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This map is obtained from the array A(k, m) given in A347834. There all positive numbers congruent to 5 modulo 8 (A004770) appear uniquely in the columns for m >= 1, and the m = 0 column gives all numbers congruent to {1, 3, 7} (mod 8) (A047529). The surjective map is f: A004770 -> A047529, with b(n) = A004770(n) -> f(b(n)) = a(n).

See also the array A178415 which has permuted rows.

This maps all entries of each row k of the array A(k, m), given in A347834, with columns m >= 1 to the entry A(k, 0) = A047529(k), for k >= 1. The numbers b(n) appear once in the array A for columns m >= 1. Column A(k, 1) = A347836(k) gives the numbers congruent to {5, 32, 29} (mod 32), and each entry for columns m >= 2 is congruent to 21 (mod 32).

The surjective map of the numbers b(n) = 5 + 8*(n-1) = A004770(n), for n >= 1, to A047529 with element a(n), is computed by switching to the companion array A347839 of A347834, with the simple recurrence, removing all factors of 4, and then going back to array A347834. See the formula below. Thanks to Antti Karttunen for motivating me to simplify the prescription, and to add in A347834 the hint for the induction proof that all 5 (mod 8) numbers appear once in the columns n >= 1.

This map f is of interest in the context of the Collatz 3*n+1 conjecture. The (modified) rooted tree with only odd labeled nodes has for each row k of the array A(k, m) (A347834) the same precursor (or (modified) Collatz map given in A075677(n+1), for 2*n+1). Therefore, all nodes with labels b(n) == 5 (mod 8) can be represented by a(n). This leads to a further restricted Collatz tree with only node labels congruent to {1, 3, 7} (mod 8) (A047529).

An even further restricted Collatz tree has only node labels congruent to 1 (mod 8) (A017077), as any positive integer can be written as m*2^(v+1)+2^v-1 or (m,v) where v is the number of trailing 1-bits in binary, and for v > 1 the next odd Collatz successor of (m,v) is (3*m+1,v-1). - Ruud H.G. van Tol, Sep 13 2023

From Klaus Purath, Aug 18 2022: (Start)

This is A028560(8*n+1), and thus a(n) + 9 is a square. (See formulas)

7 is the only prime number of this sequence in which all odd prime factors occur.

Each prime factor p appears exactly twice in any interval of p consecutive terms. If a(m) and a(n) are within such an interval containing p, then m + n == -1 (mod p). (End)

From Amiram Eldar, Sep 14 2024: (Start)

Disjoint union of A017077 and {4*A091072(n)}.

The asymptotic density of this sequence is 1/4. (End)

Hankel transform of A100096(n+1).

k, k+1, k+2, k+4, k+5, and k+6 are squarefree; k+3 is divisible by 4 but no higher power of 2 and no other prime squared.

From Amiram Eldar, Nov 28 2023: (Start)

All the terms are of the form 8*k + 1.

The numbers of terms not exceeding 10^k for k = 1, 2, ... , are 1, 2, 14, 140, 1384, 13774, 137784, 1378053, 13779491, 137794128, 1377940943, ... . Apparently, the asymptotic density of this sequence exists and equals 0.0137794... . (End)

Central terms of the triangle A033293.

Also sequence found by reading the line from 1, in the direction 1, 19, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Nov 02 2012

Apart from the first column and the first two diagonals, the rest of the elements are 8's.

For the same idea but for an isosceles triangle see A278481; for a square array see A278545, for a square spiral see A010731; and for a hexagonal spiral see A010722.