A247937 -id:A247937 - cp's OEIS frontend

A248593 Least positive integer m such that m + n divides F(m), where F(m) is the m-th Fibonacci number given by A000045.

10, 6, 84, 12, 16, 7, 27, 9, 144, 30, 28, 12, 8, 30, 14, 18, 57, 19, 342, 18, 20, 24, 66, 12, 9, 27, 144, 60, 112, 35, 16, 24, 60, 55, 20, 12, 40, 111, 24, 36, 88, 72, 80, 48, 10, 15, 72, 24, 224, 18, 50, 54, 270, 72, 54, 33, 224, 18, 28, 12

Offset: 1

Zhi-Wei Sun, Oct 09 2014

nonn

Conjecture: a(n) exists for any n > 0. Moreover, a(n) <= n*(n-1) except for n = 1, 2, 3, 9.
In contrast, it is easy to show that for any integer n > 0, there is a positive integer m such that m + n divides 2^m - 1.
a(n) exists for any n > 0. See Bloom (1998). - Amiram Eldar, Jan 15 2022

			a(1) = 10 since 10 + 1 = 11 divides F(10) = 55.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
David M. Bloom, Offset Entries, Solution to Problem B-830, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 36, No. 1 (1998), pp. 89-90.

Cf. A000045, A247937, A247940, A248588, A248590.

Do[m=1;Label[aa];If[Mod[Fibonacci[m],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]

A297573 Least positive integer m such that m*n divides F(m+n), where F(k) denotes the k-th Fibonacci number A000045(k).

Original entry on oeis.org

1, 1, 1, 2, 14170, 6, 1, 136, 207, 28340, 979, 12, 1, 322, 385, 368, 1, 306, 17, 19780, 3, 68, 1, 24, 524975, 58, 2889, 92, 13, 3570, 12749, 736, 7, 2, 165, 612, 1, 34, 633, 13160, 339, 6, 1, 1846, 5355, 2, 1, 336, 8183, 509950, 21, 116, 1, 918, 4895, 184, 51, 26, 10207, 7140

Offset: 1

Zhi-Wei Sun, Jan 01 2018

nonn

If p is a prime congruent to 2 or 3 modulo 5, then a(p) = 1 since it is known that p divides F(p+1).
Conjecture: a(n) exists for any n > 0.
See also A297574 for a similar conjecture.

			a(2) = 1 since 1*2 divides F(1+2) = F(3) = 2.
a(4) = 2 since 2*4 divides F(2+4) = 8.
a(5) = 14170 since 5*14170 = 70850 divides F(5+14170) = F(14175).
a(6) = 6 since 6*6 = 36 divides F(6+6) = F(12) = 144.

Chai Wah Wu, Table of n, a(n) for n = 1..1000 (n = 1..120 from Zhi-Wei Sun)
Zhi-Wei Sun, Introduction to Lucas sequences, a talk give in 2017.

Cf. A000045, A247937, A248123, A297574.

Do[m=1; Label[aa]; If[Mod[Fibonacci[m+n], m*n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 1, 60}]
lpi[n_]:=Module[{k=1},While[!Divisible[Fibonacci[k+n],k*n],k++];k]; Array[ lpi,60] (* Harvey P. Dale, May 05 2018 *)

a(n) = my(m=1); while(1, if(Mod(fibonacci(m+n), m*n)==0, return(m)); m++) \\ Felix Fröhlich, Jan 01 2018

A297574 Least integer m > n such that 2^m == 2^n (mod m*n).

Original entry on oeis.org

2, 6, 15, 6, 65, 8, 16, 12, 63, 30, 31, 16, 85, 26, 39, 20, 65, 72, 73, 24, 57, 32, 56, 32, 1025, 170, 513, 40, 85, 42, 91, 40, 93, 130, 155, 144, 73, 56, 111, 48, 341, 48, 127, 64, 585, 112, 2048, 60, 2107, 550, 195, 64, 157, 1026, 155, 80, 219, 86, 233, 64, 1261, 82, 171, 73, 257, 96, 595, 140, 201, 130, 281, 126

Offset: 1

Zhi-Wei Sun, Jan 01 2018

nonn

That a(n) exists for any n > 0 follows from the following theorem.
Theorem: For any integer a and positive integer n, there are infinitely many positive integers m such that a^m == a^n (mod m*n).
Proof. This is obvious for a = 0, 1, -1. Below we assume |a| > 1. Let v be the largest divisor of n coprime to a, and write n = u*v. By Dirichlet's theorem, there are infinitely many primes q > max{|a|,v} such that q == 1 (mod phi(v)), where phi(.) is Euler's totient function. Note that q, u and v are pairwise coprime. Set m = n*q. Then m*n = q*u^2*v^2. For any prime divisor p of a, clearly ord_p(a^m-a^n) >= n >= 2*ord_p(n) since p^n >= n^2 except for the case p = 2 and n = 3. So u^2 divides a^m-a^n. As q does not divide a, by Fermat's little theorem we have a^m-a^n = a^n*(a^{(q-1)n}-1) == 0 (mod q). As v is coprime to a, and phi(v^2) = v*phi(v) divides (q-1)*n = m-n, by Euler's theorem we have a^m == a^n (mod v^2). Combining the above we see that m*n = q*u^2*v^2 divides a^m-a^n. This ends the proof.
Conjecture: Let A and B be integers with A^2 not equal to 4*B. Let u(0) = 0, u(1) = 1, and u(n+1) = A*u(n) - B*u(n-1) for n > 0. Also, let v(0) = 2, v(1) = A, and v(n+1) = A*v(n) - B*v(n-1) for n > 0. Then, for any integer n > 0, there are infinitely many positive integers m such that u(m) == u(n) (mod m*n). Also, for any integer n > 0, there are infinitely many positive integers m such that v(m) == v(n) (mod m*n).
See also A297573 for a similar conjecture involving the Fibonacci sequence.

			a(1) = 2 since 2^2 - 2^1 = 2*1.
a(2) = 6 since 2^6 - 2^2 = 60 = 5*(2*6).
a(3) = 15 since 2^15 - 2^3 = 32760 = 728*(3*15).
a(4) = 6 since 2^6 - 2^4 = 48 = 2*(4*6).

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (n = 1..1000 from Zhi-Wei Sun)

Cf. A000032, A000045, A000079, A247937, A297573.

Do[m=n+1; Label[aa]; If[Mod[2^m-2^n, m*n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 1, 80}]

a(n) = my(m=n+1); while(1, if(Mod(2, m*n)^m==Mod(2, m*n)^n, return(m)); m++) \\ Felix Fröhlich, Jan 01 2018

def A297574(n):
    m = n+1
    mn = m*n
    while pow(2,m,mn) != pow(2,n,mn):
        m += 1
        mn += n
    return m # Chai Wah Wu, Jan 04 2018

A248628 Least positive integer m such that prime(m+n) divides the m-th Fibonacci number F(m) given by A000045.

Original entry on oeis.org

183, 22, 15, 56, 42, 320, 138, 852, 93, 90, 19, 39, 11, 2100, 956, 140, 921, 341, 44, 78, 644, 13160, 212, 22, 26, 855, 333330, 815, 1032, 221, 400, 28, 1188, 49897, 296, 369, 731, 13, 680, 42, 144, 651, 46, 4105, 626, 52, 204, 5529, 310, 525, 4557, 441, 128, 9768, 102, 106, 168, 442

Offset: 1

Zhi-Wei Sun, Oct 10 2014

nonn

			a(2) = 22 since prime(22+2) = 89 divides F(22) = 17711 = 89*199.

Chai Wah Wu, Table of n, a(n) for n = 1..718

Cf. A000040, A000045, A247937, A248593, A248626.

Do[m=1;Label[aa];If[Mod[Fibonacci[m],Prime[m+n]]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,58}]

A248033 Least number k such that k + n divides F(n)^2 + F(k)^2.

Original entry on oeis.org

1, 3, 2, 1, 5, 19, 18, 17, 16, 3, 14, 12, 4, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 25, 99, 98, 9, 36, 95, 94, 93, 92, 91, 90, 12, 88, 23, 50, 10, 68, 83, 22, 81, 80, 79, 11, 24, 76, 15, 74, 73, 5, 71, 18, 29, 8, 67, 66, 12, 28, 63, 62, 1, 20, 47, 58, 41, 56, 40, 54, 17, 24, 11, 50, 49, 48

Offset: 1

Derek Orr, Sep 29 2014

nonn

F(n) denotes the n-th Fibonacci number. I believe a(n) exists for all n > 0.

Cf. A000045, A247937, A248032.

a(n)=m=1;while((fibonacci(n)^2+fibonacci(m)^2)%(m+n),m++);m
vector(100,n,a(n))

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A248593 Least positive integer m such that m + n divides F(m), where F(m) is the m-th Fibonacci number given by A000045.

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A297573 Least positive integer m such that m*n divides F(m+n), where F(k) denotes the k-th Fibonacci number A000045(k).

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A297574 Least integer m > n such that 2^m == 2^n (mod m*n).

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A248628 Least positive integer m such that prime(m+n) divides the m-th Fibonacci number F(m) given by A000045.

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A248033 Least number k such that k + n divides F(n)^2 + F(k)^2.

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