A064739 -id:A064739 - cp's OEIS frontend

A045468 Primes congruent to {1, 4} mod 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

Offset: 1

N. J. A. Sloane

Rational primes that decompose in the field Q(sqrt(5)). - N. J. A. Sloane, Dec 26 2017
These are also primes p that divide Fibonacci(p-1). - Jud McCranie
Primes ending in 1 or 9. - Lekraj Beedassy, Oct 27 2003
Also primes p such that p divides 5^(p-1)/2 - 4^(p-1)/2. - Cino Hilliard, Sep 06 2004
Primes p such that the polynomial x^2-x-1 mod p has 2 distinct zeros. - T. D. Noe, May 02 2005
Same as A038872, apart from the term 5. - R. J. Mathar, Oct 18 2008
Appears to be the primes p such that p^6 mod 210 = 1. - Gary Detlefs, Dec 29 2011
Primes in A047209, also in A090771. - Reinhard Zumkeller, Jan 07 2012
Primes p such that p does not divide Sum_{i=1..p} Fibonacci(i)^2. The sum is A001654(p). - Arkadiusz Wesolowski, Jul 23 2012
Primes congruent to {1, 9} mod 10. Legendre symbol (5, a(n)) = +1. For prime 5 this symbol (5, 5) is set to 0, and (5, prime) = -1 for prime == {3, 7} (mod 10), given in A003631. - Wolfdieter Lang, Mar 05 2021

Hardy and Wright, An Introduction to the Theory of Numbers, Chap. X, p. 150, Oxford University Press, Fifth edition.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Caleb Ji, Tanya Khovanova, Robin Park, and Angela Song, Chocolate Numbers, arXiv:1509.06093 [math.CO], 2015.
Caleb Ji, Tanya Khovanova, Robin Park, and Angela Song, Chocolate Numbers, Journal of Integer Sequences, Vol. 19 (2016), #16.1.7.
Index to sequences related to decomposition of primes in quadratic fields

Cf. A030430, A030433, A064739, A038872, A010051, A003631.

Subsequence of A123976.

a045468 n = a045468_list !! (n-1)
a045468_list = [x | x <- a047209_list, a010051 x == 1]
-- Reinhard Zumkeller, Jan 07 2012

[ p: p in PrimesUpTo(1000) | p mod 5 in {1,4} ]; // Vincenzo Librandi, Aug 13 2012

for n from 1 to 500 do if(isprime(n)) and (n^6 mod 210=1) then print(n) fi od;  # Gary Detlefs, Dec 29 2011

lst={};Do[p=Prime[n];If[Mod[p,5]==1||Mod[p,5]==4,AppendTo[lst,p]],{n,6!}];lst (* Vladimir Joseph Stephan Orlovsky, Feb 26 2009 *)
Select[Prime[Range[200]],MemberQ[{1,4},Mod[#,5]]&] (* Vincenzo Librandi, Aug 13 2012 *)

list(lim)=select(n->n%5==1||n%5==4,primes(primepi(lim))) \\ Charles R Greathouse IV, Jul 25 2011

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

A069107 Composite numbers k that divide Fibonacci(k+1).

Original entry on oeis.org

323, 377, 2834, 3827, 5777, 6479, 10877, 11663, 18407, 19043, 20999, 23407, 25877, 27323, 34943, 35207, 39203, 44099, 47519, 50183, 51983, 53663, 60377, 65471, 75077, 78089, 79547, 80189, 81719, 82983, 84279, 84419, 86063, 90287, 94667

Offset: 1

Benoit Cloitre, Apr 06 2002

Primes p congruent to +2 or -2 (mod 5) divide Fibonacci(p+1) (cf. A003631 and [Hardy and Wright]).

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers (Fifth edition), Oxford Univ. Press (Clarendon), 1979, Chap. X, p. 150.

Chai Wah Wu, Table of n, a(n) for n = 1..2000 (n = 1..250 from Reinhard Zumkeller, n = 251..1000 from Giovanni Resta)

Cf. A045468, A003631, A064739, A081264 (Fibonacci pseudoprimes).

Cf. A000045, A010051, A023172, A069104.

a069107 n = a069107_list !! (n-1)
a069107_list = h 2 $ drop 3 a000045_list where
   h n (fib:fibs) = if fib `mod` n > 0 || a010051 n == 1
       then h (n+1) fibs else n : h (n+1) fibs
-- Reinhard Zumkeller, Oct 13 2011

Select[Range[2,100000],!PrimeQ[#]&&Divisible[Fibonacci[#+1],#]&] (* Harvey P. Dale, Sep 18 2011 *)

is(n)=((Mod([1,1;1,0],n))^(n+1))[1,2]==0 && !isprime(n) && n>1 \\ Charles R Greathouse IV, Oct 07 2016

Fibonacci(2*a(n)) mod a(n) = a(n) - 1. - Gary Detlefs, May 26 2014

Corrected by Ralf Stephan, Oct 17 2002

A122869 Primes p that divide Lucas((p-1)/2), where Lucas is A000032.

Original entry on oeis.org

11, 19, 31, 59, 71, 79, 131, 139, 151, 179, 191, 199, 211, 239, 251, 271, 311, 331, 359, 379, 419, 431, 439, 479, 491, 499, 571, 599, 619, 631, 659, 691, 719, 739, 751, 811, 839, 859, 911, 919, 971, 991, 1019, 1031, 1039, 1051, 1091, 1151, 1171, 1231, 1259

Offset: 1

Alexander Adamchuk, Sep 16 2006

nonn

Final digit of a(n) is 1 or 9.
A002145 is the union of this sequence and A122870, Primes p that divide Lucas((p+1)/2).
Conjecture: This sequence is just the primes congruent to 11 or 19 mod 20. - Charles R Greathouse IV, May 25 2011 [The conjecture is correct. - Jianing Song, Jun 20 2025]
Note that F(p-1) = F((p-1)/2)*Lucas((p-1)/2), where F = A000045. Since gcd(F(n),Lucas(n)) = 1 or 2 (because Lucas(n)^2 - 5*F(n)^2 = 4*(-1)^n), this sequence lists primes p such that p divides F(p-1) but does not divides F((p-1)/2). By Propositions 1.1 and 1.2 (the k = 3 case) of my link below, this is primes p == 11, 19 (mod 20). - Jianing Song, Jun 20 2025

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Vincenzo Librandi)
Jianing Song, Lucas sequences and entry point modulo p
Eric Weisstein's World of Mathematics, Gaussian Prime.
Eric Weisstein's World of Mathematics, Lucas Number.

Cf. A000032, A000045, A053032, A076518, A122870.

Subsequence of A002145, A003626, A040105, A040147 and A064739.

Select[Prime[Range[1000]],IntegerQ[(Fibonacci[(#1-1)/2-1]+Fibonacci[(#1-1)/2+1])/#1]&]

lista(kmax) = {my(lucas1 = 1, lucas2 = 3, lucas3, p); for(k = 3, kmax, lucas3 = lucas1 + lucas2; p = 2*k + 1; if(isprime(p) && !(lucas3 % p), print1(p, ", ")); lucas1 = lucas2; lucas2 = lucas3);} \\ Amiram Eldar, Jun 06 2024

A069106 Composite numbers k such that k divides F(k-1) where F(j) are the Fibonacci numbers.

Original entry on oeis.org

442, 1891, 2737, 4181, 6601, 6721, 8149, 13201, 13981, 15251, 17119, 17711, 30889, 34561, 40501, 51841, 52701, 64079, 64681, 67861, 68101, 68251, 78409, 88601, 88831, 90061, 96049, 97921, 115231, 118441, 138601, 145351, 146611, 150121, 153781, 163081, 179697, 186961, 191351, 194833

Offset: 1

Benoit Cloitre, Apr 06 2002

Primes p congruent to 1 or 4 (mod 5) divide F(p-1) (cf. A045468 and [Hardy and Wright]).

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers (Fifth edition), Oxford Univ. Press (Clarendon), 1979, Chap. X, p. 150.

Giovanni Resta, Table of n, a(n) for n = 1..1000

Subsequence of A123976.
Cf. A045468, A003631, A064739, A081264 (Fibonacci pseudoprimes).
Cf. A002808, A000045.

#include  #include  #define STARTN 10 #define N_OF_MILLER_RABIN_TESTS 5 int main() { mpz_t n, f1, f2; int flag=0; /* flag? 0: f1 contains current F[n-1] 1: f2 = F[n-1] */ mpz_set_ui (n, STARTN); mpz_init (f1); mpz_init (f2); mpz_fib2_ui (f1, f2, STARTN-1); for (;;) { if (mpz_probab_prime_p (n, N_OF_MILLER_RABIN_TESTS)) goto next_iter; if (mpz_divisible_p (!flag? f1:f2, n)) { mpz_out_str (stdout, 10, n); printf (" "); fflush (stdout); } next_iter: mpz_add_ui (n, n, 1); mpz_add (!flag? f2:f1, f1, f2); flag = !flag; } }

a069106 n = a069106_list !! (n-1)
a069106_list = [x | x <- a002808_list, a000045 (x-1) `mod` x == 0]
-- Reinhard Zumkeller, Jul 19 2013

A069106[nn_] := Select[Complement[Range[2,nn],Prime[Range[2,PrimePi[ nn]]]],Divisible[ Fibonacci[ #-1],#]&] (* Harvey P. Dale, Jul 05 2011 *)

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

Corrected and extended (with C program) by Ralf Stephan, Oct 13 2002

a(35)-a(40) added by Reinhard Zumkeller, Jul 19 2013

A119996 Numerator of Sum_{k=1..n} 1/(Fibonacci(k)*Fibonacci(k+2)).

Original entry on oeis.org

1, 5, 14, 39, 103, 272, 713, 1869, 4894, 12815, 33551, 87840, 229969, 602069, 1576238, 4126647, 10803703, 28284464, 74049689, 193864605, 507544126, 1328767775, 3478759199, 9107509824, 23843770273, 62423800997, 163427632718

Offset: 1

Alexander Adamchuk, Aug 03 2006

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A000045, A059248, A064831, A001654, A045468, A064739, A091729.

F:=Fibonacci;; List([1..30], n-> F(n+1)*F(n+2)-1); # G. C. Greubel, Jul 23 2019

[Fibonacci(n+1)* Fibonacci(n+2)-1: n in [1..30]]; // Vincenzo Librandi, Aug 14 2012

with(combinat): seq(fibonacci(n+1)*fibonacci(n+2)-1, n=1..30); # Zerinvary Lajos, Jan 31 2008

Numerator[Table[Sum[1/(Fibonacci[k]*Fibonacci[k+2]),{k,n}],{n,30}]]
LinearRecurrence[{3,0,-3,1},{1,5,14,39},30] (* Harvey P. Dale, Aug 22 2011 *)

vector(30, n, f=fibonacci; f(n+1)*f(n+2)-1) \\ G. C. Greubel, Jul 23 2019

f=fibonacci; [f(n+1)*f(n+2)-1 for n in (1..30)] # G. C. Greubel, Jul 23 2019

a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4); a(0)=1, a(1)=5, a(2)=14, a(3)=39. - Harvey P. Dale, Aug 22 2011
G.f.: ((x-2)*x-1)/(x^4 - 3*x^3 + 3*x - 1). - Harvey P. Dale, Aug 22 2011
a(n) = Fibonacci(n+1)*Fibonacci(n+2) - 1. - Gary Detlefs, Mar 31 2012
a(n) = Sum_{k=1..n} Fibonacci(k+1)^2. Can be proved by induction from Gary Detlefs formula. - Joel Courtheyn, Mar 15 2021

A121567 Fibonacci[ (p - 1) ], where p = Prime[n].

Original entry on oeis.org

1, 1, 3, 8, 55, 144, 987, 2584, 17711, 317811, 832040, 14930352, 102334155, 267914296, 1836311903, 32951280099, 591286729879, 1548008755920, 27777890035288, 190392490709135, 498454011879264, 8944394323791464

Offset: 1

Alexander Adamchuk, Aug 07 2006

nonn

p = Prime[n] divides a(n) = Fibonacci(p-1) for p = {11,19,29,31,41,59,61,71,79,89,101,109,131,...} = A045468[n] Primes congruent to {1, 4} mod 5.

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Cf. A000045, A045468, A121568, A064739.

Mathematica
```
Table[Fibonacci[Prime[n]-1],{n,1,30}]
```

a(n) = Fibonacci[ Prime[n] - 1 ].

A121568 Fibonacci[ (p - 1)/2 ], where p = Prime[n].

Original entry on oeis.org

1, 1, 2, 5, 8, 21, 34, 89, 377, 610, 2584, 6765, 10946, 28657, 121393, 514229, 832040, 3524578, 9227465, 14930352, 63245986, 165580141, 701408733, 4807526976, 12586269025, 20365011074, 53316291173, 86267571272, 225851433717

Offset: 2

Alexander Adamchuk, Aug 07 2006

nonn

p = Prime[n] divides a(n) = Fibonacci[(p-1)/2] for p = {29,41,61,89,101,109,149,181,229,241,269,281,349,381,...} = A033205[n] Primes of form x^2+5*y^2 excluding A033205[1] = 5.

Cf. A000045, A121567, A033205, A045468, A064739.

Table[Fibonacci[(Prime[n]-1)/2],{n,2,50}]

a(n) = Fibonacci[ (Prime[n]-1)/2 ], n>1.

A066047 Numbers k that divide A001045(k-1).

Original entry on oeis.org

5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 341, 347, 349, 353, 359

Offset: 1

Benoit Cloitre, Dec 29 2001

nonn

The preponderance of entries are primes. The first which is not prime is 341.

All composite entries are 2-pseudoprimes. - Charles R Greathouse IV, Nov 04 2016

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A001045, A064739.

A066047 := proc(n)
    option remember ;
    if n = 1 then
        5 ;
    else
        for a from procname(n-1)+1 do
            if modp(A001045(a-1),a) =0 then
                return a;
            end if;
        end do:
    end if;
end proc:
seq(A066047(n),n=1..100) ; # R. J. Mathar, May 25 2023

a[0] = 0; a[1] = 1; a[n_] := a[n] = a[n - 1] + 2a[n - 2]; Select[ Range[500], IntegerQ[a[ # - 1]/# ] && # != 1 & ]

is(n)=Mod(2,3*n)^(n-1) == 1 \\ Charles R Greathouse IV, Nov 04 2016

More terms from Robert G. Wilson v, Jan 03 2002

cp's OEIS Frontend

A045468 Primes congruent to {1, 4} mod 5.

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

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A032357 Convolution of Catalan numbers and powers of -1.

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A069107 Composite numbers k that divide Fibonacci(k+1).

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A122869 Primes p that divide Lucas((p-1)/2), where Lucas is A000032.

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PARI

A069106 Composite numbers k such that k divides F(k-1) where F(j) are the Fibonacci numbers.

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C

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PARI

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A119996 Numerator of Sum_{k=1..n} 1/(Fibonacci(k)*Fibonacci(k+2)).

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