Shane Findley - cp's OEIS frontend

A262708 a(n) = p-(p/5) where p = prime(n) and (p/5) is a Legendre symbol.

8, 10, 14, 18, 18, 24, 28, 30, 38, 40, 44, 48, 54, 58, 60, 68, 70, 74, 78, 84, 88, 98, 100, 104, 108, 108, 114, 128, 130, 138, 138, 148, 150, 158, 164, 168, 174, 178, 180, 190, 194, 198, 198, 210, 224, 228, 228, 234, 238, 240, 250, 258, 264, 268, 270, 278, 280

Offset: 4

Shane Findley, Sep 27 2015

nonn

The sequence lists Fibonacci indices q that are conjectured to produce Fibonacci numbers divisible by p^2, where p is a Fibonacci-Wieferich prime.

			For n=4, prime(4)=7, and a(4)=8.

Paulo Ribenboim, My Numbers, My Friends, Springer-Verlag, 2000.
Steven Vajda, Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications, Dover. (See p. 73.)

U. Alfred, On the form of primitive factors of Fibonacci numbers, Volume 1, Fibonacci Quarterly, vol 1 (1963), page 1.
Andreas-Stephan Elsenhans, The Fibonacci sequence modulo p^2., pages 1-6.
Shane Findley, Discussion of Sequence A262708.
Richard J. McIntosh and Eric L. Roettger, A search for Fibonacci-Wieferich and Wolstenholme primes, Mathematics of Computation, vol 76 (260), Oct 2007.
John Vinson, The Relation of the Period Modulo m to the Rank of Apparition of m in the Fibonacci Sequence, Fibonacci Quarterly, vol 1 (1963), pages 37-45.

Cf. A001602, A051694.

Table[Prime@ n - JacobiSymbol[Prime@ n, 5], {n, 4, 60}] (* Michael De Vlieger, Oct 04 2015 *)

lista(nn)=forprime(p=3, nn, print1(p-kronecker(p, 5), ", ");); \\ Michel Marcus, Sep 29 2015

Edited by N. J. A. Sloane, Sep 29 2015

Edited by Jon E. Schoenfield, Oct 09 2015

A172074 Continued fraction expansion of sqrt(12500)+50 = 100*phi, where phi=(sqrt(5)+1)/2 is the golden ratio.

Original entry on oeis.org

161, 1, 4, 11, 1, 1, 3, 6, 1, 13, 8, 1, 6, 1, 4, 1, 1, 2, 1, 1, 1, 1, 13, 2, 1, 3, 8, 1, 2, 19, 1, 54, 1, 19, 2, 1, 8, 3, 1, 2, 13, 1, 1, 1, 1, 2, 1, 1, 4, 1, 6, 1, 8, 13, 1, 6, 3, 1, 1, 11, 4, 1, 222

Offset: 0

Shane Findley, Jan 25 2010

The 62 trailing terms are repeated infinitely.
This is just one of an infinite set of continued fractions, related to the golden ratio, and more specifically to the square root of 125, 12500, 1250000...
Taking phi*10^k, one can look at sqrt(125) + 5, sqrt(12500) + 50 (this sequence), sqrt(1250000) + 500, etc.
This is not an efficient way to calculate phi. - Franklin T. Adams-Watters, Sep 10 2011
Periodic with a period of length 62, starting right after the initial term. Moreover, the sequence is symmetric when any 54 or 222 is taken as central value (cf. formula). - M. F. Hasler, Sep 09 2011

Cf. A001622, A010186.

ContinuedFraction[N[Sqrt[12500], 50000], 63]
ContinuedFraction[100*GoldenRatio,100] (* Harvey P. Dale, Dec 30 2018 *)

default(realprecision, 199); contfrac((sqrt(5)+1)/.02)  \\ M. F. Hasler, Sep 09 2011

a(n)=[222-61*!n, 1, 4, 11, 1, 1, 3, 6, 1, 13, 8, 1, 6, 1, 4, 1, 1, 2, 1, 1, 1, 1, 13, 2, 1, 3, 8, 1, 2, 19, 1, 54][32-abs(n%62-31)]  \\ M. F. Hasler, Sep 09 2011

a(31*k - n) = a(31*k + n), for all n < 31k, k > 0. - M. F. Hasler, Sep 09 2011

Clarified the definition, following an observation by Franklin T. Adams-Watters. M. F. Hasler, Sep 09 2011

A135500 Generating function for Viswanath's constant, using the golden string.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0

Offset: 0

Shane Findley, Feb 19 2008

For each bit of the golden string that is a 1, write seven consecutive bits and for each 0 of the golden string write eight consecutive bits.

Alternate between 0 and 1. Exclude the first seven bits since we are based at zero, just like the golden string. Heads = 1, Tails = 0.

			Bit 1 of the golden string is a 1, so bits 8-14(seven consecutive) are the same. Bit 2 of the golden string is 0, so bits 15-22(eight consecutive) are the same. Odd bits of the golden string mean to write 1 and even bits of the golden string mean to write 0.

S. Findley, Viswanath's curve

Cf. A036299, A078416, A115064.

Each power of Viswanath's constant (1.131988248...) And V^3 has a corresponding coin flip. Simply take the absolute values of the two possible(+Heads or -Tails) outcomes and choose the flip closest to the value of the power.

A115064 Continued fraction expansion of Viswanath's constant A078416.

Original entry on oeis.org

1, 7, 1, 1, 2, 1, 3, 2, 1, 2, 1, 8, 1, 5

Offset: 0

Shane Findley, Mar 01 2006

Cf. A078416 (decimal expansion).

Values from a(12) on replaced consistent with A078416; invalid comments removed - R. J. Mathar, Sep 09 2009

Offset changed by Andrew Howroyd, Aug 03 2024

A084924 Let t(x) be the highest power of 2 which divides x+1. Then a(1)=3; a(n) is the least prime p for which t(p) > t(a(n-1)).

Original entry on oeis.org

3, 7, 31, 127, 1279, 3583, 5119, 6143, 8191, 81919, 131071, 524287, 14680063, 109051903, 654311423, 738197503, 2147483647, 21474836479, 51539607551, 824633720831, 13743895347199, 26388279066623, 246290604621823

Offset: 1

Shane Findley, Jul 15 2003

nonn

			a(5)=1279 because t(a(4))=7 and 1279 is the least prime with t(p)>7.

M. F. Hasler, Table of n, a(n) for n = 1..100

a=vector(50); a[1]=3;for(i=2,length(a), j=k=2^(factor(a[i-1]+1,2)[1,2]+1); while(! isprime(j-1),j+=k);a[i]=j-1); a \\ M. F. Hasler, Mar 15 2007

Edited by Don Reble, May 08 2004

More terms from M. F. Hasler, Mar 15 2007

A074281 Primes of the form Lucas(2*n)/3.

Original entry on oeis.org

41, 281, 90481, 29134601, 3020733700601, 313195711516578281, 5280544535667472291277149119296546201, 547497418496144666543167613835090178297001

Offset: 1

Shane Findley, Sep 21 2002

nonn

The next term has 96 digits. - Harvey P. Dale, Apr 29 2011.

			Lucas(2*5)=3*41, Lucas(2*7)=3*281, Lucas(2*13)=3*90481.

Vincenzo Librandi, Table of n, a(n) for n = 1..16
Ron Knott, The Mathematics of the Fibonacci series.
Ron Knott, Pi and the Fibonacci Numbers.

Values of n are given in A074304.

Select[Table[LucasL[2n]/3,{n,400}],PrimeQ]  (* Harvey P. Dale, Apr 29 2011 *)

A064535 a(n) = (2^prime(n)-2)/prime(n); a(0) = 0 by convention.

Original entry on oeis.org

0, 1, 2, 6, 18, 186, 630, 7710, 27594, 364722, 18512790, 69273666, 3714566310, 53634713550, 204560302842, 2994414645858, 169947155749830, 9770521225481754, 37800705069076950, 2202596307308603178, 33256101992039755026, 129379903640264252430

Offset: 0

Shane Findley, Oct 09 2001

nonn

As a corollary to Fermat's little theorem, (2^p - 2)/p is always an integer for p prime. - Alonso del Arte, May 04 2013

			a(3) = 6, because prime(3) = 5, and (2^5 - 2)/5 = 30/5 = 6.
a(4) = 18, because prime(4) = 7, and (2^7  - 2)/7 = 126/7 = 18.

Harry J. Smith, Table of n, a(n) for n = 0..100

Cf. A007663, A056743, A225101 (superset).

[0] cat [(2^NthPrime(n)-2)/NthPrime(n): n in [1..25]]; // Vincenzo Librandi, Sep 14 2018

A064535 := proc(n) ( 2^ithprime(n) - 2 )/ithprime(n); end;

Table[(2^Prime[n] - 2)/Prime[n], {n, 50}] (* Alonso del Arte, Apr 28 2013 *)

{ for (n=0, 100, if (n, a=(2^prime(n) - 2)/prime(n), a=0); write("b064535.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 17 2009

a(n) = A001037(prime(n)) for n >= 1. - Hilko Koning, Sep 10 2018

a(n) = 2*A007663(n) for n > 1. - Jeppe Stig Nielsen, May 16 2021

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

A074304 Numbers k such that Lucas(2k)/3 is prime.

Original entry on oeis.org

5, 7, 13, 19, 31, 43, 89, 101, 229, 293, 457, 541, 653, 659, 1553, 2003, 2707, 2749, 7159, 10289, 26267, 59581, 63421, 80911

Offset: 1

Shane Findley, Sep 21 2002

Some of the larger entries may only correspond to probable primes.

Ron Knott, The Mathematics of the Fibonacci series.
Ron Knott, Pi and the Fibonacci Numbers.

Cf. A049685, A074281, A355980.

a(n) = 2*A355980(n) + 1. - Jinyuan Wang, Jul 22 2022

More terms found by David Broadhurst, Bouk de Water, and Shane Findley, Oct 31 2002

a(22)-a(24) from Michael S. Branicky, Nov 05 2024

A076697 Indices of record values in A079451, largest prime factor of Lucas numbers A000032.

Original entry on oeis.org

0, 2, 4, 5, 7, 8, 11, 13, 16, 17, 19, 26, 31, 37, 41, 47, 53, 61, 68, 71, 76, 79, 86, 113, 136, 164, 172, 178, 202, 218, 229, 262, 278, 284, 307, 313, 328, 353, 373, 436, 443, 458, 487, 503, 557, 577, 586, 613, 617, 746, 751, 758, 863, 914

Offset: 0

Shane Findley, Oct 25 2002

nonn

From  M. F. Hasler, Apr 09 2025: (Start)
Original name: Next-to-largest factor of Lucas(n).
The offset 0 is coherent with the fact that the initial term is a starting value rather than a record value.
When A000032(n) is prime (<=> n is in A001606), it necessarily sets a new record for the largest prime factor, since A000032 is increasing from the second term on. Therefore, A001606 is a subsequence. (End)

Ron Knott, Lucas numbers.
Douglas S. McNeil, in reply to Robert Israel, A076697, SeqFan google group, April 9, 2025.

Cf. A000042 (Lucas numbers, starting with 2), A079451 (largest prime factor of these).

Cf. A001606 (Indices of prime Lucas numbers: a subsequence).

A076697_first(n, m=0)=vector(n,i, i>1 || n=-1; until(mA079451(n++), m), );n) \\ M. F. Hasler, Apr 09 2025

def A076697(n):
    try: terms, M = A076697.terms, A076697.M
    except AttributeError: A076697.terms = terms = [0]; A076697.M = M = 2
    while len(terms) <= n: terms.append(next(i for i in range(terms[-1]+1, 1<<59)
        if M < (M:=max(A079451(i),M)))); A076697.M = M
    return terms[n] # M. F. Hasler, Apr 10 2025

New definition and data corrected and extended by M. F. Hasler, Apr 09 2025

cp's OEIS Frontend

User: Shane Findley

A262708 a(n) = p-(p/5) where p = prime(n) and (p/5) is a Legendre symbol.

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A172074 Continued fraction expansion of sqrt(12500)+50 = 100*phi, where phi=(sqrt(5)+1)/2 is the golden ratio.

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A135500 Generating function for Viswanath's constant, using the golden string.

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A115064 Continued fraction expansion of Viswanath's constant A078416.

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A084924 Let t(x) be the highest power of 2 which divides x+1. Then a(1)=3; a(n) is the least prime p for which t(p) > t(a(n-1)).

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A074281 Primes of the form Lucas(2*n)/3.

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A064535 a(n) = (2^prime(n)-2)/prime(n); a(0) = 0 by convention.

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

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