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

1,2

Comments

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

Links

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

Example

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.

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]

Prog

(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); }

Crossrefs

Cf. A000045, A001906, A087445, A291711.

Subsequence of A381581 and A381582.

Subsequences: A381584, A381585.

Similar sequences: A154701, A328210, A330932, A351721.

Sequence in context: A059211 A331090 A139271 * A133217 A001504 A192351

Adjacent sequences: A381580 A381581 A381582 * A381584 A381585

Keyword

nonn,easy,base,new

Author

Amiram Eldar, Feb 28 2025

Status

approved