V. Barbera - cp's OEIS frontend

A383046 Smallest initial value for shortcut form of the Collatz function (3x+1)/2 sequence that begins with exactly n-1 increases and decreases once before the last increment.

11, 7, 175, 287, 2239, 4223, 19199, 12799, 52223, 559103, 372735, 5840895, 9486335, 6324223, 48955391, 32636927, 1095499775, 730333183, 11940134911, 19413336063, 35848716287, 298777051135, 932192452607, 621461635071, 6278369771519, 4185579847679

Offset: 3

V. Barbera, Apr 20 2025

nonn

Using shortcut form of the Collatz sequence, a(n) represents the residue class (mod 2^y) with y=floor(1+n*log(3)/log(2)) so that for x = a(n)+k*2^y, with k>=0, we have n-1 successive rises x -> (3x+1)/2 and one fall x -> x/2 before the last rise x -> (3x+1)/2 followed by successive falls x -> x/2 until dropping below its starting x.

			For n=4, the residue class (mod 2^7) or smallest starting x = a(4) = 7 has trajectory
 7 - 11 -> 17 -> 26         -> 13-> 20                  -> 10 -> 5
     \-----------/              \--/                      \------/
     n-1=3 increases      1 decreases and 1 increases      decreases to < initial
For x = 135 = a(4) + 128 = 7 + 128 has trajectory
 135 - 203 -> 305 -> 458    -> 229-> 344                 -> 172 -> 84
     \-----------/              \--/                      \------/
     n-1=3 increases      1 decreases and 1 increases      decreases to < initial

Cf. A381707.

a(n)={my(y=floor(1+n*log(3)/log(2))+1-n); 2^(n-1)*lift(Mod(1/3^n, 2^y)) - 1}

a(n) = 2^(n-1) * (3^(-n) mod 2^(floor(1+n*log2(3))+1-n)) - 1.

A382437 a(n) = a(n-1)^2 + 4 * a(n-1), with a(0) = 2.

Original entry on oeis.org

2, 12, 192, 37632, 1416317952, 2005956546822746112, 4023861667741036022825635656102100992, 16191462721115671781777559070120513664958590125499158514329308740975788032

Offset: 0

V. Barbera, Mar 25 2025

nonn

Paolo Xausa, Table of n, a(n) for n = 0..10

Cf. A002812, A003010.

NestList[#*(4 + #) &, 2, 8] (* Paolo Xausa, Apr 01 2025 *)

a(n)=if(n, a(n-1)^2 + 4*a(n-1), 2);
vector(8, i, a(i-1))

a(n) = A003010(n) - 2.
a(n)/2 = A002812(n) - 1.
For n > 1: a(n) = 3 * 2^(2*n) * Product_{i = 0..n-2} A002812(i)^2.
Conjecture: a(n) = Sum_{k=1..2^n} (2^n * 2^k * binomial(2^n + k - 1, 2*k - 1) / k).

A382393 Positive integers k such that 6k - 1 is prime for k != 1 (mod 5) and (6k - 1)/5 is prime for k == 1 (mod 5).

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 22, 23, 25, 26, 28, 29, 30, 31, 32, 33, 36, 38, 39, 40, 42, 43, 44, 45, 47, 49, 51, 52, 53, 56, 58, 59, 60, 61, 64, 65, 66, 67, 70, 72, 74, 75, 77, 78, 80, 81, 82, 84, 85, 86, 87, 91, 93, 94, 95, 98, 99, 100

Offset: 1

V. Barbera, Mar 23 2025

nonn

For all elements of this sequence, there are no pairs (x,y) of positive integers with x > 1 such that a(n)=6*x*y+x-y.

Cf. A024898, A024899.

select(k->isprime((6*k-1)/(5-4*!(k==Mod(1,5)))), vector(100,i,i))

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User: V. Barbera

A383046 Smallest initial value for shortcut form of the Collatz function (3x+1)/2 sequence that begins with exactly n-1 increases and decreases once before the last increment.

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A382437 a(n) = a(n-1)^2 + 4 * a(n-1), with a(0) = 2.

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A382393 Positive integers k such that 6k - 1 is prime for k != 1 (mod 5) and (6k - 1)/5 is prime for k == 1 (mod 5).

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Author

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PARI

cp's OEIS Frontend

User: V. Barbera

A383046 Smallest initial value for shortcut form of the Collatz function (3x+1)/2 sequence that begins with exactly n-1 increases and decreases once before the last increment.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Formula

A382437 a(n) = a(n-1)^2 + 4 * a(n-1), with a(0) = 2.

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Author

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Programs

Mathematica

PARI

Formula

A382393 Positive integers k such that 6*k - 1 is prime for k != 1 (mod 5) and (6*k - 1)/5 is prime for k == 1 (mod 5).

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Author

Keywords

Comments

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PARI

A382393 Positive integers k such that 6k - 1 is prime for k != 1 (mod 5) and (6k - 1)/5 is prime for k == 1 (mod 5).