Jules Beauchamp - cp's OEIS frontend

A386314 a(1) = 1 and thereafter a(n) is the smallest number k of the form 6*x+-1 not already in the sequence but where the reduced Collatz step A139391(k) is in the sequence.

1, 5, 13, 17, 11, 7, 29, 19, 25, 37, 49, 53, 35, 23, 61, 65, 43, 77, 85, 101, 67, 89, 59, 113, 133, 149, 157, 173, 115, 181, 197, 131, 205, 209, 139, 185, 229, 241, 245, 163, 217, 269, 179, 119, 79, 277, 289, 301, 305, 203, 317, 211, 281, 187, 325, 341, 227, 151, 349, 373

Offset: 1

Jules Beauchamp, Jul 18 2025

These numbers are the Collatz pre-images in the form 6*x +- 1 of all previous terms not already in the sequence.
The pre-images of a term t are all p which reach t by a single odd to odd step A139391(p) = t.
These pre-images are those p = (t*2^k-1)/3 with k>=0 which are odd integers, and with here t != 0 (mod 3) there are infinitely many p != 0 (mod 3) for each t.
Multiples of 3 have no odd pre-images and are excluded here in order to have the essential part of the tree of odd to odd descents.
The trajectory of a term t reaches 1 by steps to successively earlier terms in this sequence (at various distances apart).
If the Collatz conjecture is true, then this sequence is permutation of the numbers of the form 6x +- 1 (A007310).

			a(3) = 13, since 13 (a pre-image of a(2) = 5) is the smallest unused pre-image of a(1) and a(2).
a(10) = 37 since 37 (a pre-image of a(6) = 7) is the smallest unused pre-image of all previous terms.

David A. Corneth, Table of n, a(n) for n = 1..10000 (using Michel Marcus' code)
Sean A. Irvine, Java program (github)

Cf. A007310, A178414, A178415, A139391.

lista(nn) = my(va=List(1), vs = Map(), imin=1, i=imin, nb=1); mapput(vs, 1, 1); while(#vaMichel Marcus, Aug 25 2025

A384918 If k is in the sequence, so is k*2^m + 3, for all m > 0, a(1) = 2, ordered.

Original entry on oeis.org

2, 7, 11, 17, 19, 25, 31, 35, 37, 41, 47, 53, 59, 65, 67, 71, 73, 77, 79, 85, 91, 97, 103, 109, 115, 121, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 167, 173, 179, 185, 191, 197, 203, 209, 215, 221, 227, 233, 239, 245, 251, 257, 259, 263

Offset: 1

Jules Beauchamp, Jun 12 2025

nonn

Apart from a(1) = 2, all terms are congruent to 1 or 5 mod 6. It is therefore a subset of A007310.

			Since 7 is in the sequence, so is 7*2^3 + 3 = 59.

Cf. A007310.

A383225 a(n) = sqrt(1 + P(n)P(n+1)P(n+2)*P(n+3)) where P(n) = A000129(n) are the Pell numbers.

Original entry on oeis.org

1, 11, 59, 349, 2029, 11831, 68951, 401881, 2342329, 13652099, 79570259, 463769461, 2703046501, 15754509551, 91824010799, 535189555249, 3119313320689, 18180690368891, 105964828892651, 617608282987021, 3599684869029469, 20980500931189799, 122283320718109319, 712719423377466121

Offset: 0

Jules Beauchamp, Apr 26 2025

The ratios a(n+1)/a(n) converge to 2*sqrt(2)+3 (A156035).

			a(5) = sqrt(1 + 29*70*169*408) = 11831.

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A000129, A156035.

LinearRecurrence[{5, 5, -1}, {1, 11, 59}, 25] (* Amiram Eldar, Apr 26 2025 *)

Vec((1+6*x-x^2)/((1-6*x+x^2)*(1+x))+O(x^25)) \\ Joerg Arndt, Apr 26 2025

pell(n) = ([2, 1; 1, 0]^n)[2, 1];
a(n) = pell(n+1)*pell(n+2)-(-1)^n; \\ Seiichi Manyama, May 25 2025

a(n) = P(n+1)*P(n+2) - (-1)^n. [Corrected by Seiichi Manyama, May 25 2025]

G.f.: (1+6*x-x^2)/((1-6*x+x^2)*(1+x)). - Joerg Arndt, Apr 26 2025

A375852 Numbers congruent to {0, 1, 3, 6, 7, 9, 12, 15} mod 18.

Original entry on oeis.org

0, 1, 3, 6, 7, 9, 12, 15, 18, 19, 21, 24, 25, 27, 30, 33, 36, 37, 39, 42, 43, 45, 48, 51, 54, 55, 57, 60, 61, 63, 66, 69, 72, 73, 75, 78, 79, 81, 84, 87, 90, 91, 93, 96, 97, 99, 102, 105, 108, 109, 111, 114, 115, 117, 120, 123, 126, 127, 129, 132, 133, 135, 138, 141, 144, 145, 147, 150

Offset: 1

Jules Beauchamp, Aug 31 2024

Appears to be the union of A061641 (pure numbers in the Collatz (3x+1) iteration, also called pure hailstone numbers) and A309180 (unsuspected numbers to check in the Collatz conjecture).

The differences are periodic: 1, 2, 3, 1, 2, 3, 3, 3.

Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,2,-2,2,-1).

Cf. A061641, A309180.

Select[Range[0, 200], MemberQ[{0, 1, 3, 6, 7, 9, 12, 15}, Mod[#, 18]] &] (* Amiram Eldar, Aug 31 2024 *)

From Stefano Spezia, Sep 03 2024: (Start)
G.f.: x^2*(1 + x + 2*x^2 - x^3 + 3*x^4 + 3*x^6)/((1 - x)^2*(1 + x^2 + x^4 + x^6)).
E.g.f.: ((9*x - 14)*cosh(x) + sin(x) + 2*sqrt(2)*cosh(x/sqrt(2))*sin(x/sqrt(2)) + (9*x - 14)*sinh(x) + 2*(6 + cos(x) + (sqrt(2)*cos(x/sqrt(2)) + sin(x/sqrt(2)))*sinh(x/sqrt(2))))/4. (End)

A347823 Triangle read by rows: T(n,k) = (n+k+1)*binomial(n,k), 0 <= k <= n.

Original entry on oeis.org

1, 2, 3, 3, 8, 5, 4, 15, 18, 7, 5, 24, 42, 32, 9, 6, 35, 80, 90, 50, 11, 7, 48, 135, 200, 165, 72, 13, 8, 63, 210, 385, 420, 273, 98, 15, 9, 80, 308, 672, 910, 784, 420, 128, 17, 10, 99, 432, 1092, 1764, 1890, 1344, 612, 162, 19, 11, 120, 585, 1680, 3150, 4032, 3570, 2160, 855, 200, 21

Offset: 0

Jules Beauchamp, Jan 23 2022

			Triangle begins:
  1;
  2,  3;
  3,  8,   5;
  4, 15,  18,   7;
  5, 24,  42,  32,   9;
  6, 35,  80,  90,  50,  11;
  7, 48, 135, 200, 165,  72, 13;
  8, 63, 210, 385, 420, 273, 98, 15;
  ...

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)

Row sums give A053220.
Columns give A000027, A005563, A212343.
Diagonals give A005408, A001105, A059270, A112742.
Cf. A007318, A094727.

T(n,k) = (n+k+1)*binomial(n,k) \\ Andrew Howroyd, Jan 23 2022

T(n,k) = A094727(n+1,k)*A007318(n,k).

Row g.f.: (1 + x)^(n-1)*(1 + n + x + 2*n*x). - Stefano Spezia, Jan 23 2022

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User: Jules Beauchamp

A386314 a(1) = 1 and thereafter a(n) is the smallest number k of the form 6*x+-1 not already in the sequence but where the reduced Collatz step A139391(k) is in the sequence.

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A384918 If k is in the sequence, so is k*2^m + 3, for all m > 0, a(1) = 2, ordered.

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A383225 a(n) = sqrt(1 + P(n)*P(n+1)*P(n+2)*P(n+3)) where P(n) = A000129(n) are the Pell numbers.

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A375852 Numbers congruent to {0, 1, 3, 6, 7, 9, 12, 15} mod 18.

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A347823 Triangle read by rows: T(n,k) = (n+k+1)*binomial(n,k), 0 <= k <= n.

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Formula

A383225 a(n) = sqrt(1 + P(n)P(n+1)P(n+2)*P(n+3)) where P(n) = A000129(n) are the Pell numbers.