A045468 -id:A045468 - cp's OEIS frontend

A003463 a(n) = (5^n - 1)/4.

0, 1, 6, 31, 156, 781, 3906, 19531, 97656, 488281, 2441406, 12207031, 61035156, 305175781, 1525878906, 7629394531, 38146972656, 190734863281, 953674316406, 4768371582031, 23841857910156, 119209289550781, 596046447753906, 2980232238769531

Offset: 0

N. J. A. Sloane

5^a(n) is the highest power of 5 dividing (5^n)!. - Benoit Cloitre, Feb 04 2002
n such that A002294(n) is not divisible by 5. - Benoit Cloitre, Jan 14 2003
Without leading zero, i.e., sequence {a(n+1) = (5*5^n-1)/4}, this is the binomial transform of A003947. - Paul Barry, May 19 2003 [Edited by M. F. Hasler, Oct 31 2014]
Numbers n such that a(n) is prime are listed in A004061(n) = {3, 7, 11, 13, 47, 127, 149, 181, 619, 929, ...}. Corresponding primes a(n) are listed in A086122(n) = {31, 19531, 12207031, 305175781, 177635683940025046467781066894531, ...}. 3^(m+1) divides a(2*3^m*k). 31 divides a(3k). 13 divides a(4k). 11 divides a(5k). 71 divides a(5k). 7 divides a(6k). 19531 divides a(7k). 313 divides a(8k). 19 divides a(9k). 829 divides a(9k). 71 divides a(10k). 521 divides a(10k). 17 divides a(16k). p divides a(p-1) for all prime p except p = {2,5}. p^(m+1) divides a(p^m*(p-1)) for all prime p except p = {2,5}. p divides a((p-1)/2) for prime p = {11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, ...} = A045468, Primes congruent to {1, 4} mod 5. p divides a((p-1)/3) for prime p = {13, 67, 127, 163, 181, 199, 211, 241, 313, 337, 367, 379, 409, 457, ...}. p divides a((p-1)/4) for prime p = {101, 109, 149, 181, 269, 389, 401, 409, 449, 461, 521, 541, ...} = A107219, Primes of the form x^2+100y^2. p divides a((p-1)/5) for prime p = {31, 191, 251, 271, 601, 641, 761, 1091, 1861, ...}. p divides a((p-1)/6) for prime p = {181, 199, 211, 241, 379, 409, 631, 691, 739, 769, 1039, ...}. - Alexander Adamchuk, Jan 23 2007
Starting with 1 = convolution square of A026375: (1, 3, 11, 45, 195, 873, ...). - Gary W. Adamson, May 17 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Jan 27 2010
This is the sequence A(0,1;4,5;2) = A(0,1;6,-5;0) of the family of sequences [a,b:c,d:k] considered by Gary Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
It is the Lucas sequence U(6,5). - Felix P. Muga II, Mar 21 2014
a(2*n+1) is the sum of the numerators and denominators of the reduced fractions 0 < b/5^n < 1 plus 1, with b < 5^n. - J. M. Bergot, Jul 24 2015
The sequence multiplied by 10 (0, 10, 60, 310, 1560, ...) is the maximum number of coins which can be decided by n weighings on 2 balances in the counterfeit coin problem with undecided under/overweight. [Halbeisen and Hungerbuhler, Disc. Math. 147 (1995) 139 Theorem 1]. - R. J. Mathar, Sep 10 2015
Order of the rank-n projective geometry PG(n-1,5) over the finite field GF(5). - Anthony Hernandez, Oct 05 2016
Number of zeros in the substitution system {0 -> 11100, 1 -> 11110} at step n from initial string "1" (1 -> 11110 -> 1111011110111101111011100 -> ...). - Ilya Gutkovskiy, Apr 10 2017
a(n) is the numerator of Sum_{k=1..n} 1/5^k, which approaches a limit of 1/4. The denominators are 5^n. In general, Sum_{k=1..n} 1/x^k approaches a limit of 1/(x-1). It is of interest to note that as x increases, so does the rate of convergence. See Crossrefs for numerators for other values of x which have the general form (x^n-1)/(x-1). - Gary Detlefs, Aug 31 2021

			Base 5...........decimal
0......................0
1......................1
11.....................6
111...................31
1111.................156
11111................781
111111..............3906
1111111............19531
11111111...........97656
111111111.........488281
1111111111.......2441406
etc. ...............etc.
- _Zerinvary Lajos_, Jan 14 2007

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 282.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Joseph E. Bonin and Joseph P. S. Kung The Number of Points In A Combinatorial Geometry With No 8-Point-Line Minors, Mathematical Essays in Honor of Gian-Carlo Rota, B. Sagan and R. P. Stanley, eds., Birkhäuser, 1998, 271-284.
Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
Roger B. Eggleton, Maximal Midpoint-Free Subsets of Integers, International Journal of Combinatorics Volume 2015, Article ID 216475, 14 pages.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 374
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Repunit
Index entries for linear recurrences with constant coefficients, signature (6,-5).

Cf. A004061, A026375, A045468, A060458, A074479, A086122, A107219.

Cf. A003462, A002450, A003464, A023000, A023001, A002452, A002275.

[(5^n-1)/4 : n in [0..30]]; // Wesley Ivan Hurt, Sep 25 2014

a:=n->sum(5^(n-j),j=1..n): seq(a(n), n=0..23); # Zerinvary Lajos, Jan 04 2007
A003463:=1/(5*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation
a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=6*a[n-1]-5*a[n-2]od: seq(a[n], n=0..23); # Zerinvary Lajos, Feb 21 2008

lst={}; Do[p=(5^n-1)/4; AppendTo[lst, p], {n, 0, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Sep 29 2008 *)
Table[((5^n-1)/4),{n,0,25}] (* Vincenzo Librandi, Aug 20 2012 *)
NestList[5 # + 1 &, 0, 23] (* Bruno Berselli, Feb 06 2013 *)
LinearRecurrence[{6,-5},{0,1},30] (* Harvey P. Dale, Sep 20 2023 *)

A003463(n):=floor((5^n-1)/4)$ makelist(A003463(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=5^n\4; \\ Charles R Greathouse IV, Jul 15 2011

[lucas_number1(n,6,5) for n in range(0, 24)] # Zerinvary Lajos, Apr 22 2009

[gaussian_binomial(n,1,5) for n in range(0,24)] # Zerinvary Lajos, May 28 2009

Second binomial transform of A015518; binomial transform of A000302 (preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=1..n} binomial(n,k)*4^(k-1). - Paul Barry, Mar 28 2003
a(n) = (-1)^n times the (i, j)-th element of M^n (for all i and j such that i is not equal to j), where M = ((1, -1, 1, -2), (-1, 1, -2, 1), (1, -2, 1, -1), (-2, 1, -1, 1)). - Simone Severini, Nov 25 2004
a(n) = A125118(n,4) for n>3. - Reinhard Zumkeller, Nov 21 2006
a(n) = ((3+sqrt(4))^n - (3-sqrt(4))^n)/4. - Al Hakanson (hawkuu(AT)gmail.com), Dec 31 2008
a(n) = 6*a(n-1) - 5*a(n-2) n>1, a(0)=0, a(1)=1. - Philippe Deléham, Jan 01 2009
From Wolfdieter Lang, Oct 18 2010: (Start)
O.g.f.: x/((1-5*x)*(1-x)).
a(n) = 4*a(n-1) + 5*a(n-2) + 2, a(0)=0, a(1)=1.
a(n) = 5*a(n-1) + a(n-2) - 5*a(n-3) = 7*a(n-1) - 11*a(n-2) + 5*a(n-3), a(0)=0, a(1)=1, a(2)=6. Observation by G. Detlefs. See the W. Lang comment and link. (End)
a(n) = 5*a(n-1) + 1 with n>0, a(0)=0. - Vincenzo Librandi, Nov 17 2010
a(n) = a(n-1) + A000351(n-1) n>0, a(0)=0. - Felix P. Muga II, Mar 19 2014
a(n) = a(n-1) + 20*a(n-2) + 5 for n > 1, a(0)=0, a(1)=1. - Felix P. Muga II, Mar 19 2014
a(n) = A060458(n)/2^(n+2), for n > 0. - R. J. Cano, Sep 25 2014
From Ilya Gutkovskiy, Oct 05 2016: (Start)
E.g.f.: (exp(4*x) - 1)*exp(x)/4.
Convolution of A000351 and A057427. (End)

A038872 Primes congruent to {0, 1, 4} mod 5.

Original entry on oeis.org

5, 11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, 131, 139, 149, 151, 179, 181, 191, 199, 211, 229, 239, 241, 251, 269, 271, 281, 311, 331, 349, 359, 379, 389, 401, 409, 419, 421, 431, 439, 449, 461, 479, 491, 499, 509, 521, 541, 569, 571, 599, 601, 619

Offset: 1

N. J. A. Sloane

Also odd primes p such that 5 is a square mod p: (5/p) = +1 for p > 5.
Primes of the form x^2 + x*y - y^2 (as well as of the form x^2 + 3*x*y + y^2), both with discriminant = 5 and class number = 1. Binary quadratic forms a*x^2 + b*x*y + c*y^2 have discriminant d = b^2 - 4ac and gcd(a, b, c) = 1. [This was originally a separate entry, submitted by Laura Caballero Fernandez, Lourdes Calvo Moguer, Maria Josefa Cano Marquez, Oscar Jesus Falcon Ganfornina and Sergio Garrido Morales, Jun 06 2008. R. J. Mathar proved on Jul 22 2008 that this coincides with the present sequence.]
Also primes of the form 5x^2 - y^2 (cf. A031363). - N. J. A. Sloane, May 30 2014
Also primes of the form x^2 + 4*x*y - y^2. Every binary quadratic primitive form of discriminant 20 or 5 has proper solutions for positive integers N given in A089270, including the present primes. Proof from computing the corresponding representative parallel primitive forms, which leads to x^2 - 5 == 0 (mod N) or x^2 + x - 1 == 0 (mod N) which have solutions precisely for these positive N values, including these primes. - Wolfdieter Lang, Jun 19 2019
For a Pythagorean triple a, b, c, these primes (and 2) are the possible prime factors of 2a + b, |2a - b|, 2b + a, and 2b - a. - J. Lowell, Nov 05 2011
The prime factors of A028387(n^2+3n+1). - Richard R. Forberg, Dec 12 2014
Except for p = 5, these are primes p that divide Fibonacci(p-1). - Dmitry Kamenetsky, Jul 27 2015
Apart from the first term, these are rational primes that decompose in the field Q[sqrt(5)]. For example, 11 = ((7 + sqrt(5))/2)*((7 - sqrt(5))/2), 19 = ((9 + sqrt(5))/2)*((9 - sqrt(5))/2). - Jianing Song, Nov 23 2018
The possible prime factors of x^2 - x - 1. - Charles R Greathouse IV, Mar 18 2022

Z. I. Borevich and I. R. Shafarevich, Number Theory.

T. D. Noe, Table of n, a(n) for n = 1..1000
C. Banderier, Calcul de (5/p)
Henri Darmon, Andrew Wiles's Marvelous Proof, Notices of the AMS (2017), Volume 64, Number 3 pp. 209-216. See p. 211.
Tamara M. Lavshuk, Regular polygons and polyhedra over finite field, Mathematical Notes of NEFU, Vol 22 No 4 (2015). Mentions this sequence.
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)
D. B. Zagier, Zetafunktionen und quadratische Körper, Springer, 1981.

Cf. A045468, A089270.
Cf. A038872 (d=5); A038873 (d=8); A068228, A141123 (d=12); A038883 (d=13). A038889 (d=17); A141111, A141112 (d=65).
Cf. A003631 (complement with respect to A000040).

Filtered(Concatenation([5],Flat(List([1..140],k->[5*k-1,5*k+1]))),IsPrime); # Muniru A Asiru, Nov 24 2018

[ p: p in PrimesUpTo(700) | p mod 5 in {0,1,4}]; // Vincenzo Librandi, Aug 21 2012

select(isprime, [5, seq(op([5*k-1,5*k+1]),k=1..1000)]); # Robert Israel, Dec 22 2014

Join[{5}, Select[Prime[Range[4, 100]], Mod[#, 5] == 1 || Mod[#, 5] == 4 &]] (* Alonso del Arte, Nov 27 2011 *)

forprime(p=2,1e3,if(kronecker(5,p)>=0,print1(p", "))) \\ Charles R Greathouse IV, Jun 16 2011

a(n) = A045468(n-1) for n > 1. - Robert Israel, Dec 22 2014

a(n) ~ 2n*log(n). - Charles R Greathouse IV, Nov 29 2016

Corrected and extended by Peter K. Pearson, May 29 2005

Edited by N. J. A. Sloane, Jul 28 2008 at the suggestion of R. J. Mathar

A047209 Numbers that are congruent to {1, 4} mod 5.

Original entry on oeis.org

1, 4, 6, 9, 11, 14, 16, 19, 21, 24, 26, 29, 31, 34, 36, 39, 41, 44, 46, 49, 51, 54, 56, 59, 61, 64, 66, 69, 71, 74, 76, 79, 81, 84, 86, 89, 91, 94, 96, 99, 101, 104, 106, 109, 111, 114, 116, 119, 121, 124, 126, 129, 131, 134, 136, 139, 141, 144, 146, 149, 151, 154

Offset: 1

N. J. A. Sloane

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 72 ).
Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in our case, a(n)^2 - 1 == 0 (mod 5). - Bruno Berselli, Nov 17 2010
The sum of the alternating series (-1)^(n+1)/a(n) from n=1 to infinity is (Pi/5)*cot(Pi/5), that is (1/5)*sqrt(1 + 2/sqrt(5))*Pi. - Jean-François Alcover, May 03 2013
These numbers appear in the product of a Rogers-Ramanujan identity. See A003114 also for references. - Wolfdieter Lang, Oct 29 2016
Let m be a product of any number of terms of this sequence. Then m - 1 or m + 1 is divisible by 5. Closed under multiplication. - David A. Corneth, May 11 2018

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
William A. Stein, The modular forms database.
Eric Weisstein's World of Mathematics, Determined by Spectrum.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000566, A036666, A001622, A003114, A203776, A047336, A047522, A056020, A090771, A175885, A091998, A175886, A175887, A352324.
Cf. A005408 (n=1 or 3 mod 4), A007310 (n=1 or 5 mod 6).
Cf. A045468 (primes), A032527 (partial sums).

a047209 = (flip div 2) . (subtract 2) . (* 5)
a047209_list = 1 : 4 : (map (+ 5) a047209_list)
-- Reinhard Zumkeller, Jul 19 2013, Jan 05 2011

seq(floor(5*k/2)-1, k=1..100); # Wesley Ivan Hurt, Sep 27 2013

Select[Range[0, 200], MemberQ[{1, 4}, Mod[#, 5]] &] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
LinearRecurrence[{1,1,-1},{1,4,6},70] (* Harvey P. Dale, Jul 19 2024 *)

a(n)=(10*n+(-1)^n-5)/4 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: (1+3x+x^2)/((1-x)(1-x^2)).
a(n) = floor((5n-2)/2). [corrected by Reinhard Zumkeller, Jul 19 2013]
a(1) = 1, a(n) = 5(n-1) - a(n-1). - Benoit Cloitre, Apr 12 2003
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = (10*n + (-1)^n - 5)/4.
a(n) - a(n-1) - a(n-2) + a(n-3) = 0 for n > 3.
a(n) = a(n-2) + 5 for n > 2.
a(n) = 5*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1.
a(n)^2 = 5*A036666(n) + 1 (cf. also Comments). (End)
a(n) = 5*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
E.g.f.: 1 + ((10*x - 5)*exp(x) + exp(-x))/4. - David Lovler, Aug 23 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = phi (A001622).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/5) * cosec(Pi/5) (A352324). (End)

Edited by Michael Somos, Sep 22 2002

A090771 Numbers that are congruent to {1, 9} mod 10.

Original entry on oeis.org

1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69, 71, 79, 81, 89, 91, 99, 101, 109, 111, 119, 121, 129, 131, 139, 141, 149, 151, 159, 161, 169, 171, 179, 181, 189, 191, 199, 201, 209, 211, 219, 221, 229, 231, 239, 241, 249, 251, 259, 261, 269, 271, 279, 281

Offset: 1

Giovanni Teofilatto, Feb 07 2004

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n - h)/4 (h, n natural numbers), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 10). - Bruno Berselli, Nov 17 2010

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000217, A000796, A090298, A005408, A047209, A007310, A019970, A047336, A047522, A091998, A094888, A113801, A175886, A175887, A188593.
Cf. A056020 (n = 1 or 8 mod 9), A175885 (n = 1 or 10 mod 11).
Cf. A045468 (primes), A195142 (partial sums).

a090771 n = a090771_list !! (n-1)
a090771_list = 1 : 9 : map (+ 10) a090771_list
-- Reinhard Zumkeller, Jan 07 2012

Flatten[Table[10n - {9, 1}, {n, 30}]] (* Alonso del Arte, Sep 02 2014 *)
LinearRecurrence[{1,1,-1},{1,9,11},60] (* Harvey P. Dale, Jul 05 2020 *)

a(n)=n\2*10-(-1)^n \\ Charles R Greathouse IV, Sep 24 2015

a(n) = sqrt(40*A057569(n) + 1). - Gary Detlefs, Feb 22 2010
From Bruno Berselli, Sep 16 2010 - Nov 17 2010: (Start)
  G.f.: x*(1 + 8*x + x^2)/((1 + x)*(1 - x)^2).
  a(n) = (10*n + 3*(-1)^n - 5)/2.
  a(n) = -a(-n + 1) = a(n-1) + a(n-2) - a(n-3) = a(n-2) + 10.
  a(n) = 10*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
a(n) = 10*n - a(n-1) - 10 (with a(1) = 1). - Vincenzo Librandi, Nov 16 2010
a(n) = sqrt(10*A132356(n-1) + 1). - Ivan N. Ianakiev, Nov 09 2012
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/10)*cot(Pi/10) = A000796 * A019970 / 10 = sqrt(5 + 2*sqrt(5))*Pi/10. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((10*x - 5)*exp(x) + 3*exp(-x))/2. - David Lovler, Sep 03 2022
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(phi+2) (A188593).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi*phi/5 = A094888/10. (End)

Edited and extended by Ray Chandler, Feb 10 2004

A005968 Sum of cubes of first n Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 10, 37, 162, 674, 2871, 12132, 51436, 217811, 922780, 3908764, 16558101, 70140734, 297121734, 1258626537, 5331629710, 22585142414, 95672204155, 405273951280, 1716768021816, 7272346018247, 30806152127640, 130496954475672, 552793970116297, 2341672834801754

Offset: 0

N. J. A. Sloane

From Alexander Adamchuk, Aug 07 2006: (Start)
The only two prime terms are a(2) = 2 and a(4) = 37.
The prime p divides a(p-1) iff p is in A045468.
The prime p divides a((p-1)/2) iff p is in A047650.
3^4 divides a(p) iff p is in A003628.
3^5 divides a(p) for p = {37,53,109,181,197,269,397,431,541,...}.
3^6 divides a(p) for p = {109,541,...}.
3^7 divides a(p) for p = {557,...}. (End)

Art Benjamin, Timothy A. Carnes, and Benoit Cloitre, Recounting the Sums of Cubes of Fibonacci Numbers, Congressus Numerantium, Proceedings of the Eleventh International Conference on Fibonacci Numbers and their Applications, (William Webb, ed.), Vol 194, pp. 45-51, 2009.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 14.
A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 18.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Art Benjamin, Timothy A. Carnes, and Benoit Cloitre, Counting the Sums of Cubes of Fibonacci Numbers.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
David Treeby, Hidden Formulas in Fibonacci Tilings, Fibonacci Quart. 54 (2016), no. 1, 23-30.
Index entries for linear recurrences with constant coefficients, signature (4,3,-9,2,1).

Partial sums of A056570.  Cf. A119284 (alternating sum).
Cf. A045468, A047650, A003628.
Sums of other powers: A000071, A001654, A005969, A098531, A098532, A098533, A128697.

[(1/10)*( Fibonacci(3*n+2)-(-1)^(n)*6*Fibonacci(n-1)+5 ): n in [0..30]]; // G. C. Greubel, Jan 17 2018

with(combinat): l[0] := 0: for i from 1 to 50 do l[i] := l[i-1]+fibonacci(i)^3; printf(`%d,`,l[i]) od: # James Sellers, May 29 2000
A005968:=(-1+2*z+z**2)/(z-1)/(z**2+4*z-1)/(z**2-z-1); # conjectured by Simon Plouffe in his 1992 dissertation

f[n_]:=(Fibonacci[n]*Fibonacci[n+1]^2+(-1)^(n-1)*Fibonacci[n-1]+1)/2;Table[f[n],{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Nov 22 2010 *)
Accumulate[Fibonacci[Range[0,20]]^3]
CoefficientList[Series[x*(1-2*x-x^2)/((1-x)*(1+x-x^2)*(1-4*x-x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)

a(n)=(fibonacci(n)*fibonacci(n+1)^2+(-1)^(n-1)*fibonacci(n-1)+1)/2

a(n)=(fibonacci(3*n+2)-(-1)^(n)*6*fibonacci(n-1)+5)/10

PARI
```
a(n)=sum(i=1,n,fibonacci(i)^3)
```

G.f.: x*(1-2*x-x^2)/((1-x)*(1+x-x^2)*(1-4*x-x^2)). - Ralf Stephan, Apr 23 2004
a(n) = (1/2)*(F(n)*F(n+1)^2 + (-1)^(n-1)*F(n-1) + 1). - Benoit Cloitre, Aug 06 2004
a(n) = Sum_{i=1..n} A000045(i)^3.
a(n) = (1/10)*(F(3*n+2) - (-1)^(n)*6*F(n-1) + 5). - Art Benjamin and Timothy A. Carnes
a(n+5) = 4*a(n+4) + 3*a(n+3) - 9*a(n+2) + 2*a(n+1) + a(n). - Benoit Cloitre, Sep 12 2004

More terms from James Sellers, May 29 2000

A064739 Primes p such that Fibonacci(p)-1 is divisible by p.

Original entry on oeis.org

2, 11, 19, 29, 31, 41, 59, 61, 71, 79, 89, 101, 109, 131, 139, 149, 151, 179, 181, 191, 199, 211, 229, 239, 241, 251, 269, 271, 281, 311, 331, 349, 359, 379, 389, 401, 409, 419, 421, 431, 439, 449, 461, 479, 491, 499, 509, 521, 541, 569, 571, 599, 601, 619

Offset: 1

Shane Findley and N. J. A. Sloane, Oct 17 2001

nonn

Harry J. Smith, Table of n, a(n) for n = 1..1000

{2} union A045468. Complement is A003631 minus {2}.

lst={};Do[p=Prime[n];If[Mod[(Fibonacci[p]-1),p]==0,AppendTo[lst,p]],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, Apr 03 2009 *)
Select[Prime[Range[150]],Divisible[Fibonacci[#]-1,#]&] (* Harvey P. Dale, Sep 24 2022 *)

forprime(p=2,700, if((fibonacci(p)-1)%p==0,print1(p,", ")))

{ n=0; for (m=1, 10^9, p=prime(m); if ((fibonacci(p) - 1)%p==0, write("b064739.txt", n++, " ", p); if (n==1000, break)) ) } \\ Harry J. Smith, Sep 24 2009

Presumably this consists of 2 together with the primes congruent to +-1 mod 5.

More terms from Klaus Brockhaus, Oct 18 2001

A023173 Numbers k such that Fibonacci(k) == 1 (mod k).

Original entry on oeis.org

1, 2, 11, 19, 22, 29, 31, 38, 41, 58, 59, 61, 62, 71, 79, 82, 89, 101, 109, 118, 122, 131, 139, 142, 149, 151, 158, 178, 179, 181, 191, 199, 202, 211, 218, 229, 239, 241, 251, 262, 269, 271, 278, 281, 298, 302, 311, 323, 331, 349, 358, 359, 362, 379, 382, 389, 398, 401

Offset: 1

David W. Wilson

Sequence contains 1, A064739(k) for all k>=0, 2*A064739(k) for k>1. - Benoit Cloitre, Apr 06 2002

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A000045, A045468, A003631, A064739.

lst={};Do[If[Mod[Fibonacci[n],n]==1,AppendTo[lst,n]],{n,1,7!}];Take[Union[lst],5! ] (* Vladimir Joseph Stephan Orlovsky, Apr 03 2009 *)
Select[Range[500],Mod[Fibonacci[#]-1,#]==0&] (* Harvey P. Dale, Sep 22 2021 *)

fibmod(n,m)=((Mod([1,1;1,0],m))^n)[1,2]
is(n)=fibmod(n,n)==1 \\ Charles R Greathouse IV, Oct 06 2016

A032357 Convolution of Catalan numbers and powers of -1.

Original entry on oeis.org

1, 0, 2, 3, 11, 31, 101, 328, 1102, 3760, 13036, 45750, 162262, 580638, 2093802, 7601043, 27756627, 101888163, 375750537, 1391512653, 5172607767, 19293659253, 72188904387, 270870709263, 1019033438061, 3842912963391, 14524440108761

Offset: 0

Wolfdieter Lang

Absolute value of the alternating sum of Catalan Numbers. - Alexander Adamchuk, Jul 03 2006
Sums of two consecutive terms are a(n-1) + a(n) = 1, 2, 5, 14, 42, ... = A000108(n) (Catalan Numbers). The prime p divides a((p-3)/2) for p = 11, 19, 29, 31, 41, 59, 61, 71, ... = A045468 (Primes congruent to {1, 4} mod 5). Prime p divides a(2*p+1) for p = 5, 11, 19, 29, 31, 41, 59, 61, 71, ... = A038872 (Primes congruent to {0, 1, 4} mod 5). Also odd primes where 5 is a square mod p. - Alexander Adamchuk, Jul 03 2006
Hankel transform is F(2*n+1), where F = A000045. - Paul Barry, Jul 22 2008
Equals INVERTi transform of A000958. - Gary W. Adamson, Apr 10 2009
Inverse binomial transform of A002212. - Philippe Deléham, Sep 17 2009
Number of singleton and plus-decomposable (2143, 2413, 3142)-avoiding permutations with no +bonds (ascents by 1), with offset 1. Equivalently, number of (2143, 2413, 3142)-avoiding permutations that start with 1 or end with n (top entry). E.g., 132 and 213 for n = 3; 1324, 1432, 3214 for n = 4. - Alexander Burstein, May 22 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Shalosh B. Ekhad and Mingjia Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, 2017.

Cf. A000045, A000108, A000958, A001622, A002212, A014137, A014138, A033297, A038872, A045468, A064739.

rec:= (n+1)*a(n) +3*(-n+1)*a(n-1) +2*(-2*n+1)*a(n-2)=0:
A:= gfun:-rectoproc({rec,a(0)=1,a(1)=0},a(n),remember):
seq(A(n),n=0..50); # Robert Israel, May 22 2015

Table[Sum[(-1)^(k+n)*CatalanNumber[k],{k,0,n}],{n,0,60}] (* Alexander Adamchuk, Jul 03 2006 *)
Round@Table[(-1)^n/GoldenRatio + CatalanNumber[n + 1] Hypergeometric2F1[1, n + 3/2, n + 3, -4], {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster - Vladimir Reshetnikov, Oct 02 2016 *)
Table[(CatalanNumber[n] (2 + (n + 1) Hypergeometric2F1[1, -n, 1/2, 5/4]) - (-1)^n)/2, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)

def A032357():
    f, c, n = 1, 1, 1
    while True:
        yield f
        n += 1
        c = c * (4*n - 6) // n
        f = c - f
a = A032357()
print([next(a) for  in range(27)]) # _Peter Luschny, Nov 30 2016

G.f.: c(x)/(1 + x), where c(x) is the g.f. for the Catalan numbers A000108.
a(n) = Sum_{k=0..n} (-1)^(n-k)*C(k), where C(k) = A000108(k).
a(n) = ((-1)^(n+1) - binomial(2*(n+1), n+1)*Sum_{k=0..n+1} (-5)^k*binomial(n+1, k)/binomial(2*k, k))/2.
a(n) = C(2*n, n)/(n+1) - a(n-1) = A000108(n) - a(n-1) with a(0) = 1. - Labos Elemer, Apr 26 2003
Conjecture: (n+1)*a(n) + 3*(-n+1)*a(n-1) + 2*(-2*n+1)*a(n-2) = 0. - R. J. Mathar, Nov 30 2012
Conjecture is true since the g.f. satisfies (x - 3*x^2 - 4*x^3)*g'(x) + (1 - 6*x^2)*g(x) = 1. - Robert Israel, May 22 2015
a(n) = (-1)^n/A001622 + A000108(n+1)*hypergeom([1, n + 3/2], [n + 3], -4). - Vladimir Reshetnikov, Oct 02 2016
a(n) ~ 2^(2*n + 2) / (5*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 03 2016
a(n) = (A000108(n) * (2 + (n + 1)*hypergeom([1,-n], [1/2], 5/4)) - (-1)^n)/2. - Vladimir Reshetnikov, Oct 03 2016

More terms from Christian G. Bower, Apr 15 1998

More terms from Alexander Adamchuk, Jul 03 2006

cp's OEIS Frontend

A094808 Numbers of the form Fibonacci(p-1)/p, where p are primes ending in 1 or 9 (i.e., A045468).

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A336403 Multiplicative closure of A045468: numbers which are the product of zero or more primes which are 1 or 4 mod 5.

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A003463 a(n) = (5^n - 1)/4.

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A038872 Primes congruent to {0, 1, 4} mod 5.

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A047209 Numbers that are congruent to {1, 4} mod 5.

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A090771 Numbers that are congruent to {1, 9} mod 10.

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