A102871 -id:A102871 - cp's OEIS frontend

A001570 Numbers k such that k^2 is centered hexagonal.

1, 13, 181, 2521, 35113, 489061, 6811741, 94875313, 1321442641, 18405321661, 256353060613, 3570537526921, 49731172316281, 692665874901013, 9647591076297901, 134373609193269601, 1871582937629476513, 26067787517619401581, 363077442309042145621

Offset: 1

N. J. A. Sloane

Chebyshev T-sequence with Diophantine property. - Wolfdieter Lang, Nov 29 2002
a(n) = L(n,14), where L is defined as in A108299; see also A028230 for L(n,-14). - Reinhard Zumkeller, Jun 01 2005
Numbers x satisfying x^2 + y^3 = (y+1)^3. Corresponding y given by A001921(n)={A028230(n)-1}/2. - Lekraj Beedassy, Jul 21 2006
Mod[ a(n), 12 ] = 1. (a(n) - 1)/12 = A076139(n) = Triangular numbers that are one-third of another triangular number. (a(n) - 1)/4 = A076140(n) = Triangular numbers T(k) that are three times another triangular number. - Alexander Adamchuk, Apr 06 2007
Also numbers n such that RootMeanSquare(1,3,...,2*n-1) is an integer. - Ctibor O. Zizka, Sep 04 2008
a(n), with n>1, is the length of the cevian of equilateral triangle whose side length is the term b(n) of the sequence A028230. This cevian divides the side (2*x+1) of the triangle in two integer segments x and x+1. - Giacomo Fecondo, Oct 09 2010
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(12)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Beal's conjecture would imply that set intersection of this sequence with the perfect powers (A001597) equals {1}. In other words, existence of a nontrivial perfect power in this sequence would disprove Beal's conjecture. - Max Alekseyev, Mar 15 2015
Numbers n such that there exists positive x with x^2 + x + 1 = 3n^2. - Jeffrey Shallit, Dec 11 2017
Given by the denominators of the continued fractions [1,(1,2)^i,3,(1,2)^{i-1},1]. - Jeffrey Shallit, Dec 11 2017
A near-isosceles integer-sided triangle with an angle of 2*Pi/3 is a triangle whose sides (a, a+1, c) satisfy Diophantine equation (a+1)^3 - a^3 = c^2. For n >= 2, the largest side c is given by a(n) while smallest and middle sides (a, a+1) = (A001921(n-1), A001922(n-1)) (see Julia link). - Bernard Schott, Nov 20 2022

			G.f. = x + 13*x^2 + 181*x^3 + 2521*x^4 + 35113*x^5 + 489061*x^6 + 6811741*x^7 + ...

E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - N. J. A. Sloane, Mar 03 2022
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).

G. C. Greubel, Table of n, a(n) for n = 1..870 (terms 1..101 from T. D. Noe)
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
G. Julia, Triangles dont un angle mesure 120 degrés, Problème Capes, part C (in French).
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Sociedad Magic Penny Patagonia, Leonardo en Patagonia
V. Thebault, Consecutive cubes with difference a square, Amer. Math. Monthly, 56 (1949), 174-175.
Eric Weisstein's World of Mathematics, Hex Number
Wikipedia, Beal's conjecture
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Bisection of A003500/4. Cf. A006051, A001921, A001922.
One half of odd part of bisection of A001075. First differences of A007655.
Cf. A077417 with companion A077416.
Row 14 of array A094954.
Cf. A076139, A076140, A102871.
A122571 is another version of the same sequence.
Row 2 of array A188646.
Cf. similar sequences listed in A238379.
Cf. A028231, which gives the corresponding values of x in 3n^2 = x^2 + x + 1.
Similar sequences of the type cosh((2*m+1)*arccosh(k))/k are listed in A302329. This is the case k=2.

[((2 + Sqrt(3))^(2*n - 1) + (2 - Sqrt(3))^(2*n - 1))/4: n in [1..50]]; // G. C. Greubel, Nov 04 2017

A001570:=-(-1+z)/(1-14*z+z**2); # Simon Plouffe in his 1992 dissertation.

NestList[3 + 7*#1 + 4*Sqrt[1 + 3*#1 + 3*#1^2] &, 0, 24] (* Zak Seidov, May 06 2007 *)
f[n_] := Simplify[(2 + Sqrt@3)^(2 n - 1) + (2 - Sqrt@3)^(2 n - 1)]/4; Array[f, 19] (* Robert G. Wilson v, Oct 28 2010 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
  ] (* Complement of A041017 *)
a[12, 20] (* Gerry Martens, Jun 07 2015 *)
LinearRecurrence[{14, -1}, {1, 13}, 19] (* Jean-François Alcover, Sep 26 2017 *)
CoefficientList[Series[x (1-x)/(1-14x+x^2),{x,0,20}],x] (* Harvey P. Dale, Sep 18 2024 *)

{a(n) = real( (2 + quadgen( 12)) ^ (2*n - 1)) / 2}; /* Michael Somos, Feb 15 2011 */

a(n) = ((2 + sqrt(3))^(2*n - 1) + (2 - sqrt(3))^(2*n - 1)) / 4. - Michael Somos, Feb 15 2011
G.f.: x * (1 - x) / (1 -14*x + x^2). - Michael Somos, Feb 15 2011
Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i) then a(n) = q(n, 12). - Benoit Cloitre, Dec 10 2002
a(n) = S(n, 14) - S(n-1, 14) = T(2*n+1, 2)/2 with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 14)=A007655(n+1) and T(n, 2)=A001075(n). - Wolfdieter Lang, Nov 29 2002
a(n) = A001075(n)*A001075(n+1) - 1 and thus (a(n)+1)^6 has divisors A001075(n)^6 and A001075(n+1)^6 congruent to -1 modulo a(n) (cf. A350916). - Max Alekseyev, Jan 23 2022
4*a(n)^2 - 3*b(n)^2 = 1 with b(n)=A028230(n+1), n>=0.
a(n)*a(n+3) = 168 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = 14*a(n-1) - a(n-2), a(0) = a(1) = 1. a(1 - n) = a(n) (compare A122571).
a(n) = 12*A076139(n) + 1 = 4*A076140(n) + 1. - Alexander Adamchuk, Apr 06 2007
a(n) = (1/12)*((7-4*sqrt(3))^n*(3-2*sqrt(3))+(3+2*sqrt(3))*(7+4*sqrt(3))^n -6). - Zak Seidov, May 06 2007
a(n) = A102871(n)^2+(A102871(n)-1)^2; sum of consecutive squares. E.g. a(4)=36^2+35^2. - Mason Withers (mwithers(AT)semprautilities.com), Jan 26 2008
a(n) = sqrt((3*A028230(n+1)^2 + 1)/4).
a(n) = A098301(n+1) - A001353(n)*A001835(n).
a(n) = A000217(A001571(n-1)) + A000217(A133161(n)), n>=1. - Ivan N. Ianakiev, Sep 24 2013
a(n)^2 = A001922(n-1)^3 - A001921(n-1)^3, for n >= 1. - Bernard Schott, Nov 20 2022
a(n) = 2^(2*n-3)*Product_{k=1..2*n-1} (2 - sin(2*Pi*k/(2*n-1))). Michael Somos, Dec 18 2022
a(n) = A003154(A101265(n)). - Andrea Pinos, Dec 19 2022

A109437 a(-1) = a(0) = 0, a(1) = 1; a(n) = 5a(n-1) - 5a(n-2) + a(n-3) + 2*(-1)^(n+1), alternatively a(n) = 3a(n-1) + 3a(n-2) - a(n-3).

Original entry on oeis.org

0, 1, 3, 12, 44, 165, 615, 2296, 8568, 31977, 119339, 445380, 1662180, 6203341, 23151183, 86401392, 322454384, 1203416145, 4491210195, 16761424636, 62554488348, 233456528757, 871271626679, 3251629977960, 12135248285160, 45289363162681, 169022204365563, 630799454299572

Offset: 0

Creighton Dement, Jun 28 2005

See A105968 for a similar sequence. Observe the four periodic sequences (1,1,1,1,); (-1,-1,-1,-1); (1,-1,1,-1,); (-1,1,-1,1,); (a(n)) is the (Type 1A) jbasejfor-transform of the periodic sequence (1,1,1,1) with respect to the floretion given in the program code. A109438 is the (Type 1A) jbasejfor-transform of the periodic sequence (-1,-1,-1,-1) with respect to the floretion given in the program code. A001834 is the (Type 1A) jbasejfor-transform of the periodic sequence (1,-1,1,-1) with respect to the floretion given in the program code. A102871 is the (Type 1A) jbasejfor-transform of the periodic sequence (-1,1,-1,1) with respect to the floretion given in the program code.

Floretion Algebra Multiplication Program, FAMP Code: (-1)^(n+1)jbasejfor[ + .5'ii' + .5'kk' + .5'ij' + .5'ji' + .5'jk' + .5'kj'] 1vesfor = (1,1,1,1,)

R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6
Charles H. Jepsen, Packing a box with bricks, Math. Mag. 64 (2) (1991) 92-97. b(n) there is 0, 4, 12, 48,... = 4*a(n) here.
Index entries for linear recurrences with constant coefficients, signature (3,3,-1).

Cf. A001353, A001834, A002530, A011916, A102871, A105968, A109438.

with(numtheory):a := cfrac (tan(Pi/3),60): > b := cfrac (tan(Pi/6),60): > seq(nthnumer (b,i)*nthdenom (a,i), i=0..24 ); # Zerinvary Lajos, Feb 08 2007

LinearRecurrence[{3,3,-1},{0,1,3},40] (* Harvey P. Dale, Apr 21 2018 *)

{a(n) = local(s=1); if( n<0, n = -1 - n; s=-1); s * polcoeff( x / ((x + 1) * (x^2 -4*x + 1)) + x * O(x^n), n)} /* Michael Somos, Jul 27 2012 */

G.f.: x/((x+1)*(x^2-4*x+1)).
a(n) = A002530(n)*A002530(n+1). - Zerinvary Lajos, Feb 08 2007
a(-1 - n) = -a(n). a(2*n) = A011916(n). a(2*n + 1) = -A011916(-1 -n). - Michael Somos, Jul 27 2012
6*a(n) = A001353(n)+A001353(n+1)-(-1)^n. - R. J. Mathar, Sep 07 2016
a(n) = ((1 + sqrt(3))*(2 + sqrt(3))^n + (1 - sqrt(3))*(2 - sqrt(3))^n - 2*(-1)^n)/12. - Stefano Spezia, Sep 19 2023
a(n)+a(n+1) = A001353(n+1). - R. J. Mathar, Aug 31 2025

A109438 a(n) = 5a(n-1) - 5a(n-2) + a(n-3) + 2*(-1)^(n+1), alternatively a(n) = 3a(n-1) + 3a(n-2) - a(n-3).

Original entry on oeis.org

1, 5, 18, 68, 253, 945, 3526, 13160, 49113, 183293, 684058, 2552940, 9527701, 35557865, 132703758, 495257168, 1848324913, 6898042485, 25743845026, 96077337620, 358565505453, 1338184684193, 4994173231318, 18638508241080

Offset: 0

Creighton Dement, Jun 28 2005

See A109437 for comments.

Floretion Algebra Multiplication Program, FAMP Code: (-1)^(n)jbasejfor[ + .5'ii' + .5'kk' + .5'ij' + .5'ji' + .5'jk' + .5'kj'] 1vesfor = (-1,-1,-1,-1,)

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,3,-1).

Cf. A109437, A001834, A102871.

LinearRecurrence[{3,3,-1},{1,5,18},30] (* Harvey P. Dale, Sep 07 2021 *)

Vec((1 + 2*x) / ((1 + x)*(1 - 4*x + x^2)) + O(x^30)) \\ Colin Barker, May 12 2019

G.f.: (1+2*x) / ((x+1)*(x^2-4*x+1)).

a(n) = (-2*(-1)^n + (7-5*sqrt(3))*(2-sqrt(3))^n + (2+sqrt(3))^n*(7+5*sqrt(3))) / 12. - Colin Barker, May 12 2019

A133161 Indices of the triangular numbers which are also centered triangular number.

Original entry on oeis.org

1, 4, 16, 61, 229, 856, 3196, 11929, 44521, 166156, 620104, 2314261, 8636941, 32233504, 120297076, 448954801, 1675522129, 6253133716, 23337012736, 87094917229, 325042656181, 1213075707496, 4527260173804, 16895964987721

Offset: 1

Richard Choulet, Oct 09 2007

Also, indices of the triangular numbers which are sums of three consecutive triangular numbers (see A129803).

A. Kozikowska, Topological classes of statically determinate beams with arbitrary number of supports, JOURNAL OF THEORETICAL AND APPLIED MECHANICS 49, 4, pp. 1079-1100, Warsaw 2011; (see Eq. 5.18). - _N. J. A. Sloane_, Dec 17 2011.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001834, A102871, A128862, A129803, A001571.

LinearRecurrence[{5,-5,1},{1,4,16},30] (* Harvey P. Dale, Aug 29 2017 *)

a(n+2)=4*a(n+1)-a(n)+1.
a(n+1)=2*a(n)+0.5+0.5*(12*a(n)^2+12*a(n)-15)^0.5.
G.f.: x*(1-x+x^2)/(1-x)/(1-4*x+x^2). - R. J. Mathar, Oct 24 2007
a(n)-a(n-1)= A005320(n-1). - R. J. Mathar, Mar 14 2016

A107401 a(n) = -a(n-1)+4a(n-2)+4a(n-3)-a(n-4)-a(n-5).

Original entry on oeis.org

0, 1, 1, 2, 3, 8, 10, 31, 36, 117, 133, 438, 495, 1636, 1846, 6107, 6888, 22793, 25705, 85066, 95931, 317472, 358018, 1184823, 1336140, 4421821, 4986541, 16502462, 18610023, 61588028, 69453550, 229849651, 259204176, 857810577, 967363153

Offset: 0

Roger L. Bagula, May 25 2005

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1, 4, 4, -1, -1).

n = 4 M = {{0, 1, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}, {0, 0, 0, 0, 1}, {-1, -1, m, m, -1}} Expand[Det[M - x*IdentityMatrix[5]]] NSolve[Det[M - x*IdentityMatrix[5]] == 0, x] v[1] = {0, 1, 1, 2, 3} digits = 50 v[n_] := v[n] = M.v[n - 1] a = Table[v[n][[1]], {n, 1, digits}]
LinearRecurrence[{-1,4,4,-1,-1},{0,1,1,2,3},40] (* Harvey P. Dale, Aug 16 2011 *)

G.f.:-x*(3*x^3+x^2-2*x-1)/((x+1)*(x^4-4*x^2+1)). [From Maksym Voznyy (voznyy(AT)mail.ru), Aug 12 2009]

a(2n) = A102871(n+1).

Definition replaced by recurrence by the Associate Editors of the OEIS, Sep 28 2009

cp's OEIS Frontend

A001570 Numbers k such that k^2 is centered hexagonal.

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A109437 a(-1) = a(0) = 0, a(1) = 1; a(n) = 5a(n-1) - 5a(n-2) + a(n-3) + 2*(-1)^(n+1), alternatively a(n) = 3a(n-1) + 3a(n-2) - a(n-3).

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A109438 a(n) = 5a(n-1) - 5a(n-2) + a(n-3) + 2*(-1)^(n+1), alternatively a(n) = 3a(n-1) + 3a(n-2) - a(n-3).

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A133161 Indices of the triangular numbers which are also centered triangular number.

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A107401 a(n) = -a(n-1)+4*a(n-2)+4*a(n-3)-a(n-4)-a(n-5).

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A107401 a(n) = -a(n-1)+4a(n-2)+4a(n-3)-a(n-4)-a(n-5).