cp's OEIS Frontend

Essentially the same as A001045, and perhaps also A152046.

Also the binomial transform of the sequence with terms (-1)^(n+1)*A128209(n).

Exponents of 2 in -1+p^s if the exponent s[u]=(2^u)k-(2^(u-1) comes from other sequence generated with s[u-1] exponent by adding 1 to terms of the "previous" sequence. E.g. s=256k-128 needed an addition of 6 to the terms of A091282.

This deals with an aspect of the Josephus problem.

Contribution from Paul Curtz, May 30 2011: (Start)

For comparison with A000265, one can arrange the sequence in blocks of length (and with row sum) 2^k, like

1;

1;

1, 3;

1, 5, 3, 7;

1, 9, 5, 13, 3, 7, 11, 15;

1, 17, 9, 25, 5, 13, 21, 29, 3, 7, 11, 15, 19, 23, 27, 31;

or

1, 1,

1, 3,

1, 5, 3, 7,

1, 9, 5, 13, 3, 7, 11, 15,

The even numbers of A181733 are essentially A152423:

2,

2,4,

2,4,6,8,

2,4,6,8,10,12,14,16,

2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32.

(End)

The number of digits of a(n) is 1, 1, 5, 17, 54, 167, 506, 1525, ... .

Integer 2*a(n) implies that 5^delta(3^n) == -1 (mod 3^n), n>=1, with the degree delta(3^n) = phi(2*3^n)/2 = 3^(n-1) of the minimal polynomial C(3^n,x) of the algebraic number 2*cos(pi/3^n). For delta and the coefficient array of C see A055034 and A187360, respectively. That 2*a(n) is indeed an even integer can be shown by analyzing the terms of the binomial expansion of (6-1)^(3^(n-1)) + 1.

This congruence implies that floor(5^(3^(n-1))/3^n) = 2*a(n) - 1, i.e., it is odd. Hence 5^delta(3^n) == +1 (Modd 3^n), n>=2. For Modd n (not to be confused with mod n) see a comment on A203571. One can show that 5 is the smallest positive primitive root Modd 3^n for n>=2 (for n=1 one has 1^1 == +1 (Modd 3)). See A206550. The proof uses the fact that the order of 5 using multiplication Modd 3^n has to be a divisor of delta(3^n)=3^(n-1), i.e., a power of 3. This is because the multiplicative group Modd 3^n has order delta(3^n) and the subgroup formed by the cycle has the order of 5 considered Modd 3^n. Then apply Lagrange's theorem. That for n>=2 no number 5^(3^(n-1-j)), with j=1, 2..., n-1, is congruent +1 (Modd 3^n) follows from the above established congruence and an analysis of the relevant expansion for a given smaller power.

The above statements show that for n>=1 the multiplicative group Modd 3^n is cyclic (for n=1 the cycle is [1], and for n>=2 the cycle is generated by 5). For the cyclic moduli see A206551.

Inverse binomial transform of this sequence is A090129.

Conjecture: Last digit of a(n) has repeating pattern of 20 digits as follows: {1, 3, 7, 7, 5, 7, 1, 1, 9, 1, 5, 5, 3, 5, 9, 9, 7, 9, 3, 3}, with an equal frequency of the five odd digits.