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%I #28 Jul 17 2017 02:17:48
%S 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,
%T 259,155,21,13,22,26,187,364,220,84,46,14,28,41,199,377,238,87,184,
%U 1362,16,29,45,237,413,467,189,414,1981,1654,23,32,54,262,479,495,309,445,2378,3055,1419
%N 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.
%C The array is read by antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.
%C The topmost row, which has row number 1, is the sequence A246261 (numbers n such that A003961(n) is of the form 4k+1), as these are exactly the numbers that do not require any additional iterations of A003961 for the result to be of the form 4k+1 (because it already is of that form).
%C If a number k occurs in any particular row, then 4k, 9k, 16k, etc. also occur on the same row, because multiplying a number by a perfect square does not affect the result when it is reduced modulo 4.
%C The array has an infinite number of rows, provided that A246271 is not bounded, because if A246271(n) = k > 0 for some n, then A246271(A003961(n)) = k-1, thus all the preceding rows would also have terms. However, if A246271 really had an absolute maximum m, then we can consider the array to have just m rows. In either case, all the natural numbers occur in it, each just once, thus the sequence forms a permutation of natural numbers.
%C Applying A003961 to the terms of any row below the first row gives a subsequence of the immediately preceding row.
%H Antti Karttunen, <a href="/A246259/b246259.txt">Table of n, a(n) for n = 1..120; the first 15 antidiagonals of the array</a>
%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%e The top left corner of the array:
%e 1, 3, 4, 9, 10, 11, 12, 13, 14, 16, 23, 25, ...
%e 2, 7, 8, 15, 18, 19, 22, 28, 29, 32, 43, 50, ...
%e 5, 6, 17, 20, 24, 26, 41, 45, 54, 57, 61, 68, ...
%e 66, 70, 94, 186, 187, 199, 237, 262, 264, 278, 280, 286, ...
%e 91 197, 259, 364, 377, 413, 479, 627, 665, 669, 705, 763, ...
%e 55, 155, 220, 238, 467, 495, 497, 526, 535, 543, 620, 880, ...
%e 21, 84, 87, 189, 309, 336, 348, 358, 463, 525, 679, 756, ...
%e ...
%e 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.
%e 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.
%e 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.
%e 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.
%o (Scheme, with _Antti Karttunen_'s IntSeq-library)
%o (define (A246259 n) (A246259bi (A002260 n) (A004736 n)))
%o (define (A246259bi row col) ((rowfun-for-A246259 row) col))
%o (definec (rowfun-for-A246259 row) (MATCHING-POS 1 1 (lambda (k) (= (- row 1) (A246271 k)))))
%Y Transpose: A246258.
%Y First row: A246261.
%Y First column: A246280.
%Y Cf. A246278, A246271, A003961, A002260, A004736.
%K nonn,tabl
%O 1,2
%A _Antti Karttunen_, Aug 22 2014
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