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A345297
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a(n) is the least k >= 0 such that A331835(k) = n.
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4
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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)
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OFFSET
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0,3
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COMMENTS
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Sequence A200947 gives the position of the last occurrence of a number in A331835.
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LINKS
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FORMULA
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EXAMPLE
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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.
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PROG
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(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):
if fk not in adict: adict[fk] = k
n, alst = 0, []
while n in adict: alst.append(adict[n]); n += 1
return alst
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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