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A345297 a(n) is the least k >= 0 such that A331835(k) = n. 2
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 (list; graph; refs; listen; history; text; internal format)
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 A235991 A327906

Adjacent sequences:  A345294 A345295 A345296 * A345298 A345299 A345300

KEYWORD

nonn,base

AUTHOR

Rémy Sigrist, Jun 13 2021

STATUS

approved

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Last modified September 24 05:29 EDT 2021. Contains 347623 sequences. (Running on oeis4.)