A297573 -id:A297573 - cp's OEIS frontend

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

cp's OEIS Frontend

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

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