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A228085
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a(n) = number of distinct k which satisfy n = k + wt(k), where wt(k) (A000120) gives the number of 1's in binary representation of a nonnegative integer k.
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23
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1, 0, 1, 1, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 2, 1, 1, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 2, 1, 1, 2, 0, 1, 1, 1, 1, 1, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0
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OFFSET
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0,6
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COMMENTS
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wt(k) is also called bitcount(k).
a(n) = number of times n occurs in A092391.
The first 2 occurs at n=5 (as we have two solutions A092391(3) = A092391(4) = 5).
The first 3 occurs at n=129 (as we have three solutions A092391(123) = A092391(124) = A092391(128) = 129).
A number with five inverses was found by Donovan Johnson, Oct 19 2013, namely 2^136 + 6, which has inverses 2^136 - 129, 2^136 - 125, 2^136 - 124, 2^136 + 3, 2^136 + 4. - N. J. A. Sloane, Oct 20 2013.
I wrote a new program that is more efficient than the previous one I used. The new program only checks the last 20 bits for each inverse because when the inverse is < 2^m, all of the most significant bits are 1's. When the inverse is >= 2^m, the most significant bits are a 1 followed by all 0's.
Here is the smallest I found:
5 inverses: 2^136 + 6 (same as what I sent previously)
6 inverses: 2^260 + 130
7 inverses: 2^4233 + 130
8 inverses: 2^8206 + 4103
I have checked m values (exponents) up to 10^6 and did not find a solution for 9 inverses. (End)
For n>=1, a(2^n) = a(n-1) since an integer k = m is a solution to n-1 = m + wt(m) if and only if k = 2^n - 1 - m is a solution to 2^n = k + wt(k). - Max Alekseyev, Feb 23 2021
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LINKS
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Max A. Alekseyev and N. J. A. Sloane, On Kaprekar's Junction Numbers, arXiv:2112.14365, 2021; Journal of Combinatorics and Number Theory, 2022 (to appear).
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MAPLE
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# Find all inverses of m under x -> x + wt(x) - N. J. A. Sloane, Oct 19 2013
A000120 := proc(n) local w, m, i; w := 0; m := n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120;
F:=proc(m) local ans, lb, n, i;
lb:=m-ceil(log(m+1)/log(2)); ans:=[];
for n from max(1, lb) to m do if (n+wt(n)) = m then ans:=[op(ans), n]; fi; od:
[seq(ans[i], i=1..nops(ans))];
end;
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MATHEMATICA
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nmax = 8191; Clear[a]; a[_] = 0;
Scan[Set[a[#[[1]]], #[[2]]]&, Tally[Table[n + DigitCount[n, 2, 1], {n, 0, nmax}]]];
a[n_] := Module[{k, cnt = 0}, For[k = n - Floor[Log[2, n]] - 1, k < n, k++, If[n == k + DigitCount[k, 2, 1], cnt++]]; cnt];
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PROG
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(Haskell)
a228085 n = length $ filter ((== n) . a092391) [n - a070939 n .. n]
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CROSSREFS
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A010061 gives the position of zeros, A228082 the positions of nonzeros, A228088 the positions of ones. Cf. A000120, A010062, A092391, A228086, A228087, A228091 (positions of 2's), A227643, A230058, A230092 (positions of 3's), A230093, A227915 (positions of 4's), A070939.
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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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