A028231 -id:A028231 - 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

A307745 Perfect powers y^m with y > 1 and m > 1 which are Brazilian repdigits with three or more digits > 1 in some base.

Original entry on oeis.org

1521, 1600, 2401, 2744, 6084, 17689, 61009, 244036, 294849, 1179396, 1483524, 2653641, 2725801, 2989441, 4717584, 5239521, 7371225, 9591409, 10614564, 11957764, 14447601, 17397241, 18870336, 20277009, 20958084, 23882769, 26904969, 29484900, 38365636, 38825361

Offset: 1

Bernard Schott, Apr 26 2019

The terms of this sequence are solutions y^m of the Diophantine equation a * (b^q - 1)/(b-1) = y^m with 1 < a < b, y >= 2, q >= 3, m >= 2. This equation has been studied by Kustaa A. Inkeri in Acta Arithmetica; some terms of this sequence come from his article where the author limits the study of this equation to bases b <= 100.
The case a = 1 is clarified in A208242; it corresponds to the Nagell-Ljunggren equation.
The sequence has infinitely many terms because the Diophantine equation 3*(x^2+x+1) = y^2 has infinitely many solutions. - Giovanni Resta, Apr 26 2019
The corresponding solutions (x, y) of this Diophantine equation are (A028231, A341671).
The integers y such that y^2 (m = 2) satisfies this equation are in A158235, except 11 and 20 corresponding to a = 1. - Bernard Schott, Apr 27 2019

			3 * (22^3-1)/(22-1) = 39^2 and (333)_22 = 39^2 = 1521.
58 * (99^4-1)/(99-1) = 7540^2 and (AAAA)_99 = 7540^2 = 56851600 where A is the symbol for 58 in base 99.

Kustaa A. Inkeri, On the Diophantine equation a(x^n-1)/(x-1) = y^m, Acta Arithmetica, XXI, 1972.
Index entries for sequences related to Brazilian numbers.

Subsequence of A001597 and of A125134.
Cf. A158235, A208242 (a=1, that is, with repunits).
Cf. A028231, A341671.

rupQ[n_, mx_] := Block[{t, x, p}, p = x^2 + x + 1; While[(t = p /. x -> mx) <= n && Reduce[p == n && x >= mx, x, Integers] === False, p = x*p + 1]; t <= n]; repdQ[n_] := AnyTrue[ Rest@ Most@ Divisors@ n, rupQ[n/#, #+1] &]; ex = 2; up = 10^7; L = {}; While[2^ex <= up, L = Union[L, Parallelize@ Select[ Range[2, Floor[ up^(1/ex)] ]^ex, repdQ]]; ex = NextPrime@ ex]; L (* Giovanni Resta, Apr 27 2019 *)

isokb(n) = for(b=2, n-2, d=digits(n, b); if((#d > 2) && (vecmin(d)==vecmax(d)) && (d[1] > 1), return (1))); 0;
isok(n) = ispower(n) && isokb(n); \\ Michel Marcus, Apr 28 2019

More terms from Giovanni Resta, Apr 26 2019

A341671 Solutions y of the Diophantine equation 3*(x^2+x+1) = y^2.

Original entry on oeis.org

3, 39, 543, 7563, 105339, 1467183, 20435223, 284625939, 3964327923, 55215964983, 769059181839, 10711612580763, 149193516948843, 2077997624703039, 28942773228893703, 403120827579808803, 5614748812888429539, 78203362552858204743, 1089232326927126436863, 15171049214426911911339

Offset: 1

Bernard Schott, Feb 17 2021

nonn

Corresponding x are in A028231.
This equation belongs to the family of equations studied by Kustaa A. Inkeri, y^m = a * (x^q-1)/(x-1) with here: m=2, a=3, q=3. This equation is exhibed in A307745 by Giovanni Resta to prove that this sequence has infinitely many terms.
This Diophantine equation 3*(x^2+x+1) = y^2 has infinitely many solutions because the Pell-Fermat equation u^2 - 3*v^2 = -2 also has infinitely many solutions. The corresponding (u,v) are in (A001834, A001835) and for each pair (u,v), the corresponding solutions of 3*(x^2+x+1) = y^2 are x = (3*u*v-1)/2 and y = 3*(u^2+1)/2.
Note that if y = 3*z, this equation becomes 3*z^2 = x^2+x+1 with solutions (x, z) = (A028231, A001570).

			The first few values for (x,y) are (1,3), (22,39), (313,543), (4366,7563), (60817,105339), ...

Kustaa A. Inkeri, On the Diophantine equation a(x^n-1)/(x-1) = y^m, Acta Arithmetica, Vol. 21, No. 1 (1972), pp. 299-311.

Cf. A001834, A001835, A001570, A028231, A307745.

Subsequence of A158235, for a(n)>3.

f[x_] := Sqrt[3*(x^2 + x + 1)]; f /@ LinearRecurrence[{15, -15, 1}, {1, 22, 313}, 20] (* Amiram Eldar, Feb 17 2021 *)

a(n) = 3*A001570(n). - Hugo Pfoertner, Feb 17 2021

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).

More terms from Amiram Eldar, Feb 17 2021

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

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A307745 Perfect powers y^m with y > 1 and m > 1 which are Brazilian repdigits with three or more digits > 1 in some base.

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A341671 Solutions y of the Diophantine equation 3*(x^2+x+1) = y^2.

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