A345297 a(n) is the least k >= 0 such that A331835(k) = n.

0, 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 22, 23, 26, 27, 29, 30, 31, 43, 45, 46, 47, 54, 55, 58, 59, 61, 62, 63, 94, 95, 107, 109, 110, 111, 118, 119, 122, 123, 125, 126, 127, 187, 189, 190, 191, 222, 223, 235, 237, 238, 239, 246, 247, 250, 251, 253, 254, 255

Offset

0,3

Comments

Sequence A200947 gives the position of the last occurrence of a number in A331835.

Links

Rémy Sigrist, Table of n, a(n) for n = 0..2000

Rémy Sigrist, C program for A345297

Formula

a(A014284(n)) = 2^n - 1.

a(n) <= A200947(n).

Example

We have:
           n|  0  1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18
  ----------+------------------------------------------------------------------
  A331835(n)|  0  1  2  3  3  4  5  6  5  6   7   8   8   9  10  11   7   8   9
So a(0) = 0,
   a(1) = 1,
   a(2) = 2,
   a(3) = 3,
   a(4) = 5,
   a(5) = 6,
   a(6) = 7,
   a(7) = 10,
   a(8) = 11,
   a(9) = 13,
   a(10) = 14,
   a(11) = 15.

Prog

(C) See Links section.
(Python)
from sympy import prime
def p(n): return prime(n) if n >= 1 else 1
def A331835(n): return sum(p(i)*int(b) for i, b in enumerate(bin(n)[:1:-1]))
def adict(klimit):
    adict = dict()
    for k in range(klimit+1):
        fk = A331835(k)
        if fk not in adict: adict[fk] = k
    n, alst = 0, []
    while n in adict: alst.append(adict[n]); n += 1
    return alst
print(adict(255)) # Michael S. Branicky, Jun 13 2021

Crossrefs

Cf. A014284, A200947, A331835.

Sequence in context: A293427 A293430 A087006 * A336533 A359782 A235991

Adjacent sequences: A345294 A345295 A345296 * A345298 A345299 A345300

Keyword

nonn,base

Author

Rémy Sigrist, Jun 13 2021

Status

approved