A342013 Position of the n-th colossally abundant number in A329886, the primorial inflation of Doudna-tree.

1, 2, 5, 9, 19, 21, 37, 75, 139, 267, 535, 539, 555, 1067, 2091, 4139, 8279, 16471, 32855, 32919, 32923, 65691, 131227, 262299, 524599, 1048887, 2097463, 4194615, 4194647, 8388951, 16777559, 33554775, 67109207, 67109463, 134218327, 268436655, 536872111, 536872119, 1073743031, 2147484855, 2147485879, 4294969527, 8589936823

Offset: 1

Author

Antti Karttunen and Michael De Vlieger, Mar 08 2021

Keywords

• nonn

Comments

a(n) is the unique integer k such that A329886(k) = A004490(n).

Like A342012, also this sequence is monotonic. Proof: the doubling step corresponds here to step *2 + 1, and "bumping up" some of the prime factors likewise results a larger A156552-code, thus both steps keep the result growing.

The binary length of these numbers (A070939, = 1+A000523) grows by 0 or 1 at each step, thus the next colossally abundant number is always found on either on the same row (right of the current CA-number), or the next row of A329886, the row immediately below. The next CA-number will be on the same row only when its factorization contains neither a new prime nor yet another instance of prime 2.

Links

• Index entries for sequences computed from indices in prime factorization
• Index entries for sequences related to sigma(n)

Crossrefs

Cf. A004490, A005940, A073751, A108951, A156552, A319626, A329886, A329900, A342000, A342010 (the binary weight), A342011, A342012.

Programs

PARI

A156552(n) = {my(f = factor(n), p, p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res};
A342013(n) = A156552(A342012(n)); \\ Uses also code from A342012.

Formula

a(n) = A156552(A342012(n)) = A156552(A319626(A004490(n))).