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.

1, 7, 5, 3, 15, 9, 27, 19, 23, 13, 11, 59, 35, 31, 17, 107, 75, 91, 51, 39, 21, 43, 235, 139, 123, 67, 47, 25, 427, 299, 363, 203, 155, 83, 55, 29, 171, 939, 555, 491, 267, 187, 99, 63, 33, 1707, 1195, 1451, 811, 619, 331, 219, 115, 71, 37

Offset

1,2

Comments

Conjecture (now Lemma 1): The sequence is a permutation of the odd natural numbers.

Proof from Max Alekseyev, Apr 29 2015:

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

(1) 2^n*(6*k - 3 + 2*(-1)^n) = 6*m + 2.

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

(2) 6*k = (6*m + 2)/2^n + 3 - 2*(-1)^n.

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

(6*m + 2)/2^n + 3 - 2*(-1)^n == 2/(-1)^n + 0 - 2*(-1)^n == 0 (mod 3);

so the r.h.s. of (2) is also divisible by 3. Therefore k is uniquely defined by

k = ((6*m + 2)/2^n + 3 - 2*(-1)^n)/6.

Finally, it is easy to see that (6*m + 2)/2^n >= 1, so k >= 1.

QED

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.

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.

Links

Table of n, a(n) for n=1..55.

Max Alekseyev, Proof of conjecture in A257499, Sequence fanatics mailing list, April 29 and May 01, 2015

Example

Array A begins:
.       1     5     9    13    17     21     25     29     33     37
.       7    15    23    31    39     47     55     63     71     79
.       3    19    35    51    67     83     99    115    131    147
.      27    59    91   123   155    187    219    251    283    315
.      11    75   139   203   267    331    395    459    523    587
.     107   235   363   491   619    747    875   1003   1131   1259
.      43   299   555   811  1067   1323   1579   1835   2091   2347
.     427   939  1451  1963  2475   2987   3499   4011   4523   5035
.     171  1195  2219  3243  4267   5291   6315   7339   8363   9387
.    1707  3755  5803  7851  9899  11947  13995  16043  18091  20139

Mathematica

(* Array: *)
Grid[Table[(1 + 2^n*(6*k - 3 + 2*(-1)^n))/3, {n, 10}, {k, 10}]]
(* Array antidiagonals flattened: *)
Flatten[Table[(1 + 2^(n - k + 1)*(6*k - 3 + 2*(-1)^(n - k + 1)))/3, {n, 10}, {k, n}]]

Crossrefs

Cf. A007814, A075677, A254067, A257480.

Cf. A255138 (column 1).

Sequence in context: A255196 A345736 A206643 * A144846 A090289 A160670

Adjacent sequences: A257496 A257497 A257498 * A257500 A257501 A257502

Keyword

nonn,tabl

Author

L. Edson Jeffery, Apr 27 2015

Status

approved