cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Wolfdieter Lang

Wolfdieter Lang's wiki page.

Wolfdieter Lang has authored 3222 sequences. Here are the ten most recent ones:

A386894 Markoff numbers that are powers of one odd prime or twice powers of one odd prime.

Original entry on oeis.org

1, 2, 5, 13, 29, 34, 89, 169, 194, 233, 433, 1597, 2897, 5741, 7561, 28657, 33461, 43261, 96557, 426389, 514229, 646018, 1686049, 2012674, 2922509, 3276509, 11485154, 21531778, 94418953, 253191266, 321534781, 433494437, 780291637, 1405695061, 1475706146, 2971215073, 6684339842, 19577194573
Offset: 1

Author

Wolfdieter Lang, Aug 07 2025

Keywords

Comments

A subsequence of A000961 without numbers divisible by 4.
The powers of odd primes are given in A061345 (with offset 0).
These Markoff numbers (see A002559) have been proved to obey the Frobenius-Markoff uniqueness conjecture. See Aigner, Corollary 3.20, p. 59, and there the references [4] A. Baragar, [18] J. O. Button, and [119] Ying Zhang.

Examples

			26 = 2*13 is not a Markoff number, hence not in the present sequence.
610 = 2*5*61 is a Markoff number but not a prime power nor is 305 a prime power.
		

References

  • Martin Aigner, Markov's theorem and 100 years of the uniqueness conjecture. A mathematical journey from irrational numbers to perfect matchings. Springer, 2013.

Crossrefs

Programs

  • Mathematica
    MAX=10^11; data=NestWhile[Select[Union[Sort/@Flatten[Table[{a, b, 3a b -c}/.MapThread[Rule, {{a, b, c}, #}]&/@Map[RotateLeft[ii, #]&, Range[3]], {ii, #}], 1]], Max[#]James C. McMahon, Aug 12 2025 *)
  • SageMath
    def A386894List(len: int = 50, MAX: int = 10**10) -> list[int]:
        # Using function 'MarkovNumbers' from A002559.
        M = MarkovNumbers(len, MAX)
        U = set([1])
        for m in M:  # if m is a Markov number and ...
            z = ZZ(m)
            if is_prime_power(z) or (is_even(z) and is_prime_power(z//2)):
                U.add(m)
        return sorted(U)
    # Balance required sequence length and search depth.
    print(A386894List(len=120, MAX=10**12))  # Peter Luschny, Aug 12 2025

Formula

Markoff numbers of the form 2^j*p^k, with an odd prime p, j = 0 or 1, and k >= 0, ordered strictly increasing.

A385449 Irregular triangle, read by rows: row n gives the pair of proper positive fundamental solutions (x, y) of the form x^2 - 2*y^2 representing -A057126(n).

Original entry on oeis.org

1, 1, 4, 3, 1, 2, 5, 4, 2, 3, 6, 5, 1, 3, 9, 7, 3, 4, 7, 6, 1, 4, 13, 10, 4, 5, 8, 7, 3, 5, 11, 9, 2, 5, 14, 11, 5, 6, 9, 8, 1, 5, 17, 13, 6, 7, 10, 9, 1, 6, 21, 16, 5, 7, 13, 11, 7, 8, 11, 10, 4, 7, 16, 13, 3, 7, 19, 15, 2, 7, 22, 17, 1, 7, 25, 19, 8, 9, 12, 11, 5, 8, 17, 14, 7, 9, 15, 13, 3, 8, 23, 18, 9, 10, 13, 12
Offset: 1

Author

Wolfdieter Lang, Jul 11 2025

Keywords

Comments

The number of (x, y) pairs in row n is 1 for n = 1 and 2, and 2^P, with P = P1 + P7, where P1 and P7 are the number of prime factors 1 modulo 8 and 7 modulo 8, respectively, of A057126(n), for n >= 3.
See A057126 for comments concerning its representation by x^2 - 2*y^2.
The numbers A057126 are given by 2^e_2 * Product_{i=1..P1} p_{1,i}^e_{1,i} * Product_{j=1..P7} p_{7,j}^e_{7,j}, with the odd primes p_{1,i} and p_{7,j} congruent to 1 and 7 modulo 8, respectively. See A007519 and A007522 for these odd primes. Together with 2 these primes are given in A038873, and without 2 in A001132. The exponents are e_2 = 0 or 1, and e_{1,i} and e_{7,j} are nonnegative. The a(1) = 1 is obtained if all exponents vanish. For the proof see Lemma 18 of the linked W. Lang paper, pp. 22 - 23.
The general solutions are obtained from each fundamental solution by application of integer powers of the matrix Auto' = Mat([3,4], [2,3]). See the linked paper eq (28), p. 14, and eq. (40), p. 17 for D = 2, and k = A057126(n). For the explicit form of the powers of Auto' in terms of Chebyshev polynomials S(n, 6) = A001109(n+1) see there eq. (38), and Lemma 10, eq. (43), p. 17.
The conversion to the pair of proper solutions (X, Y) of X^2 - 2*Y^2 = A057126(n) is given by (X, Y) = (2*y - x, x - y). This may result in solutions with negative Y values. They are then transformed to the fundamental positive proper solutions via the mentioned matrix Auto'. See the right part of the example below. For this conversion see also the Nov 09 2009 comment in A035251 by Franklin T. Adams-Watters.

Examples

			n, A057126(n) /k  1  2   3  4 ...   2^P | (X, Y) = (2*y - x, x - y)
-------------------------------------------------------------------
1,  1           | 1  1               1  |  1   0 (3   2)
2,  2           | 4  3               1  |  2   1
3,  7           | 1  2,  5  4        2  |  3  -1 (5   3),  3  1
4, 14 = 2*7     | 2  3,  6  5        2  |  4  -1 (8   5),  4  1
5, 17           | 1  3,  9  7        2  |  5  -2 (7   4),  5  2
6, 23           | 3  4,  7  6        2  |  5  -1 (11  7),  5, 1
7, 31           | 1  4, 13 10        2  |  7  -3 (9   5),  7  3
8, 34 = 2*17    | 4  5,  8  7        2  |  6  -1 (14  9),  6  1
9, 41           | 3  5, 11  9        2  |  7  -2 (13  8),  7  2
10, 46 = 2*23   | 2  5, 14 11        2  |  8  -3 (12  7),  8  3
11, 47          | 5  6,  9  8        2  |  7  -1 (17 11),  7  1
12, 49 = 7^2    | 1  5, 17 13        2  |  9  -4 (11  6),  9  4
13, 62 = 2*31   | 6  7, 10  9        2  |  8  -1 (20 13),  8  1
14, 71          | 1  6, 21 16        2  | 11  -5 (13  7), 11  5
15, 73          | 5  7, 13 11        2  |  9  -2 (19 12),  9  2
16, 79          | 7  8, 11 10        2  |  9  -1 (23 15),  9  1
17, 82 = 2*41   | 4  7, 16 13        2  | 10  -3 (18 11), 10  3
18, 89          | 3  7, 19 15        2  | 11  -4 (17 10), 11  4
19, 94 = 2*47   | 2  7, 22 17        2  | 12  -5 (16  9), 12  5
20, 97          | 1  7, 25 19        2  | 13  -6 (15  8), 13  6
21, 98 = 2*7^2  | 8  9, 12 11        2  | 10  -1 (26 17), 10  1
...
The corresponding fundamental positive proper solutions of X^2 - 2*Y^2 = +119 are: [13 -5 (19 11), 13, 5] and [11 -1 (29 19), 11 1].
		

A385447 Decimal expansion of sqrt(8 + 9*phi), with the golden section A001622.

Original entry on oeis.org

4, 7, 4, 9, 9, 7, 9, 5, 6, 8, 2, 4, 5, 4, 3, 1, 2, 6, 7, 2, 7, 2, 0, 3, 6, 9, 2, 7, 0, 3, 7, 1, 5, 4, 8, 9, 2, 7, 7, 4, 6, 6, 1, 6, 7, 4, 6, 8, 8, 0, 8, 9, 8, 0, 6, 1, 0, 3, 4, 2, 6, 0, 3, 9, 5, 7, 4, 1, 8, 8, 3, 2, 4, 0, 1, 1, 6, 5, 9, 9, 4, 0, 9, 5
Offset: 1

Author

Wolfdieter Lang, Jul 01 2025

Keywords

Comments

This constant c gives sqrt(11)*rho_1, where rho_1 = length(Z, D_1), with the fixed point Z = -(A385445 + A385446*i)/(2*11) of the complex map w given in A385445 and D_1 = i.
See A385445 for details and the linked paper, eq. (7b).

Examples

			4.7499795682454312672720369270371548927...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[Sqrt[8 + 9*GoldenRatio], 10, 120][[1]] (* Amiram Eldar, Jul 02 2025 *)

Formula

Equals sqrt(8 + 9*phi) = sqrt(5 + 7*phi)*s, with s = A182007, the length of the side of a regular pentagon circumcribed by a unit circle.

A385445 Decimal expansion of (-1 + 3*phi)*sqrt(3 - phi), with the golden section phi = A001622.

Original entry on oeis.org

4, 5, 3, 0, 7, 6, 8, 5, 9, 3, 1, 8, 5, 9, 7, 5, 1, 7, 4, 3, 6, 1, 2, 2, 4, 0, 9, 0, 9, 9, 8, 1, 4, 7, 3, 2, 3, 2, 3, 8, 8, 8, 6, 9, 2, 9, 4, 6, 8, 2, 0, 9, 3, 5, 2, 5, 3, 9, 2, 8, 8, 9, 0, 5, 0, 6, 6, 3, 6, 2, 0, 7, 2, 1, 8, 6, 4, 5, 7, 0, 9, 5, 2, 9
Offset: 1

Author

Wolfdieter Lang, Jul 01 2025

Keywords

Comments

This constant c gives the real part of -2*11*Z = (c + d*i), where Z is the (finite) fixed point of a complex function w (of the loxodromic type) mapping iteratively the vertices of golden triangles, starting with vertices (D_1, D_2, D_3), circumscribed by the unit circle with center at the origin, and D_1 = i, D_2 = (s - phi*i)/2 and D_3 = (-s - phi*i)/2. This function is w(z) = a*z + b, with a = (-1 + phi) * exp(-(3*Pi/5)*i) = -((2 - phi) + s*i)/2 and b = (1 - phi)*i, where s = sqrt(3 - phi) = A182007 (the length of the base (D_2, D_3) of the first triangle, also called s_1).
The imaginary part of -2*11*Z is d = -7 + 10*phi = A385446.
If the fixed point Z = -(0.20594... + 0.41728...*i) is chosen as origin then the loxodromic map is W(z') = a*z' (where z' = z - Z and W(z') = w(z'+Z) - Z).
For details see the linked paper, eqs.(5a,b) for w(z), eq.(6) for Z and eq.(7) for W(z'). (In eq.(5b) the i is missing in the exponent.) The nesting of golden triangles as shown in Fig. 1 of the link leads to the fixed point Z.
The vertices of the nested golden triangles can be connected by a spiral built of circular sections with angle 108 degrees, centered at vertices D_{n+3} and shrinking radii s_n =(- 1 + phi)^(n-1)*s. Note that the curvature of this spiral is not continuous.
The length(Z, D_n) =: rho_n of the spokes of the spiral is (-1 + phi)^(n-1)*rho_1, with sqrt(11)*rho_1 = sqrt(8 + 9*phi) = sqrt(5 + 7*phi)*s = A385447.
For the length ratio rho_1/s see A385448.
The logarithmic spiral connecting the vertices D_{n+1} is given in polar coordinates by rho(Phi) = rho_1 * exp((-(5/(3*Pi)) * log(phi)*Phi), with the vertices obtained in polar coordinates for Phi = Phi_n = (3*Pi/5)*n, namely rho(Phi_n) = rho_{n+1}, for n >= 0. For log(phi) see A002390. Note that the nonnegative x-axis is now along Z, D_1. The angle(Z, D1, D4) =: gamma is given by arctan((18 - 11*phi)/s) = arcsin(rho_4 / 2) = 0.169860704... or 9.7323... degrees. See Fig. 3 of the linked paper.

Examples

			4.5307685931859751743612240909981473232388869294682093...
		

Programs

  • Mathematica
    RealDigits[Sqrt[27 - 4*GoldenRatio], 10, 120][[1]] (* Amiram Eldar, Jul 02 2025 *)

Formula

Equals (-1 + 3*phi)*sqrt(3 - phi) = (A090550 - 2)*A182007.
Equals sqrt(27 - 4*phi).

A385446 Decimal expansion of -7 + 10*phi, with the golden section phi = A001622.

Original entry on oeis.org

9, 1, 8, 0, 3, 3, 9, 8, 8, 7, 4, 9, 8, 9, 4, 8, 4, 8, 2, 0, 4, 5, 8, 6, 8, 3, 4, 3, 6, 5, 6, 3, 8, 1, 1, 7, 7, 2, 0, 3, 0, 9, 1, 7, 9, 8, 0, 5, 7, 6, 2, 8, 6, 2, 1, 3, 5, 4, 4, 8, 6, 2, 2, 7, 0, 5, 2, 6, 0, 4, 6, 2, 8, 1, 8, 9, 0, 2, 4, 4, 9, 7, 0, 7
Offset: 1

Author

Wolfdieter Lang, Jul 01 2025

Keywords

Comments

This constant d gives the imaginary part of -2*11*Z = c + d*i, where Z is the fixed point of a complex function w (of the loxodromic type) mapping vertices of golden triangles, starting with vertices (D_1, D_2, D_3), circumcribed by the unit circle with center at the origin, and D_1 = i (the complex unit), D_2 = (s - phi*i)/2 and D_3 = (-s - phi*i)/2. This function is w(z) = a*z + b, with a = (-1 + phi) * exp(-3*Pi*i/5) = -((2 - phi) + s*i)/2 and b = (1 - phi)*i, where s = sqrt(3 - phi) = A182007 (the length of the base (D2, D3) of the first triangle).
The real part c = (-1 + 3*phi)*s is given in A385445.
For details see A385445, and eqs.(5a,b) of the linked paper there.

Examples

			9.18033988749894848204586834365638117720309179805762862...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[10*GoldenRatio - 7, 10, 120][[1]] (* Amiram Eldar, Jul 02 2025 *)

Formula

Equals -7 + 10*phi, an integer in the quadratic number field Q(sqrt(5)).
Equals 10*A176055-12 = 10*A104457-17 = 10*A001622-7 . - R. J. Mathar, Jul 06 2025

A385448 Decimal expansion of sqrt(5 + 7*phi)/sqrt(11), with the golden section phi = A001622.

Original entry on oeis.org

1, 2, 1, 8, 2, 7, 8, 8, 8, 7, 3, 5, 9, 6, 6, 2, 2, 9, 1, 5, 3, 5, 4, 6, 0, 2, 6, 7, 9, 1, 7, 2, 7, 4, 7, 4, 5, 2, 0, 3, 6, 8, 7, 4, 0, 0, 5, 3, 1, 5, 5, 4, 3, 5, 6, 6, 6, 6, 9, 9, 1, 9, 0, 4, 7, 5, 6, 9, 3, 9, 7, 6, 5, 7, 4, 7, 5, 7, 2, 2, 2, 0, 5, 8
Offset: 1

Author

Wolfdieter Lang, Jul 01 2025

Keywords

Comments

This equals the ratio length(Z, D_1)/s, with the fixed point of a complex loxodromic map w mapping iteratively golden triangles, starting with the one inscribed in a circumcircle with center ot the origin of the complex plane, the top vertex D_1 = i (the complex unit) and the base D_2 = (s - phi*i)/2, D_3 = (-s - phi*i)/2, with s = A182007.
See A385445 for details and a linked paper.

Examples

			1.218278887359662291535460267917274745203687400531554...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[Sqrt[(5 + 7*GoldenRatio)/11], 10, 120][[1]] (* Amiram Eldar, Jul 02 2025 *)

Formula

Equals sqrt(5 + 7*phi)/sqrt(11) = sqrt(5 + 7*phi)/A010468.

A384921 Number of permutations [p_1, p_2, ..., p_n], for n >= 1, with |p_{i+1} - p_i| >= 2, for i = 1..n-1, and |p_n - p_1| = 0 or 1.

Original entry on oeis.org

1, 0, 0, 2, 4, 30, 184, 1322, 10668, 96566, 969280, 10690146, 128527348, 1673257262, 23451539784, 352079626010, 5637207651004, 95886993887142, 1726775043225808, 32821564079286866, 656647922936247300, 13793480376190668446
Offset: 1

Author

Wolfdieter Lang, Jun 17 2025

Keywords

Comments

This sequence gives the number of the so-called king permutations, for n >= 1, counted in A002464, that satisfy the additional restriction |p_n - p_1| = 0 or 1.

Examples

			n=1: The permutation is [1].
n=4: The two king permutations are [2, 4, 1, 3] and its reversal [3, 1, 4, 2].
n=5: The four permutations are [2,,4, 1, 5, 3], [3, 1, 5, 2, 4] and their reversals [3, 5, 1, 4, 2], [4, 2, 5, 1, 3]. See III of the Abramson and Moser link, p. 1254.
n=6: The 30 permutations are (in short cut version): 146352, 153642, 246153, 251463, 264153, 315264, 351624, 352614, 361524, 362514, 413625, 426135, 426315, 524136, 531426, and their reversals.
		

Crossrefs

Formula

a(n) = A002464(n) - A002493(n), for n >= 2, but A002493(1) = 1, not 0, as it is here, if instead of A002493 the definition |p_{i+1} - p_i| >= 2, for i = 1..n, for n >= 1, and p_{n+1} = p_1 is used; hence a(1) = 1, not 0.
a(n) = 2*Sum_{k=0..floor((n-2)/2)} A002464(n - (2*k+1)), for n >= 3, and a(1) = 1, a(2) = 0. (Compare this with the formula given by Vladeta Jovovic in A002493, Nov 24 2007.)

A384920 Tooth numbering sytem (deciduous), FDI and ISO 3950.

Original entry on oeis.org

55, 54, 53, 52, 51, 61, 62, 63, 64, 65, 85, 84, 83, 82, 81, 71, 72, 73, 74, 75
Offset: 1

Author

Wolfdieter Lang, Jun 12 2025

Keywords

Comments

Only two rows, each of length 10.
This array gives the deciduous (baby, milk) tooth numbering systems for humans, according to the FDI two-digit notation, and IS0 3950.
For the permanent teeth numbering for humans see A384919.

Examples

			Human deciduous teeth.
Row 1: baby's upper right, baby's upper left:
55 54 53 52 51, 61 62 63 64 65
Row 2: baby's lower right, baby's lower left:
85 84 83 82 81, 71 72 73 74 75
		

Crossrefs

A384919 Tooth numbering sytem (permanent), FDI and ISO 3950.

Original entry on oeis.org

18, 17, 16, 15, 14, 13, 12, 11, 21, 22, 23, 24, 25, 26, 27, 28, 48, 47, 46, 45, 44, 43, 42, 41, 31, 32, 33, 34, 35, 36, 37, 38
Offset: 1

Author

Wolfdieter Lang, Jun 12 2025

Keywords

Comments

Only two rows, each of length 16.
This array gives the permanent tooth numbering systems for humans, according to the FDI two-digit notation, and IS0 3950.
For the deciduous (baby, milk) tooth numbering system for humans see A384920.

Examples

			Row 1: patient's upper right, patient's upper left:
18 17 16 15 14 13 12 11, 21 22 23 24 25 26 27 28
Row 2: patient's lower right, patient's lower left:
48 47 46 45 44 43 42 41, 31 32 33 34 35 36 37 38
		

Crossrefs

A384561 One fourth of the number of permutations of [n] with |p(i+1) - p(i)| >= 2, for i = 1..(n-1) and n appears at position i = 1 or i = n.

Original entry on oeis.org

1, 6, 39, 284, 2337, 21474, 218179, 2430216, 29459301, 386182478, 5444570631, 82157021556, 1321282006249, 22562446559034, 407722012334667, 7773697259015264, 155956589714240109, 3284208113313605286, 72434065593967762831, 1669777527837108720588, 40157785493048522566641
Offset: 5

Author

Wolfdieter Lang, Jun 04 2025

Keywords

Comments

The number of such permutations of [n] is 1 for n = 1 (the p(i) condition is not needed), and 0 for n = 2, 3 and 4, hence a(1) = 1/4 and a(n) = 0 for n = 2, 3 and 4.
The number of permutations of [n] with |p(i+1) - p(i)| >= 2, for i = 1..(n-1), for n >= n is given by A002464(n), for n >= 0. See also A001266(n) = A002464(n)/2, for n >= 2. These permutations are also called king permutations, e.g., in A382644.

Examples

			n = 5: the 4 permutations are 2 4 1 3 5, 3 1 4 2 5 and their reversals 5 3 1 4 2,  5 2 4 1 3.
a(5) = 1 = A382644(4)/2 = (A001236(5) - A242522(6))/2 = (7 - 5)/2, and A382644(5)/2 - A242522(6) = 6 - 5 = 1
a(6) = a(5) + A242522(6) = 1 + 5 = 6.
		

Crossrefs

Formula

a(n) = A382644(n-1)/2, for n >= 5.
a(n) = (A001266(n) - A242522(n+1))/2, for n >= 5.
a(n) = A382644(n)/2 - A242522(n+1), for n >= 5.
a(n) = a(n-1) + A242522(n), for n >= 6, with a(5) = 1.