A091998 -id:A091998 - cp's OEIS frontend

A175885 Numbers that are congruent to {1, 10} mod 11.

1, 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87, 89, 98, 100, 109, 111, 120, 122, 131, 133, 142, 144, 153, 155, 164, 166, 175, 177, 186, 188, 197, 199, 208, 210, 219, 221, 230, 232, 241, 243, 252, 254, 263, 265, 274, 276, 285, 287, 296, 298

Offset: 1

Bruno Berselli, Oct 08 2010 - Nov 17 2010

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n + (h-4)*(-1)^n - h)/4 (h, n natural numbers), therefore ((2*h*n + (h-4)*(-1)^n - h)/4)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 11).

Bruno Berselli, Table of n, a(n) for n = 1..10000.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A090771 (n==1 or 9 mod 10), A091998 (n==1 or 11 mod 12).
Cf. A195043 (partial sums).
Cf. A000217, A005408, A047209, A007310, A047336, A047522, A056020, A113801, A175886, A175887, A195312, A195313.

a175885 n = a175885_list !! (n-1)
a175885_list = 1 : 10 : map (+ 11) a175885_list
-- Reinhard Zumkeller, Jan 07 2012

[(22*n+7*(-1)^n-11)/4: n in [1..60]]; // Vincenzo Librandi, Sep 19 2011

Rest[Flatten[{#-1,#+1}&/@(11 Range[0,50])]] (* Harvey P. Dale, Nov 05 2010 *)

a(n)=n%2*9 + 1 \\ Charles R Greathouse IV, Aug 01 2016

G.f.: x*(1+9*x+x^2)/((1+x)*(1-x)^2).
a(n) = (22*n + 7*(-1)^n - 11)/4.
a(n) = -a(-n+1) = a(n-2) + 11 = a(n-1) + a(n-2) - a(n-3).
a(n) = 11*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i) for n > 1.
a(n) = A195312(n) + A195312(n-1) = A195313(n) - A195313(n-2). - Bruno Berselli, Sep 18 2011
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/11)*cot(Pi/11). - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((22*x - 11)*exp(x) + 7*exp(-x))/4. - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/11).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/11)*cosec(Pi/11). (End)

A195143 a(n) = n-th concentric 12-gonal number.

Original entry on oeis.org

0, 1, 12, 25, 48, 73, 108, 145, 192, 241, 300, 361, 432, 505, 588, 673, 768, 865, 972, 1081, 1200, 1321, 1452, 1585, 1728, 1873, 2028, 2185, 2352, 2521, 2700, 2881, 3072, 3265, 3468, 3673, 3888, 4105, 4332, 4561, 4800, 5041, 5292, 5545, 5808, 6073, 6348

Offset: 0

Omar E. Pol, Sep 17 2011

Concentric dodecagonal numbers. [corrected by Ivan Panchenko, Nov 09 2013]
Sequence found by reading the line from 0, in the direction 0, 12,..., and the same line from 1, in the direction 1, 25,..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Main axis, perpendicular to A028896 in the same spiral.
Partial sums of A091998. - Reinhard Zumkeller, Jan 07 2012
Column 12 of A195040. - Omar E. Pol, Sep 28 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

A135453 and A069190 interleaved.
Cf. A001082, A032527, A032528, A077221, A195043, A195045, A195040, A195142, A195145, A195146, A195147, A195148, A195149.
Cf. A016921 (6n+1), A016969 (6n+5), A091998 (positive integers of the form 12*k +- 1), A092242 (positive integers of the form 12*k +- 5).

a195143 n = a195143_list !! n
a195143_list = scanl (+) 0 a091998_list
-- Reinhard Zumkeller, Jan 07 2012

[(3*n^2+(-1)^n-1): n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

Table[Sum[2*(-1)^(n - k + 1) + 6*k - 3, {k, n}], {n, 0, 47}] (* L. Edson Jeffery, Sep 14 2014 *)

From Vincenzo Librandi, Sep 27 2011: (Start)
a(n) = 3*n^2+(-1)^n-1.
a(n) = -a(n-1) + 6*n^2 - 6*n + 1. (End)
G.f.: -x*(1+10*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
a(n) = Sum_{k=1..n} (2*(-1)^(n-k+1) + 3*(2*k-1)), n>0, a(0) = 0. - L. Edson Jeffery, Sep 14 2014
Sum_{n>=1} 1/a(n) = Pi^2/72 + tan(Pi/sqrt(6))*Pi/(4*sqrt(6)). - Amiram Eldar, Jan 16 2023

A175886 Numbers that are congruent to {1, 12} mod 13.

Original entry on oeis.org

1, 12, 14, 25, 27, 38, 40, 51, 53, 64, 66, 77, 79, 90, 92, 103, 105, 116, 118, 129, 131, 142, 144, 155, 157, 168, 170, 181, 183, 194, 196, 207, 209, 220, 222, 233, 235, 246, 248, 259, 261, 272, 274, 285, 287, 298, 300, 311, 313, 324, 326, 337, 339, 350

Offset: 1

Bruno Berselli, Oct 08 2010 - Nov 17 2010

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2-1 == 0 (mod 13).

Bruno Berselli, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000217, A091998, A113801, A005408, A047209, A007310, A047336, A047522, A056020, A090771, A175885, A175887.

Cf. A195045 (partial sums).

a175886 n = a175886_list !! (n-1)
a175886_list = 1 : 12 : map (+ 13) a175886_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..350] | n mod 13 in [1, 12]]; // Bruno Berselli, Feb 29 2012

[(26*n+9*(-1)^n-13)/4: n in [1..55]]; // Vincenzo Librandi, Aug 19 2013

Select[Range[1, 350], MemberQ[{1, 12}, Mod[#, 13]]&] (* Bruno Berselli, Feb 29 2012 *)
CoefficientList[Series[(1 + 11 x + x^2) / ((1 + x) (1 - x)^2), {x, 0, 55}], x] (* Vincenzo Librandi, Aug 19 2013 *)
LinearRecurrence[{1,1,-1},{1,12,14},60] (* Harvey P. Dale, Oct 23 2015 *)

a(n)=(26*n+9*(-1)^n-13)/4 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(1+11*x+x^2)/((1+x)*(1-x)^2).
a(n) = (26*n+9*(-1)^n-13)/4.
a(n) = -a(-n+1) = a(n-1)+a(n-2)-a(n-3).
a(n) = a(n-2)+13.
a(n) = 13*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n>1.
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/13)*cot(Pi/13). - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((26*x - 13)*exp(x) + 9*exp(-x))/4. - David Lovler, Sep 04 2022
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = 2*cos(Pi/13).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/13)*cosec(Pi/13). (End)

A175887 Numbers that are congruent to {1, 14} mod 15.

Original entry on oeis.org

1, 14, 16, 29, 31, 44, 46, 59, 61, 74, 76, 89, 91, 104, 106, 119, 121, 134, 136, 149, 151, 164, 166, 179, 181, 194, 196, 209, 211, 224, 226, 239, 241, 254, 256, 269, 271, 284, 286, 299, 301, 314, 316, 329, 331, 344, 346, 359, 361, 374, 376, 389, 391, 404

Offset: 1

Bruno Berselli, Oct 08 2010 - Nov 17 2010

Cf. property described by Gary Detlefs in A113801: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1==0 (mod h); in this case, a(n)^2-1==0 (mod 15).

Bruno Berselli, Table of n, a(n) for n = 1..10000.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000217, A019693, A019976, A113801, A175886, A091998, A175885, A090771, A056020, A047522, A047336, A007310, A047209, A005408, A001651.

Cf. A132240 (primes).

a175887 n = a175887_list !! (n-1)
a175887_list = 1 : 14 : map (+ 15) a175887_list
-- Reinhard Zumkeller, Jan 07 2012

[n: n in [1..450] | n mod 15 in [1,14]];

[(30*n+11*(-1)^n-15)/4: n in [1..55]]; // Vincenzo Librandi, Aug 19 2013

Select[Range[1, 450], MemberQ[{1,14}, Mod[#, 15]]&]
CoefficientList[Series[(1 + 13 x + x^2) / ((1 + x) (1 - x)^2), {x, 0, 55}], x] (* Vincenzo Librandi, Aug 19 2013 *)

a(n)=(30*n+11*(-1)^n-15)/4 \\ Charles R Greathouse IV, Sep 28 2015

G.f.: x*(1+13*x+x^2)/((1+x)*(1-x)^2).
a(n) = (30*n+11*(-1)^n-15)/4.
a(n) = -a(-n+1) = a(n-1)+a(n-2)-a(n-3).
a(n) = 15*A000217(n-1) -2*sum(a(i), i=1..n-1) +1 for n>1.
a(n) = A047209(A047225(n+1)).
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi/15)*cot(Pi/15) = A019693 * A019976 / 10. - Amiram Eldar, Dec 04 2021
E.g.f.: 1 + ((30*x - 15)*exp(x) + 11*exp(-x))/4. - David Lovler, Sep 05 2022
From Amiram Eldar, Nov 25 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = (Pi/15)*cosec(Pi/15).
Product_{n>=2} (1 + (-1)^n/a(n)) = 2*cos(Pi/15). (End)

A188887 Decimal expansion of sqrt(2 + sqrt(3)).

Original entry on oeis.org

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

Offset: 1

Clark Kimberling, Apr 12 2011

Decimal expansion of the length/width ratio of a sqrt(2)-extension rectangle.  See A188640 for definitions of shape and r-extension rectangle.
A sqrt(2)-extension rectangle matches the continued fraction [1,1,13,1,2,15,10,1,18,1,1,21,,...] (A188888) for the shape L/W = sqrt(2 + sqrt(3)). This is analogous to the matching of a golden rectangle to the continued fraction [1,1,1,1,1,1,1,1,...].  Specifically, for the sqrt(2)-extension rectangle, 1 square is removed first, then 1 square, then 13 squares, then 1 square, ..., so that the original rectangle of shape sqrt(2 + sqrt(3)) is partitioned into an infinite collection of squares.
sqrt(2 + sqrt(3)) is also the shape of the greater sqrt(6)-contraction rectangle; see A188738.
This constant is also the length of the Steiner span of three vertices of a unit square. - Jean-François Alcover, May 22 2014
It is also the larger positive coordinate of (symmetrical) intersection points created by x^2 + y^2 = 4 circle and y = 1/x hyperbola. The smaller coordinate is A101263. - Leszek Lezniak, Sep 18 2018
Length of the shortest diagonal in a regular 12-gon with unit side. - Mohammed Yaseen, Nov 12 2020

			1.931851652578136573499486399457794735267809678016809...

G. C. Greubel, Table of n, a(n) for n = 1..10000
Burkard Polster, Irrational roots, Mathologer video (2018).
Index entries for algebraic numbers, degree 4.

Cf. A019884, A019973, A091998, A101263, A188640, A188888, A214726, A217870, A329247.

Sqrt(2 + Sqrt(3)); // G. C. Greubel, Apr 10 2018

r = 2^(1/2); t = (r + (4 + r^2)^(1/2))/2; FullSimplify[t]
N[t, 130]
RealDigits[N[t, 130]][[1]]
ContinuedFraction[t, 120]
RealDigits[Sqrt[2 + Sqrt[3]], 10, 100][[1]] (* G. C. Greubel, Apr 10 2018 *)

sqrt(2 + sqrt(3)) \\ G. C. Greubel, Apr 10 2018

Equals (sqrt(6) + sqrt(2))/2.
Equals exp(asinh(cos(Pi/4))). - Geoffrey Caveney, Apr 23 2014
Equals cos(Pi/4) + sqrt(1 + cos(Pi/4)^2). - Geoffrey Caveney, Apr 23 2014
Equals i^(1/6) + i^(-1/6). - Gary W. Adamson, Jul 07 2022
Equals the largest root of x - 1/x = sqrt(2) and of x^2 + 1/x^2 = 4. - Gary W. Adamson, Jun 12 2023
Equals Product_{k>=0} ((12*k + 2)*(12*k + 10))/((12*k + 1)*(12*k + 11)). - Antonio Graciá Llorente, Feb 24 2024
From Amiram Eldar, Nov 23 2024: (Start)
Equals A214726 / 2 = 2 * A019884 = 1 / A101263 = exp(A329247) = A217870^2 = sqrt(A019973).
Equals Product_{k>=1} (1 - (-1)^k/A091998(k)). (End)

A092260 Permutation of natural numbers generated by 6-rowed array shown below.

Original entry on oeis.org

1, 11, 2, 13, 10, 3, 23, 14, 9, 4, 25, 22, 15, 8, 5, 35, 26, 21, 16, 7, 6, 37, 34, 27, 20, 17, 12, 47, 38, 33, 28, 19, 18, 49, 46, 39, 32, 29, 24, 59, 50, 45, 40, 31, 30, 61, 58, 51, 44, 41, 36, 71, 62, 57, 52, 43, 42, 73, 70, 63, 56, 53, 48, 83, 74, 69, 64, 55, 54, 85, 82, 75

Offset: 1

Giovanni Teofilatto, Feb 19 2004

nonn

11 13 23 25 35 37 47 49 59... (A091998)
10 14 22 26 34 38 46 50 58... (A091999)
9 15 21 27 33 39 45 51 57... (A016945)
8 16 20 28 32 40 44 52 56... (A092259)
7 17 19 29 31 41 43 53 55... (A092242)
12 18 24 30 36 42 48 54 60... (A008588, excluding initial term)
For such arrays A_k, here A_6, see a W. Lang comment on A113807, the A_7 case. However, to get the array A_6 one should take the last line as the first one and add a 0 in front (thus obtaining a permutation of the nonnegative integers). - Wolfdieter Lang, Feb 02 2012

Cf. A088520, A090243, A090298, A113807.

Edited and extended by Ray Chandler, Feb 21 2004

A249916 a(n) = 4*(n - 1) - a(n-3), n >= 3, a(0) = a(1) = 1, a(2) = 5.

Original entry on oeis.org

1, 1, 5, 7, 11, 11, 13, 13, 17, 19, 23, 23, 25, 25, 29, 31, 35, 35, 37, 37, 41, 43, 47, 47, 49, 49, 53, 55, 59, 59, 61, 61, 65, 67, 71, 71, 73, 73, 77, 79, 83, 83, 85, 85, 89, 91, 95, 95, 97, 97, 101, 103, 107, 107, 109, 109, 113, 115, 119, 119, 121, 121, 125, 127

Offset: 0

L. Edson Jeffery, Jan 14 2015

Conjecture: These are the natural numbers of the form 6*j +- 1 in which those of the form 12*k +- 1 are repeated.
From Jianing Song, Jan 28 2019: (Start)
The second conjecture in Formula section is correct.
We can see this from the recurrence: a(n) = 4*(n - 1) - a(n-3)            (1)
Replace n with n+1:  a(n+1) = 4*n - a(n-2)                                (2)
Subtract (1) from (2): a(n+1) = 4 + a(n) - a(n-2) + a(n-3)                (3)
Replace n with n+1:  a(n+2) = 4 + a(n+1) - a(n-1) + a(n-2)                (4)
Subtract (3) from (4): a(n+2) = 2*a(n+1)-a(n)-a(n-1)+2*a(n-2)-a(n-3)      (5)
This also confirms the conjecture in Comment section and the conjecture on the g.f. in Formula section. (End)

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

Cf. A007310, A091998.

a[0] = a[1] = 1; a[2] = 5; a[n_] := 4*(n - 1) - a[n - 3]; Table[a[n], {n, 0, 63}]
RecurrenceTable[{a[0]==a[1]==1,a[2]==5,a[n]==4(n-1)-a[n-3]},a,{n,70}] (* Harvey P. Dale, Jan 26 2019 *)

G.f.: (1 - x + 4*x^2 - x^3 + x^4)/((1 - x)^2*(1 + x^3)). [Confirmed, see Jianing Song in Comment section.]

Recurrence: a(n) = 2*a(n-1) - a(n-2) - a(n-3) + 2*a(n-4) - a(n-5) for n > 4, a(0)=a(1)=1, a(2)=5, a(3)=7, a(4)=11. [Confirmed, see Jianing Song in Comment section.]

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A175885 Numbers that are congruent to {1, 10} mod 11.

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A195143 a(n) = n-th concentric 12-gonal number.

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A175886 Numbers that are congruent to {1, 12} mod 13.

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A175887 Numbers that are congruent to {1, 14} mod 15.

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A188887 Decimal expansion of sqrt(2 + sqrt(3)).

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A092260 Permutation of natural numbers generated by 6-rowed array shown below.

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A249916 a(n) = 4*(n - 1) - a(n-3), n >= 3, a(0) = a(1) = 1, a(2) = 5.

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