A026142 -id:A026142 - cp's OEIS frontend

A026145 a(n) = s(k), where k is the n-th number such that s(j) < s(k) for all j < k, where s = A026142. Also a(n) = 2*t(n) for n >= 2, where t = A026144.

1, 4, 8, 12, 14, 20, 24, 26, 32, 36, 38, 42, 44, 50, 56, 60, 62, 68, 72, 74, 78, 80, 86, 92, 96, 98, 104, 108, 110, 114, 116, 122, 126, 128, 132, 134, 140, 146, 150, 152, 158, 164, 168, 170, 176, 180, 182, 186, 188, 194, 200, 204, 206, 212, 216, 218, 222

Offset: 1

Clark Kimberling

nonn

F. M. Dekking, Permutations of N generated by left-right filling algorithms, arXiv:2001.08915 [math.CO], 2020, see R_rec in section 2.4.

Cf. A026142, A026144.

A026224 Numbers n such that t(n) = s(n) + 1, where s = A026136, t = A026142.

Original entry on oeis.org

2, 4, 10, 13, 22, 28, 31, 37, 40, 49, 58, 64, 67, 76, 82, 85, 91, 94, 103, 109, 112, 118, 121, 130, 139, 145, 148, 157, 166, 172, 175, 184, 190, 193, 199, 202, 211, 220, 226, 229, 238, 244, 247, 253, 256, 265, 271, 274, 280, 283

Offset: 1

Clark Kimberling

n is chosen to denote the numbers, as each n represents an index for sequences s and t.
From Peter Munn, Mar 08 2022: (Start)
2 with numbers of the form 3^i*(3k+1) + 1, i >= 1.
Proof:
n = 1 is clearly excluded by either definition, as t(1) <> s(1) + 1 and 1 is not of the form 3^i*(3k+1) + 1, i >= 1. For n >= 2 the remaining argument applies.
Considering the conditions s and t place on individual terms, and using basic arithmetic, it is easy to show that "s(n) > n, t(n) > n" is a necessary condition for t(n) = s(n) + 1. Taking into account the lexicographically earliest properties of s and t, it is then straightforward to show the condition is also sufficient. I omit the details.
Proofs in A026136 and A026142 show: s(n) > n if and only if s(n) has the form 3^i*(6k+2)+1; t(n) > n if and only if t(n) has the form (A) 3^i*4 or (B) 3^i*(6k+2), k >= 1. We consider (A) and (B) separately:
(A) s(n) + 1 = 3^i_1*(6k_1+2) + 2 = 3^i_2*4 = t(n)
Modulo 3, the left-hand side can be congruent to 1 or 2, the right-hand side to 0 or 1. Equality requires i_2 = 0, so t(n) = 4, from which we complete the solution with n = 2 and s(n) = 3.
or
(B) s(n) + 1 = 3^i_1*(6k_1+2) + 2 = 3^i_2*(6k_2+2) = t(n), k_2 >= 1
Modulo 3, the left-hand side can be congruent to 1 or 2, the right-hand side to 0 or 2. Equality requires i_1 >= 1, i_2 = 0.
So we have 3^i_1*(6k_1+2) + 2 = 6k_2+2, i_1 >= 1, k_2 >= 1. Clearly, for any i_1 >= 1 and k_1, there is a solution for k_2.
So for n to qualify under (B), s(n) must have the form 3^i*(6k+2) + 1, i >= 1, and therefore also the form 6j+1. If s(n) has the form 6j+1 and s(n) > n, then n = 3j+1 (see A026136) and also t(3j+1) = 6j+2 (see A026142, given j >= 1). So we need n to have the form 3^i*(3k+1) + 1, i >= 1, and for all such n there is a solution s(n) + 1 = 2*3^i*(3k+1) + 2 = t(n).
(End)

Cf. A026136, A026142.

A026144 Numbers k such that s(j) < s(k) for all j < k, where s = A026142.

Original entry on oeis.org

1, 2, 4, 6, 7, 10, 12, 13, 16, 18, 19, 21, 22, 25, 28, 30, 31, 34, 36, 37, 39, 40, 43, 46, 48, 49, 52, 54, 55, 57, 58, 61, 63, 64, 66, 67, 70, 73, 75, 76, 79, 82, 84, 85, 88, 90, 91, 93, 94, 97, 100, 102, 103, 106, 108, 109, 111, 112, 115, 117, 118, 120, 121

Offset: 1

Clark Kimberling

nonn

F. M. Dekking, Permutations of N generated by left-right filling algorithms, arXiv:2001.08915 [math.CO], 2020, see R_pos in section 2.4.

Cf. A026142, A026145.

A026222 Numbers k such that A026136(k) = A026142(k).

Original entry on oeis.org

1, 3, 9, 15, 24, 27, 33, 42, 45, 51, 60, 69, 72, 78, 81, 87, 96, 99, 105, 114, 123, 126, 132, 135, 141, 150, 153, 159, 168, 177, 180, 186, 195, 204, 207, 213, 216, 222, 231, 234, 240, 243, 249, 258, 261, 267, 276, 285, 288, 294, 297

Offset: 1

Clark Kimberling

nonn

Michael De Vlieger, Table of n, a(n) for n = 1..10000
F. M. Dekking, Permutations of N generated by left-right filling algorithms, arXiv:2001.08915 [math.CO], 2020.

Block[{nn = 10^3, a, b, s, t}, a[1] = b[1] = 1; Do[If[! IntegerQ[a[#1]], Set[a[#1], i], Set[a[#2], i]] & @@ {i - #, i + #} &@ Floor[i/2], {i, nn}]; s = TakeWhile[Array[a[#] &, nn], IntegerQ]; Do[If[! IntegerQ[b[#1]], Set[b[#1], i], Set[b[#2], i]] & @@ {i - #, i + #} &@ Floor[(i + 1)/2], {i, nn}]; t = TakeWhile[Array[b[#] &, nn], IntegerQ]; Select[Range@ Min[Last /@ {s, t}], s[[#]] == t[[#]] &]] (* Michael De Vlieger, Apr 21 2020 *)

A026143 a(n) = position of n in A026142.

Original entry on oeis.org

1, 3, 5, 2, 8, 9, 11, 4, 14, 15, 17, 6, 20, 7, 23, 24, 26, 27, 29, 10, 32, 33, 35, 12, 38, 13, 41, 42, 44, 45, 47, 16, 50, 51, 53, 18, 56, 19, 59, 60, 62, 21, 65, 22, 68, 69, 71, 72, 74, 25, 77, 78, 80, 81, 83, 28, 86, 87, 89, 30, 92, 31, 95, 96, 98, 99, 101, 34, 104

Offset: 1

Clark Kimberling

nonn

A026190 a(n) = (1/2)*s(n), where s(n) is the n-th even number in A026142.

Original entry on oeis.org

2, 1, 4, 6, 7, 3, 10, 12, 13, 5, 16, 18, 19, 21, 22, 8, 25, 9, 28, 30, 31, 11, 34, 36, 37, 39, 40, 14, 43, 15, 46, 48, 49, 17, 52, 54, 55, 57, 58, 20, 61, 63, 64, 66, 67, 23, 70, 24, 73, 75, 76, 26, 79, 27, 82, 84, 85, 29, 88, 90, 91, 93, 94

Offset: 1

Clark Kimberling

nonn

A026192 (1/3)*s(n), where s(n) is the n-th multiple of 3 in A026142.

Original entry on oeis.org

1, 4, 2, 8, 3, 12, 14, 5, 6, 20, 7, 24, 26, 9, 10, 32, 11, 36, 38, 13, 42, 44, 15, 16, 50, 17, 18, 56, 19, 60, 62, 21, 22, 68, 23, 72, 74, 25, 78, 80, 27, 28, 86, 29, 30, 92, 31, 96, 98, 33, 34, 104, 35, 108, 110, 37, 114, 116, 39, 40, 122

Offset: 1

Clark Kimberling

nonn

Is this a duplicate of A026142? - R. J. Mathar, Jun 24 2025

A026194 a(n) = (1/4)*s(n), where s(n) is the n-th multiple of 4 in A026142.

Original entry on oeis.org

1, 2, 3, 5, 6, 8, 9, 11, 4, 14, 15, 17, 18, 20, 7, 23, 24, 26, 27, 29, 10, 32, 33, 35, 12, 38, 13, 41, 42, 44, 45, 47, 16, 50, 51, 53, 54, 56, 19, 59, 60, 62, 21, 65, 22, 68, 69, 71, 72, 74, 25, 77, 78, 80, 81, 83, 28, 86, 87, 89, 30, 92, 31

Offset: 1

Clark Kimberling

nonn

A026229 Numbers k such that A026166(k) = A026142(k) - 2.

Original entry on oeis.org

2, 6, 12, 18, 21, 30, 36, 39, 48, 54, 57, 63, 66, 75, 84, 90, 93, 102, 108, 111, 117, 120, 129, 138, 144, 147, 156, 162, 165, 171, 174, 183, 189, 192, 198, 201, 210, 219, 225, 228, 237, 246, 252, 255, 264, 270, 273, 279, 282, 291

Offset: 1

Clark Kimberling

nonn

Cf. A026142, A026166.

Title corrected by Sean A. Irvine, Sep 22 2019

A026231 Numbers k such that A026166(k) = A026142(k) + 1.

Original entry on oeis.org

8, 17, 20, 26, 35, 44, 47, 53, 56, 62, 71, 74, 80, 89, 98, 101, 107, 116, 125, 128, 134, 137, 143, 152, 155, 161, 164, 170, 179, 182, 188, 197, 206, 209, 215, 218, 224, 233, 236, 242, 251, 260, 263, 269, 278, 287, 290, 296, 299

Offset: 1

Clark Kimberling

nonn

Cf. A026142, A026166.

Title corrected by Sean A. Irvine, Sep 22 2019

cp's OEIS Frontend

A026145 a(n) = s(k), where k is the n-th number such that s(j) < s(k) for all j < k, where s = A026142. Also a(n) = 2*t(n) for n >= 2, where t = A026144.

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A026224 Numbers n such that t(n) = s(n) + 1, where s = A026136, t = A026142.

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A026144 Numbers k such that s(j) < s(k) for all j < k, where s = A026142.

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A026222 Numbers k such that A026136(k) = A026142(k).

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A026143 a(n) = position of n in A026142.

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A026190 a(n) = (1/2)*s(n), where s(n) is the n-th even number in A026142.

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A026192 (1/3)*s(n), where s(n) is the n-th multiple of 3 in A026142.

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A026194 a(n) = (1/4)*s(n), where s(n) is the n-th multiple of 4 in A026142.

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A026229 Numbers k such that A026166(k) = A026142(k) - 2.

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A026231 Numbers k such that A026166(k) = A026142(k) + 1.

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