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Symmetric array T(n, k), n, k >= 0, read by antidiagonals; for any m >= 0 with binary expansion Sum_{i >= 0} b_i*2^i, let d(m) = Sum_{i >= 0} b_i * 2^A130472(i); let t be the inverse of d; T(n, k) = t(d(n) + d(k)).
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%I #11 Feb 23 2022 10:51:10

%S 0,1,1,2,4,2,3,3,3,3,4,6,1,6,4,5,5,4,4,5,5,6,16,6,5,6,16,6,7,7,7,7,7,

%T 7,7,7,8,18,5,18,16,18,5,18,8,9,9,16,16,17,17,16,16,9,9,10,12,10,17,

%U 18,20,18,17,10,12,10,11,11,11,11,19,19,19,19,11,11,11,11

%N Symmetric array T(n, k), n, k >= 0, read by antidiagonals; for any m >= 0 with binary expansion Sum_{i >= 0} b_i*2^i, let d(m) = Sum_{i >= 0} b_i * 2^A130472(i); let t be the inverse of d; T(n, k) = t(d(n) + d(k)).

%C The function d is a bijection from the nonnegative integers to the nonnegative dyadic rationals satisfying d(A000695(n)) = n for any n >= 0.

%H Rémy Sigrist, <a href="/A351785/b351785.txt">Table of n, a(n) for n = 0..10010</a>

%H Rémy Sigrist, <a href="/A351785/a351785.png">Colored representation of the table for n, k < 2^10</a> (where the hue is function of T(n, k))

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Dyadic_rational">Dyadic rational</a>

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%F T(A000695(n), A000695(k)) = A000695(n + k).

%F T(n, k) = T(k, n).

%F T(m, T(n, k)) = T(T(m, n), k).

%F T(n, 0) = n.

%e Array T(n, k) begins:

%e n\k| 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

%e ---- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --

%e 0| 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

%e 1| 1 4 3 6 5 16 7 18 9 12 11 14 13 24 15 26

%e 2| 2 3 1 4 6 7 5 16 10 11 9 12 14 15 13 24

%e 3| 3 6 4 5 7 18 16 17 11 14 12 13 15 26 24 25

%e 4| 4 5 6 7 16 17 18 19 12 13 14 15 24 25 26 27

%e 5| 5 16 7 18 17 20 19 22 13 24 15 26 25 28 27 30

%e 6| 6 7 5 16 18 19 17 20 14 15 13 24 26 27 25 28

%e 7| 7 18 16 17 19 22 20 21 15 26 24 25 27 30 28 29

%e 8| 8 9 10 11 12 13 14 15 2 3 1 4 6 7 5 16

%e 9| 9 12 11 14 13 24 15 26 3 6 4 5 7 18 16 17

%e 10| 10 11 9 12 14 15 13 24 1 4 3 6 5 16 7 18

%o (PARI) d(n) = { my (v=0, k); while (n, n-=2^k=valuation(n, 2); v+=2^((-1)^k*(k+1)\2)); v }

%o t(n) = { my (v=0, k); while (n, n-=2^k=valuation(n, 2); v+=2^if (k>=0, 2*k, -1-2*k)); v }

%o T(n,k) = t(d(n)+d(k))

%Y Cf. A000695, A130472, A351705, A351706, A351786 (multiplication).

%K nonn,base,tabl

%O 0,4

%A _Rémy Sigrist_, Feb 19 2022