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.

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A086272 Rectangular array T(n,k) of central polygonal numbers, by antidiagonals.

Original entry on oeis.org

1, 3, 1, 7, 4, 1, 13, 10, 5, 1, 21, 19, 13, 6, 1, 31, 31, 25, 16, 7, 1, 43, 46, 41, 31, 19, 8, 1, 57, 64, 61, 51, 37, 22, 9, 1, 73, 85, 85, 76, 61, 43, 25, 10, 1, 91, 109, 113, 106, 91, 71, 49, 28, 11, 1, 111, 136, 145, 141, 127, 106, 81, 55, 31, 12, 1, 133, 166, 181, 181, 169
Offset: 1

Views

Author

Clark Kimberling, Jul 14 2003

Keywords

Comments

In the standard notation, the offset is different: the first row are the 2-gonal, the second row the 3-gonal numbers, etc. - R. J. Mathar, Oct 07 2011

Examples

			First rows:
1,3,7,13,21,31,43,57,73,91,111,..   A002061
1,4,10,19,31,46,64,85,109,136,166,...  A005448
1,5,13,25,41,61,85,113,145,181,221,..   A001844
1,6,16,31,51,76,106,141,181,226,276,...  A005891
1,7,19,37,61,91,127,169,217,271,331,...   A003215
1,8,22,43,71,106,148,197,253,316,386,...    A069099
1,9,25,49,81,121,169,225,289,361,441,...    A016754
1,10,28,55,91,136,190,253,325,406,496,...    A060544
		

Crossrefs

Formula

T(k, n)=(k+1)*binomial(n, 2)+1.

A101946 a(n) = 6*2^n - 3*n - 5.

Original entry on oeis.org

1, 4, 13, 34, 79, 172, 361, 742, 1507, 3040, 6109, 12250, 24535, 49108, 98257, 196558, 393163, 786376, 1572805, 3145666, 6291391, 12582844, 25165753, 50331574, 100663219, 201326512, 402653101, 805306282, 1610612647, 3221225380, 6442450849, 12884901790
Offset: 0

Views

Author

Gary W. Adamson, Dec 22 2004

Keywords

Comments

Sequence generated from a 3 X 3 matrix, companion to A101945.
Characteristic polynomial of M = x^3 - 4x^2 + 5x - 2.
Sequence can also be generated by the same method as A061777 with slightly different rules. Refer to A061777, which is the "vertex to vertex" expansion version. For this case, the expandable vertices of the existing generation will contact the sides of the new ones, i.e., "vertex to side" expansion version. Let us assign the label "1" to the triangle at the origin; at n-th generation add a triangle at each expandable vertex, i.e., each vertex where the added generations will not overlap the existing ones, although overlaps among new generations are allowed. The non-overlapping triangles will have the same label value as a predecessor; for the overlapping ones, the label value will be sum of label values of predecessors. a(n) is the sum of all label values at n-th generation. The triangles count is A005448. See illustration. - Kival Ngaokrajang, Sep 26 2014
The number of ways to select 0 or more nodes of a 2 X n rectangular grid such that the selected nodes are connected, but do not fill any 2 X 2 square. This question arises in logic puzzles such as Nurikabe. - Hugo van der Sanden, Feb 22 2024

Examples

			a(4) = 79 = 4*34 - 5*13 + 2*4 = 4*a(3) - 5*a(2) + 2*a(1).
a(4) = right term in M^4 * [1 1 1], since M^4 * [1 1 1] = [1 46 a(4)], where 46 = A033484(4).
		

Crossrefs

Programs

  • Magma
    [6*2^n -3*n-5: n in [0..40]]; // G. C. Greubel, Feb 06 2022
    
  • Mathematica
    a[0]=1; a[1]=4; a[2]=13; a[n_]:= a[n]= 4a[n-1] -5a[n-2] +2a[n-3]; Table[ a[n], {n, 0, 30}] (* Or *)
    a[n_] := (MatrixPower[{{1, 0, 0}, {2, 2, 0}, {1, 2, 1}}, n].{{1}, {1}, {1}})[[3, 1]]; Table[ a[n], {n, 0, 30}] (* Robert G. Wilson v, Jan 12 2005 *)
    Table[6*2^n-3n-5,{n,0,40}] (* or *) LinearRecurrence[{4,-5,2},{1,4,13},40] (* Harvey P. Dale, Jun 03 2017 *)
  • PARI
    a(n) = if (n<1, 1, 5*(2^n-1)+a(n-1))\\ Kival Ngaokrajang, Sep 26 2014
    
  • PARI
    Vec(-(2*x^2+1)/((x-1)^2*(2*x-1)) + O(x^100)) \\ Colin Barker, Sep 26 2014
    
  • Sage
    [3*(2^(n+1) -n-2) +1 for n in (0..40)] # G. C. Greubel, Feb 06 2022

Formula

a(0)=1, a(1)=4, a(2)=13 and for n>2, a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3).
a(n) = right term in M^n * [1 1 1], where M = the 3X3 matrix [1 0 0 / 2 2 0 / 1 2 1]. M^n * [1 1 1] = [1 A033484(n) a(n)].
a(0) = 1, for n >= 1, a(n) = 3*A000225(n) + a(n-1). - Kival Ngaokrajang, Sep 26 2014
G.f.: (1+2*x^2)/((1-x)^2*(1-2*x)). - Colin Barker, Sep 26 2014
E.g.f.: 6*exp(2*x) - (5+3*x)*exp(x). - G. C. Greubel, Feb 06 2022

Extensions

New definition from Ralf Stephan, May 17 2007

A226903 Shiraishi numbers: a parametrized family of solutions c to the Diophantine equation a^3 + b^3 + c^3 = d^3 with d = c+1.

Original entry on oeis.org

5, 18, 53, 102, 197, 306, 491, 684, 989, 1290, 1745, 2178, 2813, 3402, 4247, 5016, 6101, 7074, 8429, 9630, 11285, 12738, 14723, 16452, 18797, 20826, 23561, 25914, 29069, 31770, 35375, 38448, 42533, 46002, 50597, 54486, 59621, 63954, 69659, 74460, 80765, 86058
Offset: 1

Views

Author

Jonathan Sondow, Jun 22 2013

Keywords

Comments

Shiraishi's solutions to a^3 + b^3 + c^3 = d^3 are a = 3n^2; b = 6n^2 - 3n + 1 or 6n^2 + 3n + 1; c = 9n^3 - 6n^2 + 3n - 1 or 9n^3 + 6n^2 + 3n, respectively, for n > 0; and d = c+1. See Smith and Mikami for a derivation.
Shiraishi's formulas show that the sequence is infinite. Hence the sequences A023042 (solutions to x^3 + y^3 + z^3 = w^3), A225908 (solutions to a^3 + b^3 = c^3 - d^3), A225909 (solutions to a^3 + b^3 = (c+1)^3 - c^3) and A226902 (numbers c in A225909) are also infinite.
Shiraishi's solution b = 6n^2 +/- 3n + 1 is the centered triangular numbers A005448 except 1.

Examples

			The first two terms are a(1) = 9 - 6 + 3 - 1 = 5 and a(2) = 9 + 6 + 3 = 18. Then Shiraishi's formulas give 3^3 + 4^3 + 5^3 = 6^3 and 3^3 + 10^3 + 18^3 = 19^3.
		

References

  • Shiraishi Chochu (aka Shiraishi Nagatada), Shamei Sampu (Sacred Mathematics), 1826.

Crossrefs

Formula

a(2n-1) = 9n^3 - 6n^2 + 3n - 1.
a(2n) = 9n^3 + 6n^2 + 3n.
G.f.: x*(5 + 13*x + 20*x^2 + 10*x^3 + 5*x^4 + x^5) / ((1 + x)^3*(1 - x)^4). [Bruno Berselli, Jun 22 2013]
a(n) = (18*n^3 + 27*n^2 + 27*n + 1 - (3*n^2 + 3*n + 1)*(-1)^n)/16. [Bruno Berselli, Jun 22 2013]
a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7) for n > 7. - Chai Wah Wu, Aug 05 2025

A069126 Centered 13-gonal numbers.

Original entry on oeis.org

1, 14, 40, 79, 131, 196, 274, 365, 469, 586, 716, 859, 1015, 1184, 1366, 1561, 1769, 1990, 2224, 2471, 2731, 3004, 3290, 3589, 3901, 4226, 4564, 4915, 5279, 5656, 6046, 6449, 6865, 7294, 7736, 8191, 8659, 9140, 9634, 10141, 10661, 11194
Offset: 1

Views

Author

Terrel Trotter, Jr., Apr 07 2002

Keywords

Comments

Centered tridecagonal numbers or centered triskaidecagonal numbers. - Omar E. Pol, Oct 03 2011

Examples

			a(5) = 131 because 131 = (13*5^2 - 13*5 + 2)/2 = (325 - 65 + 2)/2 = 262/2 = 131.
		

Crossrefs

Programs

  • Mathematica
    FoldList[#1 + #2 &, 1, 13 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
    LinearRecurrence[{3,-3,1},{1,14,40},60] (* Harvey P. Dale, Jan 20 2014 *)
    With[{nn=50},Total/@Thread[{PolygonalNumber[13,Range[nn]],Range[0,nn-1]^2}]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Aug 29 2016 *)
  • PARI
    a(n)=13*n(n-1)/2+1 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = (13n^2 - 13n + 2)/2.
Binomial transform of [1, 13, 13, 0, 0, 0, ...]; Narayana transform (A001263) of [1, 13, 0, 0, 0, ...]. - Gary W. Adamson, Dec 29 2007
a(n) = 13*n+a(n-1)-13 (with a(1)=1). - Vincenzo Librandi, Aug 08 2010
G.f.: -x*(1+11*x+x^2) / (x-1)^3. - R. J. Mathar, Feb 04 2011
a(n) = A152741(n-1) + 1. - Omar E. Pol, Oct 03 2011
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi*tan(sqrt(5/13)*Pi/2)/sqrt(65).
Sum_{n>=1} a(n)/n! = 15*e/2 - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 15/(2*e) - 1. (End)
E.g.f.: exp(x)*(1 + 13*x^2/2) - 1. - Stefano Spezia, May 15 2022

A125602 Centered triangular numbers that are prime.

Original entry on oeis.org

19, 31, 109, 199, 409, 571, 631, 829, 1489, 1999, 2341, 2971, 3529, 4621, 4789, 7039, 7669, 8779, 9721, 10459, 10711, 13681, 14851, 16069, 16381, 17659, 20011, 20359, 23251, 25939, 27541, 29191, 29611, 31321, 34429, 36739, 40099, 40591, 42589
Offset: 1

Views

Author

Zak Seidov, Nov 27 2006

Keywords

Comments

Prime terms in A005448, or primes of the form 3n(n-1)/2 + 1.
Primes that are the sum of 3 consecutive triangular numbers. - Vicente Izquierdo Gomez, Nov 03 2015

Crossrefs

Programs

  • Magma
    [a: n in [0..200] | IsPrime(a) where a is  (3*n^2 - 3*n + 2) div 2]; // Vincenzo Librandi, Mar 22 2013
    
  • Maple
    select(isprime, [seq(3*n*(n-1)/2+1, n=1..1000)]); # Robert Israel, Nov 03 2015
  • Mathematica
    lst={};Do[If[PrimeQ[p=3n(n-1)/2+1], (*Print[p];*)AppendTo[lst, p]], {n, 10^3}];lst (* Vladimir Joseph Stephan Orlovsky, Aug 21 2008 *)
    Select[Total/@Partition[Accumulate[Range[200]],3,1],PrimeQ] (* Harvey P. Dale, Dec 29 2020 *)
  • PARI
    is(n)=n%6==1 && ispolygonal((n-1)/3,3) && isprime(n) \\ Charles R Greathouse IV, Nov 03 2015

A179178 The number of equal-sized equilateral triangles visible (when viewed from above) in successive Genealodrons formed from 2^n -1 same size equilateral triangles.

Original entry on oeis.org

1, 3, 7, 14, 25, 40, 57, 77, 100, 126, 155, 187, 222, 260, 301, 345, 392, 442, 495, 551, 610, 672, 737, 805, 876, 950, 1027, 1107, 1190, 1276, 1365, 1457, 1552, 1650, 1751, 1855, 1962, 2072, 2185, 2301, 2420, 2542, 2667, 2795, 2926, 3060, 3197, 3337, 3480
Offset: 1

Views

Author

Elizabeth Hignell (elizabethhignell(AT)hotmail.com), Jun 30 2010

Keywords

Comments

A Genealodron represents a rooted binary tree and is composed of equal sized equilateral triangles. One edge of each triangle is attached to its parent and the other two to its child trees. The first triangle, which is the root of the tree, has a designated unattached parent edge. Triangles may overlap as needed.
The first Genealodron consists of one equilateral triangle.
The second Genealodron is formed by joining another same size equilateral triangle to the left edge and to the right edge of the first so that the second Genealodron is made up of three triangles.
The third Genealodron is formed by joining same size equilateral triangles to the left and right edges of both the second and third triangle of the second Genealodron so that the third Genealodron is made up of seven triangles.
The fourth Genealodron is formed by joining same size equilateral triangles to the left and right edges of the fourth, fifth, sixth and seventh triangles of the third Genealodron so that the fourth Genealodron has fifteen triangles. The fourth Genealodron has the first overlap so although it contains 15 triangles only 14 are seen when it is viewed from above.
The fifth Genealodron is formed by adding 16 more triangles to the edges of last eight triangles added to the fourth Genealodron so the fifth Genealodron has 31 triangles, only 25 of which are seen when it is viewed from above because of the increasing number of overlaps.
The sixth Genealodron has 63 triangles only 40 of which are visible.
Gradually within the Genealodron spirals (which are hexagonal in cross-section) are building counterclockwise on the left hand edge of every triangle and clockwise on the right hand edge of every triangle. Because of the way the triangles form into hexagonal stacks although the total number of triangles in successive Genealodrons is 2^n - 1 the rate at which the number of visible triangles increases becomes arithmetic with a common difference of 3.
A Genealodron formed from an infinite number of same size equilateral triangles creates a hyperbolic plane.
Also, the crystal ball sequence for the honeycomb point lattice with a single edge removed at the origin. Without this removal the sequence would be A005448. The sixth Genealodron when viewed from above has the shape of a hexagon (see illustration of initial terms). All subsequent generations will retain this shape and so the sequence becomes A005449, the second pentagonal numbers. - Andrew Howroyd, Mar 24 2016

Crossrefs

Cf. A179316.

Programs

  • Mathematica
    CoefficientList[Series[x (1 + x^2 + x^3 + x^4 - 2 x^6 + x^7)/(1 - x)^3, {x, 0, 49}], x] (* Michael De Vlieger, Mar 26 2016 *)
  • PARI
    Vec(x*(1+x^2+x^3+x^4-2*x^6+x^7)/(1-x)^3 + O(x^50)) \\ Colin Barker, Mar 26 2016

Formula

a(n) = A005449(n-1) for n>=6. - Andrew Howroyd, Mar 24 2016
From Colin Barker, Mar 26 2016: (Start)
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>5.
a(n) = (2-5*n+3*n^2)/2 for n>5.
G.f.: x*(1+x^2+x^3+x^4-2*x^6+x^7) / (1-x)^3.
(End)

Extensions

a(12)-a(40) from Andrew Howroyd, Mar 24 2016
Definition improved by Andrew Howroyd, Apr 16 2020

A262221 a(n) = 25*n*(n + 1)/2 + 1.

Original entry on oeis.org

1, 26, 76, 151, 251, 376, 526, 701, 901, 1126, 1376, 1651, 1951, 2276, 2626, 3001, 3401, 3826, 4276, 4751, 5251, 5776, 6326, 6901, 7501, 8126, 8776, 9451, 10151, 10876, 11626, 12401, 13201, 14026, 14876, 15751, 16651, 17576, 18526, 19501, 20501, 21526, 22576, 23651
Offset: 0

Views

Author

Bruno Berselli, Sep 15 2015

Keywords

Comments

Also centered 25-gonal (or icosipentagonal) numbers.
This is the case k=25 of the formula (k*n*(n+1) - (-1)^k + 1)/2. See table in Links section for similar sequences.
For k=2*n, the formula shown above gives A011379.
Primes in sequence: 151, 251, 701, 1951, 3001, 4751, 10151, 12401, ...

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 51 (23rd row of the table).

Crossrefs

Cf. centered polygonal numbers listed in A069190.
Similar sequences of the form (k*n*(n+1) - (-1)^k + 1)/2 with -1 <= k <= 26: A000004, A000124, A002378, A005448, A005891, A028896, A033996, A035008, A046092, A049598, A060544, A064200, A069099, A069125, A069126, A069128, A069130, A069132, A069174, A069178, A080956, A124080, A163756, A163758, A163761, A164136, A173307.

Programs

  • Magma
    [25*n*(n+1)/2+1: n in [0..50]];
  • Mathematica
    Table[25 n (n + 1)/2 + 1, {n, 0, 50}]
    25*Accumulate[Range[0,50]]+1 (* or *) LinearRecurrence[{3,-3,1},{1,26,76},50] (* Harvey P. Dale, Jan 29 2023 *)
  • PARI
    vector(50, n, n--; 25*n*(n+1)/2+1)
    
  • Sage
    [25*n*(n+1)/2+1 for n in (0..50)]
    

Formula

G.f.: (1 + 23*x + x^2)/(1 - x)^3.
a(n) = a(-n-1) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A123296(n) + 1.
a(n) = A000217(5*n+2) - 2.
a(n) = A034856(5*n+1).
a(n) = A186349(10*n+1).
a(n) = A054254(5*n+2) with n>0, a(0)=1.
a(n) = A000217(n+1) + 23*A000217(n) + A000217(n-1) with A000217(-1)=0.
Sum_{i>=0} 1/a(i) = 1.078209111... = 2*Pi*tan(Pi*sqrt(17)/10)/(5*sqrt(17)).
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=0} a(n)/n! = 77*e/2.
Sum_{n>=0} (-1)^(n+1) * a(n)/n! = 23/(2*e). (End)
E.g.f.: exp(x)*(2 + 50*x + 25*x^2)/2. - Elmo R. Oliveira, Dec 24 2024

A280950 Expansion of Product_{k>=0} 1/(1 - x^(3*k*(k+1)/2+1)).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 9, 11, 11, 12, 13, 15, 15, 16, 17, 19, 20, 22, 24, 26, 27, 29, 31, 33, 34, 37, 40, 43, 45, 48, 51, 54, 56, 60, 63, 67, 70, 76, 80, 84, 87, 93, 97, 102, 106, 113, 118, 125, 130, 138, 143, 151, 157, 166, 172, 181, 189, 200, 207, 217, 225, 237, 245, 257, 267, 280
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 11 2017

Keywords

Comments

Number of partitions of n into centered triangular numbers (A005448).

Examples

			a(8) = 3 because we have [4, 4], [4, 1, 1, 1, 1] and [1, 1, 1, 1, 1, 1, 1, 1].
		

Crossrefs

Programs

  • Maple
    N:= 100:
    kmax:= floor((sqrt(24*N-15)-3)/6):
    S:= series(mul(1/(1-x^(3*k*(k+1)/2+1)),k=0..kmax),x,N+1):
    seq(coeff(S,x,j),j=0..N); # Robert Israel, Jan 25 2017
  • Mathematica
    nmax = 78; CoefficientList[Series[Product[1/(1 - x^(3 k (k + 1)/2 + 1)), {k, 0, nmax}], {x, 0, nmax}], x]

Formula

G.f.: Product_{k>=0} 1/(1 - x^(3*k*(k+1)/2+1)).

A322908 The determinant of an n X n Toeplitz matrix M(n) whose first row consists of successive positive integer numbers 1, ..., n and whose first column consists of 1, n + 1, ..., 2*n - 1.

Original entry on oeis.org

1, -5, 38, -386, 4928, -75927, 1371808, -28452356, 666445568, -17402398505, 501297595904, -15792876550662, 540190822408192, -19937252888438459, 789770307546718208, -33422580292067020808, 1504926927960887066624, -71839548181524098808909, 3624029163661165580910592
Offset: 1

Views

Author

Stefano Spezia, Dec 30 2018

Keywords

Comments

The matrix M(n) differs from that of A318173 in using successive positive integers in place of successive prime numbers.
The trace of the matrix M(n) is A000027(n).
The sum of the first row of the matrix M(n) is A000217(n).
The sum of the first column of the matrix M(n) is A005448(n). [Corrected by Stefano Spezia, Dec 11 2019]
For n > 1, the sum of the superdiagonal of the matrix M(n) is A005843(n).

Examples

			For n = 1 the matrix M(1) is
   1
with determinant Det(M(1)) = 1.
For n = 2 the matrix M(2) is
   1, 2
   3, 1
with Det(M(2)) = -5.
For n = 3 the matrix M(3) is
   1, 2, 3
   4, 1, 2
   5, 4, 1
with Det(M(3)) = 38.
		

Crossrefs

Cf. A322909 (permanent of matrix M(n)).

Programs

  • Maple
    a:= proc(n) uses LinearAlgebra;
    Determinant(ToeplitzMatrix([seq(i, i=2*n-1..n+1, -1), seq(i, i=1..n)]))
    end proc:
    map(a, [$1..20]);
  • Mathematica
    b[n_]:=n; a[n_]:=Det[ToeplitzMatrix[Join[{b[1]}, Array[b, n-1, {n+1, 2*n-1}]], Array[b, n]]]; Array[a, 20]
  • PARI
    tm(n) = {my(m = matrix(n, n, i, j, if (i==1, j, if (j==1, n+i-1)))); for (i=2, n, for (j=2, n, m[i, j] = m[i-1, j-1]; ); ); m; }
    a(n) = matdet(tm(n)); \\ Michel Marcus, Nov 11 2020

Formula

a(n) ~ -(-1)^n * (3*exp(1) - exp(-1)) * n^n / 4. - Vaclav Kotesovec, Jan 05 2019

A001922 Numbers k such that 3*k^2 - 3*k + 1 is both a square (A000290) and a centered hexagonal number (A003215).

Original entry on oeis.org

1, 8, 105, 1456, 20273, 282360, 3932761, 54776288, 762935265, 10626317416, 148005508553, 2061450802320, 28712305723921, 399910829332568, 5570039304932025, 77580639439715776, 1080558912851088833, 15050244140475527880, 209622859053806301481
Offset: 0

Views

Author

Keywords

Comments

Also larger of two consecutive integers whose cubes differ by a square. Defined by a(n)^3 - (a(n) - 1)^3 = square.
Let m be the n-th ratio 2/1, 7/4, 26/15, 97/56, 362/209, ... Then a(n) = m*(2-m)/(m^2-3). The numerators 2, 7, 26, ... of m are A001075. The denominators 1, 4, 15, ... of m are A001353.
From Colin Barker, Jan 06 2015: (Start)
Also indices of centered triangular numbers (A005448) which are also centered square numbers (A001844).
Also indices of centered hexagonal numbers (A003215) which are also centered octagonal numbers (A016754).
Also positive integers x in the solutions to 3*x^2 - 4*y^2 - 3*x + 4*y = 0, the corresponding values of y being A156712.
(End)

Examples

			8 is in the sequence because 3*8^2 - 3*8 + 1 = 169 is a square and also a centered hexagonal number. - _Colin Barker_, Jan 07 2015
		

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    I:=[1, 8, 105]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Apr 16 2012
    
  • Maple
    seq(simplify((1 +ChebyshevU(n,7) +ChebyshevU(n-1,7))/2), n=0..30); # G. C. Greubel, Oct 07 2022
  • Mathematica
    With[{s1=3+2Sqrt[3],s2=3-2Sqrt[3],t1=7+4Sqrt[3],t2=7-4Sqrt[3]}, Simplify[ Table[(s1 t1^n+s2 t2^n+6)/12,{n,0,20}]]] (* or *) LinearRecurrence[ {15,-15,1},{1,8,105},21] (* Harvey P. Dale, Aug 14 2011 *)
    CoefficientList[Series[(1-7*x)/(1-15*x+15*x^2-x^3),{x,0,30}],x] (* Vincenzo Librandi, Apr 16 2012 *)
  • PARI
    Vec((1-7*x)/(1-15*x+15*x^2-x^3) + O(x^100)) \\ Colin Barker, Jan 06 2015
    
  • SageMath
    [(1+chebyshev_U(n,7) +chebyshev_U(n-1,7))/2 for n in range(30)] # G. C. Greubel, Oct 07 2022

Formula

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = (s1*t1^n + s2*t2^n + 6)/12 where s1 = 3 + 2*sqrt(3), s2 = 3 - 2*sqrt(3), t1 = 7 + 4*sqrt(3), t2 = 7 - 4*sqrt(3).
a(n) = A001075(n)*A001353(n+1).
G.f.: (1-7*x)/((1-x)*(1-14*x+x^2)). - Simon Plouffe (in his 1992 dissertation) and Colin Barker, Jan 01 2012
a(n) = A076139(n+1) - 7*A076139(n). - R. J. Mathar, Jul 14 2015
a(n) = (1/2)*(1 + ChebyshevU(n, 7) + ChebyshevU(n-1, 7)). G. C. Greubel, Oct 07 2022
a(n) = 1 - a(-1-n) = 1 + A001921(n) for all integers n. - Michael Somos, Jul 10 2025

Extensions

Additional comments from James R. Buddenhagen, Mar 04 2001
Name improved by Colin Barker, Jan 07 2015
Edited by Robert Israel, Feb 20 2017
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