A308669 a(1) = 1, a(2) = 3; for n > 1, a(n) = sum of the next two smallest integers > a(n-1) which are coprime to the sum s = a(1) + ... + a(n-1).

1, 3, 12, 28, 60, 124, 252, 510, 1032, 2070, 4146, 8298, 16602, 33210, 66426, 132858, 265722, 531450, 1062906, 2125818, 4251642, 8503290, 17006588, 34013180, 68026364, 136052736, 272105480, 544210964, 1088421932, 2176843872, 4353687750, 8707375506, 17414751018, 34829502042, 69659004090

Offset

1,2

Links

Petros Hadjicostas, Table of n, a(n) for n = 1..500

Formula

Conjecture 1: a(n) ~ 2^(n+1) * 1.0136719224974731811...

Conjecture 2: There are infinitely many n >= 1 such that the ratio (a(n+1) - A131357(n+1))/(a(n) - A131357(n)) equals exactly 2 (for example, for n = 10, 11, 12, 13, 18, 19, 20, 21, 24, 26, etc.).

Example

n = 3: a(1) = 1, a(2) = 3, s = 1 + 3 = 4; n1 = 5 and n2 = 7 (not 6) are both coprime to s, hence a(3) = n1 + n2 = 12;
n = 4: a(1) = 1, a(2) = 3, a(3) = 12, s = 1 + 3 + 12 = 16; n1 = 13 and n2 = 15 (not 14) are both coprime to s, hence a(4) = n1 + n2 = 28;
n = 5: s = 44; n1 = 29 and n2 = 31 are both coprime to s, hence a(5) = n1 + n2 = 60;
n = 6: s = 104; n1 = 61 and n2 = 63 are both comprime to s, hence a(6) = n1 + n2 = 124.

Prog

(PARI) findnext(last, s) = {my(x = last+1); while (gcd(x, s) != 1, x++); my(y = x+1); while (gcd(y, s) != 1, y++); x+y; }
lista(nn) = {my(v = [1, 3], val); print1(v[1], ", ", v[2], ", "); for (n=3, nn, val = findnext(v[#v], vecsum(v)); v = concat(v, val); print1(val, ", "); ); } \\ [Modification of Michel Marcus's PARI program from sequence A131357]

Crossrefs

Cf. A131357.

Sequence in context: A066643 A140065 A294418 * A115549 A005995 A351643

Adjacent sequences: A308666 A308667 A308668 * A308670 A308671 A308672

Keyword

nonn

Author

Petros Hadjicostas, Jun 25 2019

Status

approved