Chen, On Moduli For Which the Lucas Numbers Contain a Complete Residue System, Fibonacci Quart. 51 (2013), no. 2, 151-152. See p. 151.
S. A. Burr, On moduli for which the Fibonacci sequence contains a complete system of residues, The Fibonacci Quarterly, 9.5 (1971), 497-504.
Diana Savin and Elif Tan, On Companion sequences associated with Leonardo quaternions: Applications over finite fields, arXiv:2403.01592 [math.CO], 2024. See p. 11.
Minjia Shi and Patrick Solé, The largest number of weights in cyclic codes, arXiv:1807.08418 [cs.IT], 2018.
Eric Weisstein's World of Mathematics, Fibonacci Number
Index entries for linear recurrences with constant coefficients, signature (0,1,0,-1,1,1,-1,-1,1,0,-1,0,1).

Cf. A000045, A011655, A082115, A079343, A082116, A082117, A079344, A079002.

[Fibonacci(n) mod 5: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

Table[Mod[Fibonacci[n], 5], {n, 0, 125}] (* Alonso del Arte, Jul 29 2013 *)

a(n)=fibonacci(n)%5 \\ Charles R Greathouse IV, Oct 07 2015

Sequence is periodic with Pisano period 20.
a(n) = 2 + ((n mod 20) - ((n - 1) mod 20) - ((n - 3) mod 20) - ((n - 4) mod 20) + 3*((n - 5) mod 20) - 3*((n - 6) mod 20) + 2*((n - 8) mod 20) - 3*((n - 9) mod 20) + 4*((n - 10) mod 20) - 4*((n - 11) mod 20) + ((n - 13) mod 20) + ((n - 14) mod 20) + 2*((n - 15) mod 20) - 2*((n - 16) mod 20) - 2*((n - 18) mod 20) + 3*((n - 19) mod 20))/20. - Hieronymus Fischer, Jun 30 2007
G.f.: (x + x^2 + 2x^3 + 3x^4 + 3x^6 + 3x^7 + x^8 + 4x^9 + 4x^11 + 4x^12 + 3x^13 + 2x^14 + 2x^16 + 2x^17 + 4x^18 + x^19)/(1 - x^20), not reduced. - Hieronymus Fischer, Jun 30 2007
a(n) = A010073(n) mod 5. - Hieronymus Fischer, Jun 30 2007
G.f.: -x*(1 + x + x^2 + 2*x^3 + 3*x^6 - x^7 - 2*x^8 - x^4 + x^9 + 4*x^10 + x^11) / ( (x - 1) * (x^4 + x^3 + x^2 + x + 1) * (x^8 - x^6 + x^4 - x^2 + 1) ). - R. J. Mathar, Jul 14 2012

Added a(0)=0 from Vincenzo Librandi, Feb 04 2014

A079344 F(n) mod 8, where F(n) = A000045(n) is the n-th Fibonacci number.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0, 5, 5

Offset: 0

Jon Perry, Jan 04 2003

This sequence does not contain the complete set of residues modulo 8. See A079002. - Michel Marcus, Jan 31 2020

			a(8) = F(8) mod 8 = 21 mod 8 = 5.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Brandon Avila and Yongyi Chen, On Moduli For Which the Lucas Numbers Contain a Complete Residue System, Fibonacci Quart. 51 (2013), no. 2, 151-152. See p. 151.
S. A. Burr, On moduli for which the Fibonacci sequence contains a complete system of residues, The Fibonacci Quarterly, 9.5 (1971), 497-504.
P. Ribenboim, FFF (Favorite Fibonacci Flowers), Fib. Q. 43 (No. 1, 2005), 3-14.
Eric Weisstein's World of Mathematics, Fibonacci Number
Index entries for linear recurrences with constant coefficients, signature (1,0,0,-1,1,0,0,-1,1).

Cf. A000045, A011655, A082115, A079343, A082116, A082117, A079344, A079345, A111958, A079002.

[Fibonacci(n) mod 8: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

Mod[Fibonacci[Range[0,110]],8] (* or *) LinearRecurrence[ {1,0,0,-1,1,0,0,-1,1},{0,1,1,2,3,5,0,5,5},110] (* Harvey P. Dale, Jan 16 2014 *)

for (n=0,100,print1(fibonacci(n)%8","))

Sequence is periodic with Pisano period 12 = A001175(8).

G.f.: -x*(1+x^2+x^3+3*x^4+6*x^6-5*x^5+x^7) / ( (x-1)*(x^2-x+1)*(1+x+x^2)*(x^4-x^2+1) ). - R. J. Mathar, Aug 08 2012

Edited by N. J. A. Sloane, Dec 06 2008 at the suggestion of R. J. Mathar

A082117 Fibonacci sequence (mod 6).

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 2, 1, 3, 4, 1, 5, 0, 5, 5, 4, 3, 1, 4, 5, 3, 2, 5, 1, 0, 1, 1, 2, 3, 5, 2, 1, 3, 4, 1, 5, 0, 5, 5, 4, 3, 1, 4, 5, 3, 2, 5, 1, 0, 1, 1, 2, 3, 5, 2, 1, 3, 4, 1, 5, 0, 5, 5, 4, 3, 1, 4, 5, 3, 2, 5, 1, 0, 1, 1, 2, 3, 5, 2, 1, 3, 4, 1, 5, 0, 5, 5, 4, 3, 1, 4, 5, 3, 2, 5, 1, 0, 1, 1, 2, 3, 5, 2

Offset: 0

Eric W. Weisstein, Apr 03 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Glen Joyce C. Dulatre, Jamilah V. Alarcon, Vhenedict M. Florida, Daisy Ann A. Disu, On Fractal Sequences, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 109-113.
Syamsudhuha Muslim, Sri Gemawati, The Identity of Fibonacci Numbers Z6, International Mathematical Forum, Vol. 13, 2018, no. 5, 225-231.
Eric Weisstein's World of Mathematics, Fibonacci Number
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Cf. A011655, A082115, A079343, A082116, A082117, A079344.

[Fibonacci(n) mod 6: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014

Table[Mod[Fibonacci[n], 6], {n, 0, 100}] (* Alonso del Arte, Jul 29 2013 *)

a(n)=fibonacci(n)%6 \\ Charles R Greathouse IV, Oct 07 2016

Sequence is periodic with Pisano period 24 = A001175(6).

G.f.: -x*(x^22 + 5*x^21 + 2*x^20 + 3*x^19 + 5*x^18 + 4*x^17 + x^16 + 3*x^15 + 4*x^14 + 5*x^13 + 5*x^12 + 5*x^10 + x^9 + 4*x^8 + 3*x^7 + x^6 + 2*x^5 + 5*x^4 + 3*x^3 + 2*x^2 + x + 1)/((x - 1)*(x + 1)*(x^2 - x + 1)*(x^2 + 1)*(x^2 + x + 1)*(x^4 - x^2 + 1)*(x^4 + 1)*(x^8 - x^4 + 1)). - Colin Barker, Aug 15 2012

Added a(0)=0 from Vincenzo Librandi, Feb 04 2014

A280154 a(n) = 5*Lucas(n).

Original entry on oeis.org

10, 5, 15, 20, 35, 55, 90, 145, 235, 380, 615, 995, 1610, 2605, 4215, 6820, 11035, 17855, 28890, 46745, 75635, 122380, 198015, 320395, 518410, 838805, 1357215, 2196020, 3553235, 5749255, 9302490, 15051745, 24354235, 39405980, 63760215, 103166195, 166926410, 270092605, 437019015

Offset: 0

Bruno Berselli, Dec 27 2016

Fibonacci sequence beginning 10, 5.
After 5, the sequence provides the 3rd column of the rectangular array in A213590.
After 5, all terms belong to A191921 because a(n) = Lucas(n+4) - 3*Lucas(n-1).
From G. C. Greubel, Dec 27 2016: (Start)
{a(n) mod 3} yields (1,2,0,2,2,1,0,1), repeated, and is given as A082115.
{a(n) mod 6} yields (4,5,3,2,5,1,0,1,1,2,3,5,2,1,3,4,1,5,0,5,5,4,3,1) and is given as A082117. (End)

Bruno Berselli, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,1).

Subsequence of A084176.
Cf. A022088: 5*Fibonacci(n).
Cf. A022359: Lucas(n+5) + Lucas(n-5).
Cf. A000032, A000045, A191921, A213590.
Cf. sequences with formula Fibonacci(n+k) + Fibonacci(n-k): A006355 (k=0, without the initial 1), A000032 (k=1), A022086 (k=2), A022112 (k=3, with an initial 4), A022090 (k=4), this sequence (k=5), A022352 (k=6).

Magma
```
[5*Lucas(n): n in [0..40]];
```

F := n -> combinat:-fibonacci(n):
seq(F(n+5) + F(n-5), n=0..38); # Peter Luschny, Dec 29 2016

Mathematica
```
Table[5 LucasL[n], {n, 0, 40}]
```

vector(40, n, n--; fibonacci(n+5)+fibonacci(n-5))

def A280154():
    x, y = 10, 5
    while True:
        yield x
        x, y = y, x + y
a = A280154(); print([next(a) for  in range(39)]) # _Peter Luschny, Dec 29 2016

G.f.: 5*(2 - x)/(1 - x - x^2).
a(n) = a(n-1) + a(n-2) for n>1.
a(n) = Fibonacci(n+5) + Fibonacci(n-5), with Fibonacci(-k) = -(-1)^k*Fibonacci(k) for the negative indices.

A004696 a(n) = floor(Fibonacci(n)/3).

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 4, 7, 11, 18, 29, 48, 77, 125, 203, 329, 532, 861, 1393, 2255, 3648, 5903, 9552, 15456, 25008, 40464, 65472, 105937, 171409, 277346, 448756, 726103, 1174859, 1900962, 3075821, 4976784, 8052605, 13029389, 21081995, 34111385, 55193380

Offset: 0

N. J. A. Sloane

			G.f. = x^4 + x^5 + 2*x^6 + 4*x^7 + 7*x^8 + 11*x^9 + 18*x^10 + 29*x^11 + 48*x^12 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,0,0,0,1,-1,-1).

Cf. A000045 (Fibonacci(n)).
Cf. A082115 (Fibonacci(n) (mod 3)).
Cf. A293543 (ceiling(Fibonacci(n)/3)).
Cf. A293544 (round(Fibonacci(n)/3)).

[Floor(Fibonacci(n)/3): n in [0..40]]; // Vincenzo Librandi, Jul 09 2012

seq(iquo(fibonacci(n),3),n=0..40); # Zerinvary Lajos, Apr 20 2008

CoefficientList[Series[x^4 (x^4 + x^3 + 1)/((1 - x^8) (1 -x - x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Jul 09 2012 *)
Floor[Fibonacci[Range[0, 40]]/3] (* G. C. Greubel, May 22 2019 *)
Table[Floor[Fibonacci[n]/3], {n, 0, 20}] (* Eric W. Weisstein, Feb 09 2025 *)
Table[(8 Fibonacci[n] + 3 (-1)^n - 9 + Cos[Pi n/2] (6 - 4 Sin[Pi n/4]) + 4 Sqrt[2] Sin[Pi n/4] Sin[Pi n/2])/24, {n, 0, 20}] (* Eric W. Weisstein, Feb 09 2025 *)

vector(40, n, n--; fibonacci(n)\3) \\ Altug Alkan, Nov 06 2015

concat(vector(4), Vec(x^4*(x^4+x^3+1)/((1-x^8)*(1-x-x^2)) + O(x^40))) \\ Altug Alkan, Nov 06 2015

[floor(fibonacci(n)/3) for n in (0..40)] # G. C. Greubel, May 22 2019

G.f.: x^4*(1 +x^3 +x^4) / ((1-x^8)*(1-x-x^2)).
a(n) = (A000045(n) - A082115(n))/3. - R. J. Mathar, Jul 14 2012
From Vladimir Reshetnikov, Nov 05 2015: (Start)
a(n) = (8*A000045(n) + 3*(-1)^n - 9 + cos(Pi*n/2)*(6 - 4*sin(Pi*n/4)) + 4*sqrt(2)*sin(Pi*n/4)*sin(Pi*n/2))/24.
E.g.f.: (cos(x)-cosh(x)-2*sinh(x))/4 + (sqrt(2)*cos(x/sqrt(2))+sin(x/sqrt(2)))*sinh(x/sqrt(2))/6 + 2*exp(x/2)*sinh(x*sqrt(5)/2)/(3*sqrt(5)). (End)
The sequence b(n) = a(n+2) - a(n+1) - a(n) has period 8 and always 0 or 1. - Michael Somos, Nov 06 2015

A093062 a(n) = Fibonacci(prime(n)) - prime(Fibonacci(n)).

Original entry on oeis.org

-1, 0, 2, 8, 78, 214, 1556, 4108, 28518, 513972, 1345808, 24156990, 165578670, 433491846, 2971210580, 53316283380, 956722012572, 2504730758802, 44945570173074, 308061521102198, 806515532933562, 14472334024479534, 99194853094422264, 1779979416004150202

Offset: 1

Dennis S. Kluk (mathemagician(AT)ameritech.net), May 08 2004

Composition of prime( ) and Fibonacci( ) is not commutative. Does a prime p ever divide Fibonacci(prime(p)) - prime(Fibonacci(p))?
Note that a(3) = 2 is the only prime element of the sequence. This is because after 2, all primes are odd; and the Fibonacci number F(n) is even only for n = 3k for some integer k [which relates to the fact that A082115 Fibonacci sequence (mod 3) is periodic with Pisano period 8]. Hence after a(1) = -1, Fibonacci(prime(n)) - prime(Fibonacci(n)) is always the difference of two odd numbers, hence is even. - Jonathan Vos Post, Jan 23 2006
Is a(i) ever divisible by i? Answer: yes. The quotient is an integer for i = 4, 28 and 30 through 63. - Dennis S. Kluk (mathemagician(AT)ameritech.net), Aug 16 2006

			a(11) = Fibonacci(prime(11)) - prime(Fibonacci(11)) = 1345808.

Chai Wah Wu, Table of n, a(n) for n = 1..84 (terms 1..41 from Harry J. Smith)

Cf. A000040, A000045, A030427.

Cf. A082115.

[Fibonacci(NthPrime(n)) - NthPrime(Fibonacci(n)): n in [1..30]]; // Vincenzo Librandi, Apr 10 2020

For[i=1, i<61, i++, Print[i, " ", Fibonacci[Prime[i]]-Prime[Fibonacci[i]]]]
Table[Fibonacci[Prime[n]]-Prime[Fibonacci[n]],{n,30}] (* Harvey P. Dale, Jul 02 2018 *)

a(n) = { fibonacci(prime(n)) - prime(fibonacci(n)) } \\ Harry J. Smith, Jun 20 2009

a(n) = Fibonacci(prime(n)) - prime(Fibonacci(n)).

A160079 Lodumo_3 of Fibonacci numbers.

Original entry on oeis.org

0, 1, 4, 2, 3, 5, 8, 7, 6, 10, 13, 11, 9, 14, 17, 16, 12, 19, 22, 20, 15, 23, 26, 25, 18, 28, 31, 29, 21, 32, 35, 34, 24, 37, 40, 38, 27, 41, 44, 43, 30, 46, 49, 47, 33, 50, 53, 52, 36, 55, 58, 56, 39, 59, 62, 61, 42, 64, 67, 65, 45, 68, 71, 70, 48, 73, 76, 74, 51, 77, 80, 79, 54

Offset: 0

Philippe Deléham, May 01 2009

Permutation of nonnegative integers.

OEIS wiki, Lodumo transform
Index entries for sequences that are permutations of the nonnegative integers
Index entries for linear recurrences with constant coefficients, signature (0,0,0,1,0,0,0,1,0,0,0,-1).

Cf. A000045, A082115, A160016, A160051, A367651 (inverse permutation).

a(n) = lod_3(A000045(n)).
a(n) = 2*a(n-8) - a(n-16) for n >= 16. - Philippe Deléham, Mar 09 2023
a(8*n) = 6*n, a(8*n+1) = 9*n+1, a(8*n+2) = 9*n+4, a(8*n+3) = 9*n+2, a(8*n+4) = 6*n+3, a(8*n+5) = 9*n+5, a(8*n+6) = 9*n+8, a(8*n+7) = 9*n+7. - Philippe Deléham, Nov 24 2023
From Philippe Deléham, Nov 29 2023 : (Start)
a(n) = a(n-4) + a(n-8) - a(n-12) for n >= 12.
G.f. : (x + 4*x^2 + 2*x^3 + 3*x^4 + 4*x^5 + 4*x^6 + 5*x^7 + 3*x^8 + 4*x^9 + x^10 + 2*x^11) / (1 - x^4 - x^8 + x^12). (End)

A117875 Triangle a(n,m) = (A000045(m) mod 3)+ (A000045(n) mod 7) read by rows.

Original entry on oeis.org

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

Offset: 0

Roger L. Bagula, May 13 2006

			The triangle starts in row n=0 as:
0;
1, 2;
1, 2, 2;
2, 3, 3, 4;
3, 4, 4, 5, 3;
5, 6, 6, 7, 5, 7;
1, 2, 2, 3, 1, 3, 3;
6, 7, 7, 8, 6, 8, 8, 7;
0, 1, 1, 2, 0, 2, 2, 1, 0;
6, 7, 7, 8, 6, 8, 8, 7, 6, 7;
6, 7, 7, 8, 6, 8, 8, 7, 6, 7, 7;

Cf. A000045.

a[0] = 0; a[1] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] f[n_, m_] := Mod[a[n], 3] + Mod[a[m], 7] aout = Table[Table[f[n, m], {n, 0, m}], {m, 0, 10}] c = Flatten[aout]

a(n,m) = A082115(m) + A105870(n).

Fuzzy comments removed, formula added, row and column indices disentangled by Assoc. Eds. of the OEIS - Jun 15 2010

A164743 Digital root of 3*A000045(n).

Original entry on oeis.org

3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9, 3, 3, 6, 9, 6, 6, 3, 9

Offset: 1

Mark Dols, Aug 24 2009

Period 8.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,1)

Cf. A049341,A000045,A030132,A082115

a(n+1) = sum of digits of a(n) + a(n-1).
a(n) = A010888(3*A000045(n)). - R. J. Mathar, Nov 03 2016
G.f.: -3*x*(1+x+2*x^2+3*x^3+2*x^4+2*x^5+x^6+3*x^7) / ( (x-1)*(1+x)*(x^2+1)*(x^4+1) ). - R. J. Mathar, Nov 03 2016

cp's OEIS Frontend

A131295 a(n)=ds_4(a(n-1))+ds_4(a(n-2)), a(0)=0, a(1)=1; where ds_4=digital sum base 4.

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A082116 Fibonacci sequence (mod 5).

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A079344 F(n) mod 8, where F(n) = A000045(n) is the n-th Fibonacci number.

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A082117 Fibonacci sequence (mod 6).

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A280154 a(n) = 5*Lucas(n).

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A004696 a(n) = floor(Fibonacci(n)/3).

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PARI

PARI

Sage