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%I #37 Oct 27 2021 22:21:27
%S 1,1,3,2,3,5,6,4,7,9,12,10,12,9,29,16,24,22,24,17,57,36,48,40,50,36,
%T 57,65,96,92,96,64,114,72,101,161,192,144,228,136,192,178,192,129,473,
%U 288,384,320,388,304,456,258,384,353,801,520,912,576,768,676,768,576,922,512,801,1409
%N a(n) is the number whose binary representation is the concatenation of terms in the n-th row of A237048.
%C The number of ones in the n-th row of A237048 equals A001227(n), the same as the number of ones in the binary representation of a(n).
%C The number of zeros in the n-th row of A237048 equals A238005(n), the same as the number of zeros in the binary representation of a(n).
%C The number of terms in the n-th row of A237048 equals A003056(n), the same as the number of digits in the binary representation of a(n).
%H Michael De Vlieger, <a href="/A347266/b347266.txt">Table of n, a(n) for n = 1..10000</a>
%H Michael De Vlieger, <a href="/A347266/a347266.png">Bitmap of the first 2^10 terms</a>, showing 1s in black, 12X horizontal exaggeration.
%H Michael De Vlieger, <a href="/A347266/a347266_1.png">Bitmap of the first 2^12 terms</a>, showing 1s in black, rotated 90 degrees counterclockwise. [Click "magnify" to see the graph more clearly.]
%e The 15th row of the triangle A237048 is [1, 1, 1, 0, 1] and the concatenation of these terms is 11101 which can be interpreted as a binary number whose decimal value is 29, so a(15) = 29.
%t Table[FromDigits[Boole[Divisible[n - If[OddQ[#], 0, Quotient[#, 2]], #]] & /@ Range[Quotient[Sqrt[8 n + 1] - 1, 2]], 2], {n, 66}] (* _Jan Mangaldan_, Sep 13 2021 *)
%o (PARI) t(n, k) = if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0);
%o a(n) = fromdigits(vector(floor((sqrt(1+8*n)-1)/2), k, t(n, k)), 2); \\ _Michel Marcus_, Sep 12 2021
%Y Cf. A003056, A007088, A001227, A237048, A237593, A238005, A347765 (binary representation).
%K nonn,base,look
%O 1,3
%A _Omar E. Pol_, Sep 06 2021
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