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%I #26 Feb 25 2023 03:10:44
%S 1,2,2,1,1,3,2,4,2,1,3,1,5,7,2,8,4,6,1,5,7,1,5,11,6,10,12,2,4,8,7,11,
%T 1,3,13,15,4,10,14,2,8,10,1,3,7,9,1,13,17,19,2,8,10,20,4,10,16,18,1,5,
%U 7,17,7,11,13,17,6,12,22,24,2,8,16,22,1,5,11
%N a(n) = number of the row of (R(i,j)) that contains prime(n), where R(i,j) is the rectangle with descending antidiagonals 1; 2,3; 4,5,6; ...
%C The rectangle has this corner:
%C 1, 2, 4, 7, 11, 16, 22, 29, ...
%C 3, 5, 8, 12, 17, 23, 30, 38, ...
%C 6, 9, 13, 18, 24, 31, 39, 48, ...
%C 10, 14, 19, 25, 32, 40, 49, 59, ...
%C 15, 20, 26, 33, 41, 50, 60, 71, ...
%C 21, 27, 34, 42, 51, 61, 72, 84, ...
%C 28, 35, 43, 52, 62, 73, 85, 98, ...
%F a(n) = A002260(prime(n)). - _Kevin Ryde_, Feb 12 2023
%e The 8th prime, 19, is in row 4, so a(8) = 4.
%t s = Flatten[Table[Range[n], {n, 1, 40}]];
%t Table[s[[Prime[n]]], {n, 1, 100}]
%o (PARI) f(n) = n-binomial((sqrtint(8*n)+1)\2, 2); \\ A002260
%o a(n) = f(prime(n)); \\ _Michel Marcus_, Feb 24 2023
%Y Cf. A000027, A000040, A002260, A185787.
%Y See A057061 for primes in columns.
%K nonn
%O 1,2
%A _Clark Kimberling_, Jul 30 2000
%E Edited by _Clark Kimberling_, Feb 13 2023
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