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A348853
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Delete any least significant 0's from the Zeckendorf representation of n, leaving its "odd" part.
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5
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1, 1, 1, 4, 1, 6, 4, 1, 9, 6, 4, 12, 1, 14, 9, 6, 17, 4, 19, 12, 1, 22, 14, 9, 25, 6, 27, 17, 4, 30, 19, 12, 33, 1, 35, 22, 14, 38, 9, 40, 25, 6, 43, 27, 17, 46, 4, 48, 30, 19, 51, 12, 53, 33, 1, 56, 35, 22, 59, 14, 61, 38, 9, 64, 40, 25, 67, 6, 69, 43, 27, 72
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OFFSET
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1,4
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COMMENTS
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Terms are odd Zeckendorfs A003622 and the fixed points are where n is odd already so that a(n) = n iff n is in A003622.
A139764(n) is the least significant "10..00" part of n so Zeckendorf multiplication n = A101646(a(n), A139764(n)).
a(n) = 1 iff n is a Fibonacci number >= 1 (A000045) since they are Zeckendorf 100..00.
a(n) = 4 iff n is a Lucas number >= 4 (A000032) since they are Zeckendorf 10100..00 which reduces to 101.
In the Wythoff array A035513, a(n) is the term in column 0 of the row containing n, and hence the formula below using row number A019586 to select which of the odds (column 0) is a(n).
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LINKS
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FORMULA
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Sum_{k=1..n} a(k) ~ n^2/(2*phi), where phi is the golden ratio (A001622). - Amiram Eldar, Feb 17 2024
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EXAMPLE
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n = 81 = Zeckendorf 101001000.
a(n) = 19 = Zeckendorf 101001.
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PROG
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(PARI) my(phi=quadgen(5)); a(n) = my(q, r); while([q, r]=divrem(n+2, phi); r<1, n=q-1); n;
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CROSSREFS
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KEYWORD
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base,easy,nonn
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AUTHOR
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STATUS
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approved
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