A381584 Starts of runs of 4 consecutive integers that are all terms in A381581.

1, 55, 2104, 5222, 24784, 63510, 64264, 69487, 95463, 121393, 184327, 327303, 374589, 463110, 468168, 561069, 572550, 596868, 671407, 740310, 759030, 819948, 902670, 956680, 1023009, 1036230, 1065030, 1259817, 1274910, 1359552, 1683154, 1714470, 1731750, 2182023

Offset: 1

Author

Amiram Eldar, Feb 28 2025

Keywords

• nonn
• 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, A291711(3) = 1 divides 3, and A291711(4) = 2 divides 4.
55 is a term since A291711(55) = 1 divides 55, A291711(56) = 2 divides 56, A291711(57) = 3 divides 57, and A291711(58) = 2 divides 58.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000045, A001906, A087445, A291711.

Subsequence of A381581, A381582 and A381583.

A381585 is a subsequence.

Similar sequences: A141769, A328211, A328215, A330933.

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[12, 4]

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 = is1(3), q4); for(k = 4, lim, q4 = is1(k); if(q1 && q2 && q3 && q4, print1(k-3, ", ")); q1 = q2; q2 = q3; q3 = q4);}