A381583 Starts of runs of 3 consecutive integers that are all terms in A381581.

1, 2, 20, 55, 56, 110, 304, 364, 398, 846, 1024, 1084, 1744, 1854, 2044, 2104, 2105, 2527, 2824, 2862, 3870, 4374, 5222, 5223, 5243, 5718, 5928, 6488, 6784, 6844, 6894, 6978, 7142, 7924, 10590, 11240, 11889, 11975, 12248, 14284, 14915, 16638, 17710, 17714, 17824

Offset: 1

Author

Amiram Eldar, Feb 28 2025

Keywords

• nonn
• easy
• base

Comments

If k is congruent to 1 or 5 mod 12 (A087445), then A001906(k) = Fibonacci(2*k) is a term.

Examples

			1 is a term since A291711(1) = 1 divides 1, A291711(2) = 2 divides 2, and A291711(3) = 1 divides 3.
20 is a term since A291711(20) = 4 divides 20, A291711(21) = 1 divides 21, and A291711(22) = 2 divides 22.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000045, A001906, A087445, A291711.

Subsequence of A381581 and A381582.

Subsequences: A381584, A381585.

Similar sequences: A154701, A328210, A330932, A351721.

Programs

Mathematica

f[n_] := f[n] = Fibonacci[2*n]; q[n_] := q[n] = Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2; m -= 2*f[k], s++; m -= f[k]]]; Divisible[n, s]]; seq[count_, nConsec_] := Module[{cn = q /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {q[k]}]; k++]; s]; seq[45, 3]

PARI

mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
is1(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}
list(lim) = {my(q1 = is1(1), q2 = is1(2), q3); for(k = 3, lim, q3 = is1(k); if(q1 && q2 && q3, print1(k-2, ", ")); q1 = q2; q2 = q3);}