A246259 Square array: row n contains in ascending order all natural numbers k for which A246271(k)+1 = n, i.e., numbers m for which we need n-1 additional iterations of A003961, starting from A003961(m), before the result is 1 modulo 4.
1, 3, 2, 4, 7, 5, 9, 8, 6, 66, 10, 15, 17, 70, 91, 11, 18, 20, 94, 197, 55, 12, 19, 24, 186, 259, 155, 21, 13, 22, 26, 187, 364, 220, 84, 46, 14, 28, 41, 199, 377, 238, 87, 184, 1362, 16, 29, 45, 237, 413, 467, 189, 414, 1981, 1654, 23, 32, 54, 262, 479, 495, 309, 445, 2378, 3055, 1419
Offset: 1
Examples
The top left corner of the array: 1, 3, 4, 9, 10, 11, 12, 13, 14, 16, 23, 25, ... 2, 7, 8, 15, 18, 19, 22, 28, 29, 32, 43, 50, ... 5, 6, 17, 20, 24, 26, 41, 45, 54, 57, 61, 68, ... 66, 70, 94, 186, 187, 199, 237, 262, 264, 278, 280, 286, ... 91 197, 259, 364, 377, 413, 479, 627, 665, 669, 705, 763, ... 55, 155, 220, 238, 467, 495, 497, 526, 535, 543, 620, 880, ... 21, 84, 87, 189, 309, 336, 348, 358, 463, 525, 679, 756, ... ... 2 is the least number k such that A003961(k) = 3 modulo 4, but A003961(A003961(k)) = 1 modulo 4 (as indeed A003961(2) = 3, and A003961(3) = 5). Thus A(2,1) = 2. 7 is the second smallest number k satisfying the same property, as A003961(7) = 11 (= 3 mod 4) while A003961(11) = 13 (= 1 mod 4). Thus A(2,2) = 7. 8 is the third smallest number k satisfying the same property, as A003961(8)=27 ( = 3 mod 4) while A003961(27) = 125 = 1 mod 4. Thus A(2,3) = 8. 5 is the least number k such that both A003961(k) and A003961(A003961(k)) = 3 mod 4 but A003961(A003961(A003961(k))) = 1 mod 4. Indeed A003961(5) = 7, A003961(7) = 11 and only at A003961(11) = 13 = 1 mod 4. Thus A(3,1) = 5.
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