A001922 -id:A001922 - 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

A001921 a(n) = 14*a(n-1) - a(n-2) + 6 for n>1, a(0)=0, a(1)=7.

Original entry on oeis.org

0, 7, 104, 1455, 20272, 282359, 3932760, 54776287, 762935264, 10626317415, 148005508552, 2061450802319, 28712305723920, 399910829332567, 5570039304932024, 77580639439715775, 1080558912851088832, 15050244140475527879, 209622859053806301480

Offset: 0

N. J. A. Sloane

(a(n)+1)^3 - a(n)^3 is a square (that of A001570(n)).
The ratio A001570(n)/a(n) tends to sqrt(3) = 1.73205... as n increases. - Pierre CAMI, Apr 21 2005
Define a(1)=0 a(2)=7 such that 3*(a(1)^2) + 3*a(1) + 1 = j(1)^2 = 1^2 and 3*(a(2)^2) + 3*a(2) + 1 = j(2)^2 = 13^2. Then a(n) = a(n-2) + 8*sqrt(3*(a(n-1)^2) + 3*a(n-1) + 1). Another definition : a(n) such that 3*(a(n)^2) + 3*a(n) + 1 = j(n)^2. - Pierre CAMI, Mar 30 2005
a(n) = A001353(n)*A001075(n+1). For n>0, the triple {a(n), a(n)+1=A001922(n), A001570(n)} forms a near-isosceles triangle with angle 2*Pi/3 bounded by the consecutive sides. - Lekraj Beedassy, Jul 21 2006
Numbers n such that A003215(n) is a square, cf. A006051. - Joerg Arndt, Jan 02 2017

			G.f. = 7*x + 104*x^2 + 1455*x^3 + 20272*x^4 + 282359*x^5 + 3932760*x^6 + ... - _Michael Somos_, Aug 17 2018

J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.
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).

T. D. Noe, Table of n, a(n) for n = 0..200
J. Brenner and E. P. Starke, Problem E702, Amer. Math. Monthly, 53 (1946), 465.
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
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001570, A001922, A006051.

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; A233450 for k=3; A001652 for k=4; A129556 for k=5; this sequence for k=6. - Bruno Berselli, Dec 16 2013

[Round(-1/2 - (1/6)*Sqrt(3)*(7-4*Sqrt(3))^n + (1/6)*Sqrt(3)*(7+4*Sqrt(3))^n + (1/4)*(7+4*Sqrt(3))^n + (1/4)*(7-4*Sqrt(3))^n): n in [0..50]]; // G. C. Greubel, Nov 04 2017

A001921:=z*(-7+z)/(z-1)/(z**2-14*z+1); # Conjectured by Simon Plouffe in his 1992 dissertation.

t = {0, 7}; Do[AppendTo[t, 14*t[[-1]] - t[[-2]] + 6], {20}]; t (* T. D. Noe, Aug 17 2012 *)
LinearRecurrence[{15, -15, 1}, {0, 7, 104}, 19] (* Michael De Vlieger, Jan 02 2017 *)
a[ n_] := -1/2 + (ChebyshevT[n + 1, 7] - ChebyshevT[n, 7]) / 12; (* Michael Somos, Aug 17 2018 *)

concat(0, Vec(x*(x-7)/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, Jan 06 2015

{a(n) = -1/2 + (polchebyshev(n + 1, 1, 7) - polchebyshev(n, 1, 7)) / 12}; /* Michael Somos, Aug 17 2018 */

G.f.: x*(-7 + x)/(x - 1)/(x^2 - 14*x + 1) (see Simon Plouffe in Maple section).
a(n) = (A028230(n+1)-1)/2. - R. J. Mathar, Mar 19 2009
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). - Colin Barker, Jan 06 2015
a(n) = -1 - a(-1-n) for all n in Z. - Michael Somos, Aug 17 2018

More terms from James Sellers, Jul 04 2000

A006051 Square hex numbers.

Original entry on oeis.org

1, 169, 32761, 6355441, 1232922769, 239180661721, 46399815451081, 9001325016847969, 1746210653453054881, 338755865444875798921, 65716891685652451935769, 12748738231151130799740241, 2473189499951633722697670961, 479786014252385791072548426169

Offset: 1

N. J. A. Sloane

Numbers n of the form n = y^2 = 3*x^2 - 3*x + 1.

			G.f. = x + 169*x^2 + 32761*x^3 + 6355441*x^4 + 1232922769*x^5 + ...

M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 19.
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..435
M. Gardner & N. J. A. Sloane, Correspondence, 1973-74
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
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
Eric Weisstein's World of Mathematics, Hex Number.
Index entries for linear recurrences with constant coefficients, signature (195,-195,1).

Cf. A003500.
Intersection of A000290 and A003215.
Values of x are given by A001922, values of y by A001570.

[(7*Evaluate(ChebyshevSecond(n),97) - 7*Evaluate(ChebyshevU(n-1), 97) + 1)/8: n in [1..30]]; // G. C. Greubel, Nov 04 2017; Oct 07 2022

Rest@ CoefficientList[Series[x(1-26x+x^2)/((1-x)(1-194x+x^2)), {x,0,20}], x] (* Michael De Vlieger, Jan 02 2017 *)
LinearRecurrence[{195,-195,1},{1,169,32761},20] (* Harvey P. Dale, Nov 03 2017 *)

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

def A006051(n): return (7*chebyshev_U(n-1,97) - 7*chebyshev_U(n-2,97) + 1)/8
[A006051(n) for n in range(1,31)] # G. C. Greubel, Oct 07 2022

a(n) = A001570(n)^2.
a(1 - n) = a(n).
G.f.: x * (1 - 26*x + x^2) / ((1 - x) * (1 - 194*x + x^2)). - Simon Plouffe in his 1992 dissertation
a(n) = 194*a(n-1) - a(n-2) - 24, a(1)=1, a(2)=169. - James Sellers, Jul 04 2000
a(n+1) = A003215(A001921(n)). - Joerg Arndt, Jan 02 2017
a(n) = (1/8)*(1 + 7*(ChebyshevU(n-1, 97) - ChebyshevU(n-2, 97))). - G. C. Greubel, Oct 07 2022

A228946 Numbers m such that m^3 - k^3 is a square for some k < m, k > 0.

Original entry on oeis.org

8, 10, 14, 28, 32, 33, 40, 56, 57, 65, 71, 72, 74, 78, 90, 105, 112, 114, 126, 128, 130, 132, 140, 148, 154, 155, 160, 176, 193, 200, 217, 218, 224, 228, 250, 252, 260, 266, 273, 280, 284, 288, 296, 297, 305, 312, 329, 336, 344, 349, 350, 360, 392

Offset: 1

M. F. Hasler, Oct 05 2013

nonn

See A038596 = A038597^2 for the possible values of n^3-k^3.

Nathaniel Johnston, Table of n, a(n) for n = 1..5000

Cf. A001922, A038593-A038598, A014439, A014440, A014441, A066648, A071954.

isA228946 := proc (n) local k; for k to n-1 do if issqr(n^3-k^3) then return true end if end do; return false end proc; for n from 1 to 400 do if isA228946(n)then print(n): end if: end do: # Nathaniel Johnston, Oct 07 2013

k3sQ[n_]:=Count[n^3-Range[n-1]^3,?(IntegerQ[Sqrt[#]]&)]>0; Select[ Range[ 400],k3sQ] (* _Harvey P. Dale, May 07 2017 *)

for(n=1,999,for(k=1,n-1, issquare(n^3-k^3) & !print1(n",") & break))

is(n)=for(k=1,n-1, issquare(n^3-k^3) & return(1))

a(n) = A066648(n)^(1/3). - Amiram Eldar, Mar 20 2025

A227418 Array A(n,k) with all numbers m such that 3*m^2 +- 3^k is a square and their corresponding square roots, read by downward antidiagonals.

Original entry on oeis.org

0, 1, 1, 0, 2, 4, 3, 3, 7, 15, 0, 6, 12, 26, 56, 9, 9, 21, 45, 97, 209, 0, 18, 36, 78, 168, 362, 780, 27, 27, 63, 135, 291, 627, 1351, 2911, 0, 54, 108, 234, 504, 1086, 2340, 5042, 10864, 81, 81, 189, 405, 873, 1881, 4053, 8733, 18817, 40545

Offset: 0

Richard R. Forberg, Sep 02 2013

Array is analogous to A228405 in goal and structure, with key differences.
Left column is A001353.  Top row (not in OEIS) interleaves 0 with the powers of 3, as: 0, 1, 0, 3, 0, 9, 0, 27, 0, 81.
Either or both may be used as initializing values. See Formula section.
The left column is the second binomial transform of the top row. The intermediate transform sequence is A002605, not present in this array.
The columns of the array hold all values, in sequential order, of numbers m such that 3*m^2 + 3^k or 3*m^2 - 3^k are squares, and their corresponding square roots in the next column, which then form the "next round" of m values for column k+1.
For example: A(n,0) are numbers such that 3*m^2 + 1 are squares, the integer square roots of each are in A(n,1), which are then numbers m such that 3*m^2 - 3 are squares, with those square roots in A(n,2), etc.  The sign alternates for each increment of k, etc. No integer square roots exist for the opposite sign in a given column, regardless of n.
Also, A(n,1) are values of m such that floor(m^2/3) is square, with the corresponding square roots given by A(n,0).
A(n,k)/A(n,k-2) = 3; A(n,k)/A(n,k-1) converges to sqrt(3) for large n.
A(n,k)/A(n-1,k) converges to 2 + sqrt(3) for large n.
Several ways of combining the first few columns give OEIS sequences:
  A(n,0) + A(n,1) = A001835; A(n,1) + A(n,2)= A001834; A(n,2) + A(n,3) = A082841;
  A(n,0)*A(n,1)/2 = A007655(n); A(n+2,0)*A(n+1,1) = A001922(n);
  A(n,0)*A(n+1,1) = A001921(n); A(n,0)^2 + A(n,1)^2 = A103974(n);
  A(n,1)^2 - A(n,0)^2 = A011922(n); (A(n+2,0)^2 + A(n+1,1)^2)/2 = A122770(n) = 2*A011916(n).
The main diagonal (without initial 0) = 2*A090018.  The first subdiagonal =  abs(A099842). First superdiagonal = A141041.
A001353 (in left column) are the only initializing set of numbers where the recursive square root equation (see below) produces exclusively integer values, for all iterations of k.  For any other initial values only even iterations (at k = 2, 4, ...) produce integers.

			The array, A(n, k), begins as:
    0,    1,    0,    3,    0,     9,     0,    27, ... see A000244;
    1,    2,    3,    6,    9,    18,    27,    54, ... A038754;
    4,    7,   12,   21,   36,    63,   108,   189, ... A228879;
   15,   26,   45,   78,  135,   234,   405,   702, ...
   56,   97,  168,  291,  504,   873,  1512,  2619, ...
  209,  362,  627, 1086, 1881,  3258,  5643,  9774, ...
  780, 1351, 2340, 4053, 7020, 12159, 21060, 36477, ...
Antidiagonal triangle, T(n, k), begins as:
   0;
   1,  1;
   0,  2,   4;
   3,  3,   7,  15;
   0,  6,  12,  26,  56;
   9,  9,  21,  45,  97,  209;
   0, 18,  36,  78, 168,  362,  780;
  27, 27,  63, 135, 291,  627, 1351, 2911;
   0, 54, 108, 234, 504, 1086, 2340, 5042, 10864;
  81, 81, 189, 405, 873, 1881, 4053, 8733, 18817, 40545;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A001353, A001075, A005320, A038754, A090018, A099842, A141041, A151961.

Cf. A000244, A084156, A228879.

function A(n,k)
  if k lt 0 then return 0;
  elif n eq 0 then return Round((1/2)*(1-(-1)^k)*3^((k-1)/2));
  elif k eq 0 then return Evaluate(ChebyshevSecond(n), 2);
  else return 2*A(n, k-1) - A(n-1, k-1);
  end if; return A;
end function;
A227418:= func< n,k | A(k, n-k) >;
[A227418(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Oct 09 2022

A[n_, k_]:= If[k<0, 0, If[k==0, ChebyshevU[n-1, 2], 2*A[n, k-1] - A[n-1, k-1]]];
T[n_, k_]:= A[k, n-k];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 09 2022 *)

def A(n,k):
    if (k<0): return 0
    elif (k==0): return chebyshev_U(n-1,2)
    else: return 2*A(n, k-1) - A(n-1, k-1)
def A227418(n, k): return A(k, n-k)
flatten([[A227418(n,k) for k in range(n+1)] for n in range(15)]) # G. C. Greubel, Oct 09 2022

If using the left column and top row to initialize, then: A(n,k) = 2*A(n, k-1) - A(n-1, k-1).
If using only the top row to initialize, then: A(n,k) = 4*A(n-1,k) - A(n-2,k).
If using the left column to initialize, then: A(n,k) = sqrt(3*A(n,k-1) + (-3)^(k-1)), for all n, k > 0.
Other internal relationships that apply are: A(2*n-1, 2*k) = A(n,k)^2 - A(n-1,k)^2;
A(n+1,k) * A(n,k+1) - A(n+1, k+1) * A(n,k) = (-3)^k, for all n, k > 0.
A(n, 0) = A001353(n).
A(n, 1) = A001075(n).
A(n, 2) = A005320(n).
A(n, 3) = A151961(n).
A(1, k) = A038754(k).
A(n, n) = 2*A090018(n), for n > 0 (main diagonal).
A(n, n+1) = A141041(n-1) (superdiagonal).
A(n+1, n) = abs(A099842(n)) (subdiagonal).
From G. C. Greubel, Oct 09 2022: (Start)
T(n, 0) = (1/2)*(1-(-1)^n)*3^((n-1)/2).
T(n, 1) = A038754(n-1).
T(n, 2) = A228879(n-2).
T(2*n-1, n-1) = A141041(n-1).
T(2*n, n) = 2*A090018(n-1), n > 0.
T(n, n-4) = 3*A005320(n-4).
T(n, n-3) = 3*A001075(n-3).
T(n, n-2) = 3*A001353(n-2).
T(n, n-1) = A001075(n-1).
T(n, n) = A001353(n).
Sum_{k=0..n-1} T(n, k) = A084156(n).
Sum_{k=0..n} T(n, k) = A084156(n) + A001353(n). (End)

Offset corrected by G. C. Greubel, Oct 09 2022

A156712 Star numbers (A003154) that are also triangular numbers (A000217).

Original entry on oeis.org

1, 7, 91, 1261, 17557, 244531, 3405871, 47437657, 660721321, 9202660831, 128176530307, 1785268763461, 24865586158141, 346332937450507, 4823795538148951, 67186804596634801, 935791468814738257, 13033893758809700791, 181538721154521072811, 2528508202404485318557

Offset: 1

Aaron Meyerowitz, Feb 14 2009

From Colin Barker, Jan 06 2015: (Start)
Also indices of centered square numbers (A001844) which are also centered triangular numbers (A005448).
Also indices of centered octagonal numbers (A016754) which are also centered hexagonal numbers (A003215).
Also positive integers y in the solutions to 3*x^2-4*y^2-3*x+4*y = 0, the corresponding values of x being A001922.
(End)

Colin Barker, Table of n, a(n) for n = 1..875
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 30, 56.
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Wikipedia, Star Numbers
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A000217, A001570, A001844, A001922, A003154.

Cf. A003215, A005448, A007655, A016754.

[(Evaluate(ChebyshevSecond(n+1),7) - 13*Evaluate(ChebyshevU(n), 7) + 1)/2: n in [1..30]]; // G. C. Greubel, Oct 07 2022

f:= gfun[rectoproc]({a(n+3)=15*a(n+2)-15*a(n+1)+a(n),a(1)=1,a(2)=7,a(3)=91},a(n),'remember'):
seq(f(n), n=1..30); # Robert Israel, Jan 01 2015

f[n_] := (Simplify[(2 + Sqrt@3)^(2 n - 1) + (2 - Sqrt@3)^(2 n - 1)] + 4)/8; Array[f, 17] (* Robert G. Wilson v, Oct 28 2010 *)

Vec(-x*(x^2-8*x+1)/((x-1)*(x^2-14*x+1)) + O(x^100)) \\ Colin Barker, Jan 01 2015

def A156712(n): return (1 + chebyshev_U(n, 7) - 13*chebyshev_U(n-1, 7))/2
[A156712(n) for n in range(1,31)] # G. C. Greubel, Oct 07 2022

a(n+3) = 15*a(n+2) - 15*a(n+1) + a(n).
If x^2 - 3*y^2 = 1 with x even then a(y) = (y+2)/4 evidently related to A001570 by: add 1 and halve.
G.f.: x*(1 - 8*x + x^2)/((1-x)*(1 - 14*x + x^2)). - Alexander R. Povolotsky, Feb 15 2009
a(n) = (4 + (2 + sqrt(3))*(7 - 4*sqrt(3))^n + (2 - sqrt(3))*(7 + 4*sqrt(3))^n)/8. - Colin Barker, Mar 05 2016
a(n) = (1/2)*( 1 + ChebyshevU(n, 7) - 13*ChebyshevU(n-1, 7) ). - G. C. Greubel, Oct 07 2022

a(11) onwards from Robert G. Wilson v, Oct 28 2010

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

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A001921 a(n) = 14*a(n-1) - a(n-2) + 6 for n>1, a(0)=0, a(1)=7.

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A006051 Square hex numbers.

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A228946 Numbers m such that m^3 - k^3 is a square for some k < m, k > 0.

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A227418 Array A(n,k) with all numbers m such that 3*m^2 +- 3^k is a square and their corresponding square roots, read by downward antidiagonals.

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A156712 Star numbers (A003154) that are also triangular numbers (A000217).

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