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A110765
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Let n in binary be a k-digit number say abbaaa... where a = 1 and b = 0. a(n) = 2^a*3^b*5^b*7*a... primes in increasing order raised to the powers starting from the MSB.
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4
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2, 2, 6, 2, 10, 6, 30, 2, 14, 10, 70, 6, 42, 30, 210, 2, 22, 14, 154, 10, 110, 70, 770, 6, 66, 42, 462, 30, 330, 210, 2310, 2, 26, 22, 286, 14, 182, 154, 2002, 10, 130, 110, 1430, 70, 910, 770, 10010, 6, 78, 66, 858, 42, 546, 462, 6006, 30, 390, 330, 4290, 210, 2730
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OFFSET
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1,1
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COMMENTS
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All terms have 2-adic valuation equal to 1, i.e., they equal twice an odd (and squarefree) number, since the first digit in base two will always be "1". - M. F. Hasler, Mar 25 2011
2 appears at index n = 2^k for k >= 0, since such n_2 begins with "1" followed by k zeros, and 2^1 * 3^0 * ... * p_(k+1)^0 = 2. - Michael De Vlieger, Feb 28 2021
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LINKS
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EXAMPLE
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a(7) = 2*3*5 = 30. binary 7 = 111,
a(10) = 2^1*3^0*5^1*7^0 =10, binary(10) = 1010.
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MATHEMATICA
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Array[Times @@ Prime@ Flatten@ Position[#, 1] &@ IntegerDigits[#, 2] &, 61] (* Michael De Vlieger, Feb 28 2021 *)
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PROG
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(PARI) a(n)=factorback(Mat(vector(#n=binary(n), j, [prime(j), n[j]])~))
(PARI) a(n)=prod(j=1, #n=binary(n), prime(j)^n[j]) \\ M. F. Hasler, Mar 25 2011
(Haskell)
a110765 = product . zipWith (^) a000040_list . reverse . a030308_row
(Python)
from sympy import prime
from operator import mul
from functools import reduce
....return reduce(mul, (prime(i) for i, d in enumerate(bin(n)[2:], start=1) if int(d)))
# implementation using recursion
....nlen = len(n)
....return _A110765(n[:-1])*(prime(nlen) if int(n[-1]) else 1) if nlen > 1 else int(n) + 1
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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More terms from Stacy Hawthorne (shawtho1(AT)ashland.edu), Oct 31 2005
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STATUS
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approved
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