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A257499 Array A read by upward antidiagonals in which the entry in row n and column k is defined by A(n,k) = (1 + 2^n*(6*k-3+2*(-1)^n))/3, n,k >= 1.

%I A257499 #38 Dec 23 2024 14:53:44
%S A257499 1,7,5,3,15,9,27,19,23,13,11,59,35,31,17,107,75,91,51,39,21,43,235,
%T A257499 139,123,67,47,25,427,299,363,203,155,83,55,29,171,939,555,491,267,
%U A257499 187,99,63,33,1707,1195,1451,811,619,331,219,115,71,37
%N A257499 Array A read by upward antidiagonals in which the entry in row n and column k is defined by A(n,k) = (1 + 2^n*(6*k-3+2*(-1)^n))/3, n,k >= 1.
%C A257499 Conjecture (now Lemma 1): The sequence is a permutation of the odd natural numbers.
%C A257499 Proof from _Max Alekseyev_, Apr 29 2015:
%C A257499 Reformulating the conjecture, we need to prove that for any integer m >= 0, the equation (1 + 2^n*(6*k - 3 + 2*(-1)^n))/3 = 2*m + 1 has a unique solution in integers n,k >= 1. Simplifying a bit, we have
%C A257499 (1)  2^n*(6*k - 3 + 2*(-1)^n) = 6*m + 2.
%C A257499 Since the factor (6*k - 3 + 2*(-1)^n) is odd, n is uniquely defined by n = A007814(6*m+2). Since 6*m+2 is even, we have n>=1. Dividing (1) by 2^n and rearranging, we further get
%C A257499 (2)  6*k = (6*m + 2)/2^n + 3 - 2*(-1)^n.
%C A257499 To prove the uniqueness of k, it remains to prove that the r.h.s of (2) is divisible by 6. To that end, the value of n implies that (6*m + 2)/2^n is odd; hence the r.h.s. of (2) is even and thus divisible by 2. Now, taking the r.h.s. modulo 3, we get
%C A257499 (6*m + 2)/2^n + 3 - 2*(-1)^n == 2/(-1)^n + 0 - 2*(-1)^n == 0 (mod 3);
%C A257499 so the r.h.s. of (2) is also divisible by 3. Therefore k is uniquely defined by
%C A257499 k = ((6*m + 2)/2^n + 3 - 2*(-1)^n)/6.
%C A257499 Finally, it is easy to see that (6*m + 2)/2^n >= 1, so k >= 1.
%C A257499 QED
%C A257499 Let v(y) denote the 2-adic valuation of y (see A007814). For x an odd natural number, define the function F(x) = (3*x+1)/2^v(3*x+1) (see A075677). Let F^(j)(x) denote k-fold iteration of F and defined by the recurrence F^(j)(x) = F(F^(j-1)(x)), j>0, with initial condition F^(0)(x) = x.
%C A257499 Lemma 2: The following statements are equivalent. (i) Row n of A is the set of all odd m such that F^(n)(4*m-3) == 1 (mod 4); (ii) Row n of A is the set of all odd m such that v(1+F(4m-3)) = n.
%H A257499 Max Alekseyev, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2015-May/014799.html">Proof of conjecture in A257499</a>, Sequence fanatics mailing list, April 29 and May 01, 2015
%e A257499 Array A begins:
%e A257499 .       1     5     9    13    17     21     25     29     33     37
%e A257499 .       7    15    23    31    39     47     55     63     71     79
%e A257499 .       3    19    35    51    67     83     99    115    131    147
%e A257499 .      27    59    91   123   155    187    219    251    283    315
%e A257499 .      11    75   139   203   267    331    395    459    523    587
%e A257499 .     107   235   363   491   619    747    875   1003   1131   1259
%e A257499 .      43   299   555   811  1067   1323   1579   1835   2091   2347
%e A257499 .     427   939  1451  1963  2475   2987   3499   4011   4523   5035
%e A257499 .     171  1195  2219  3243  4267   5291   6315   7339   8363   9387
%e A257499 .    1707  3755  5803  7851  9899  11947  13995  16043  18091  20139
%t A257499 (* Array: *)
%t A257499 Grid[Table[(1 + 2^n*(6*k - 3 + 2*(-1)^n))/3, {n, 10}, {k, 10}]]
%t A257499 (* Array antidiagonals flattened: *)
%t A257499 Flatten[Table[(1 + 2^(n - k + 1)*(6*k - 3 + 2*(-1)^(n - k + 1)))/3, {n, 10}, {k, n}]]
%Y A257499 Cf. A007814, A075677, A254067, A257480.
%Y A257499 Cf. A255138 (column 1).
%K A257499 nonn,tabl
%O A257499 1,2
%A A257499 _L. Edson Jeffery_, Apr 27 2015