A036428 -id:A036428 - cp's OEIS frontend

A028230 Bisection of A001353. Indices of square numbers which are also octagonal.

1, 15, 209, 2911, 40545, 564719, 7865521, 109552575, 1525870529, 21252634831, 296011017105, 4122901604639, 57424611447841, 799821658665135, 11140078609864049, 155161278879431551, 2161117825702177665, 30100488280951055759, 419245718107612602961, 5839339565225625385695

Offset: 1

N. J. A. Sloane

Chebyshev S-sequence with Diophantine property.
4*b(n)^2 - 3*a(n)^2 = 1 with b(n) = A001570(n), n>=0.
y satisfying the Pellian x^2 - 3*y^2 = 1, for even x given by A094347(n). - Lekraj Beedassy, Jun 03 2004
a(n) = L(n,-14)*(-1)^n, where L is defined as in A108299; see also A001570 for L(n,+14). - Reinhard Zumkeller, Jun 01 2005
Product x*y, where the pair (x, y) solves for x^2 - 3y^2 = -2, i.e., a(n) = A001834(n)*A001835(n). - Lekraj Beedassy, Jul 13 2006
Numbers n such that RootMeanSquare(1,3,...,2*A001570(k)-1) = n. - Ctibor O. Zizka, Sep 04 2008
As n increases, this sequence is approximately geometric with common ratio r = lim(n -> oo, a(n)/a(n-1)) = (2 + sqrt(3))^2 = 7 + 4 * sqrt(3). - Ant King, Nov 15 2011

R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.
J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.

Vincenzo Librandi, Table of n, a(n) for n = 1..890
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
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.
T. N. E. Greville, Table for third-degree spline interpolations with equally spaced arguments, Math. Comp., 24 (1970), 179-183.
W. D. Hoskins, Table for third-degree spline interpolation using equi-spaced knots, Math. Comp., 25 (1971), 797-801.
Tanya Khovanova, Recursive Sequences
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 4, k=1, t=2.
Dino Lorenzini, and Z. Xiang, Integral points on variable separated curves, Preprint 2016.
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
Eric Weisstein's World of Mathematics, Octagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (14,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001353, A001570, A001834, A001835, A001921, A007655, A036428, A046184, A049310, A094347.

Cf. A077416 with companion A077417.

a:=[1,15];; for n in [3..30] do a[n]:=14*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

I:=[1,15]; [n le 2 select I[n] else 14*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 06 2019

seq(coeff(series((1+x)/(1-14*x+x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Dec 06 2019

LinearRecurrence[{14, - 1}, {1, 15}, 17] (* Ant King, Nov 15 2011 *)
CoefficientList[Series[(1+x)/(1-14x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 17 2014 *)

Vec((1+x)/(1-14*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Jun 16 2014

isok(n) = ispolygonal(n^2, 8); \\ Michel Marcus, Jul 09 2017

[(lucas_number2(n,14,1)-lucas_number2(n-1,14,1))/12 for n in range(1, 18)] # Zerinvary Lajos, Nov 10 2009

a(n) = 2*A001921(n)+1.
a(n) = 14*a(n-1) - a(n-2) for n>1.
a(n) = S(n, 14) + S(n-1, 14) = S(2*n, 4) with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x) = 0, S(n, 14) = A007655(n+1) and S(n, 4) = A001353(n+1).
G.f.: x*(1+x)/(1-14*x+x^2).
a(n) = (ap^(2*n+1) - am^(2*n+1))/(ap - am) with ap := 2+sqrt(3) and am := 2-sqrt(3).
a(n+1) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*16^(n-k), n>=0.
a(n) = sqrt((4*A001570(n-1)^2 - 1)/3).
a(n) ~ 1/6*sqrt(3)*(2 + sqrt(3))^(2*n-1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
4*a(n+1) = (A001834(n))^2 + 4*(A001835(n+1))^2 - (A001835(n))^2. E.g. 4*a(3) = 4*209 = 19^2 + 4*11^2 - 3^2 = (A001834(2))^2 + 4*(A001835(3))^2 - (A001835(2))^2. Generating floretion: 'i + 2'j + 3'k + i' + 2j' + 3k' + 4'ii' + 3'jj' + 4'kk' + 3'ij' + 3'ji' + 'jk' + 'kj' + 4e. - Creighton Dement, Dec 04 2004
a(n) = f(a(n-1),7) + f(a(n-2),7), where f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. - Marcos Carreira, Dec 27 2006
From Ant King, Nov 15 2011: (Start)
a(n) = 1/6 * sqrt(3) * ( (tan(5*Pi/12)) ^ (2n-1) - (tan(Pi/12)) ^ (2n-1) ).
a(n) = floor (1/6 * sqrt(3) * (tan(5*Pi/12)) ^ (2n-1)). (End)
a(n) = A001353(n)^2-A001353(n-1)^2. - Antonio Alberto Olivares, Apr 06 2020
E.g.f.: 1 - exp(7*x)*(3*cosh(4*sqrt(3)*x) - 2*sqrt(3)*sinh(4*sqrt(3)*x))/3. - Stefano Spezia, Dec 12 2022
a(n) = sqrt(A036428(n)). - Bernard Schott, Dec 19 2022

Additional comments from Wolfdieter Lang, Nov 29 2002
Incorrect recurrence relation deleted by Ant King, Nov 15 2011
Minor edits by Vaclav Kotesovec, Jan 28 2015

A046184 Indices of octagonal numbers which are also squares.

Original entry on oeis.org

1, 9, 121, 1681, 23409, 326041, 4541161, 63250209, 880961761, 12270214441, 170902040409, 2380358351281, 33154114877521, 461777249934009, 6431727384198601, 89582406128846401, 1247721958419651009, 17378525011746267721, 242051628206028097081

Offset: 1

Eric W. Weisstein

The equation a(t)*(3*a(t)-2) = m*m is equivalent to the Pell equation (3*a(t)-1)*(3*a(t)-1) - 3*m*m = 1. - Paul Weisenhorn, May 12 2009
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (2 + sqrt(3))^2 = 7 + 4 * sqrt(3). - Ant King, Nov 16 2011
Also numbers n such that the octagonal number N(n) is equal to the sum of two consecutive triangular numbers. - Colin Barker, Dec 11 2014
Also nonnegative integers y in the solutions to 2*x^2 - 6*y^2 + 4*x + 4*y + 2 + 2 = 0, the corresponding values of x being A251963. - Colin Barker, Dec 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Eric Weisstein's World of Mathematics, Octagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A028230, A036428, A251963.

Cf. A006253.

I:=[1, 9, 121]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Nov 17 2011

LinearRecurrence[ {15, -15, 1}, {1, 9, 121}, 17 ] (* Ant King, Nov 16 2011 *)
CoefficientList[Series[x (1-6x+x^2)/((1-x)(1-14x+x^2)),{x,0,30}],x] (* Harvey P. Dale, Sep 01 2021 *)

Vec(x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)) + O(x^100)) \\ Colin Barker, Dec 11 2014

{n: A000567(n) in A000290}.
Nearest integer to (1/6) * (2+sqrt(3))^(2n-1). - Ralf Stephan, Feb 24 2004
a(n) = A045899(n-1) + 1 = A051047(n+1) + 1 = A003697(2n-2). - N. J. A. Sloane, Jun 12 2004
a(n) = A001835(n)^2. - Lekraj Beedassy, Jul 21 2006
From Paul Weisenhorn, May 12 2009: (Start)
With A=(2+sqrt(3))^2=7+4*sqrt(3) the equation x*x-3*m*m=1 has solutions
x(t) + sqrt(3)*m(t) = (2+sqrt(3))*A^t and the recurrences
x(t+2) = 14*x(t+1) - x(t) with  = 2, 26, 362, 5042
m(t+2) = 14*m(t+1) - m(t) with  = 1, 15, 209, 2911
a(t+2) = 14*a(t+1) - a(t) - 4 with  = 1, 9, 121, as above. (End)
From Ant King, Nov 15 2011: (Start)
a(n) = 14*a(n-1) - a(n-2) - 4.
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = (1/6)*( (2+sqrt(3))^(2n-1) + (2-sqrt(3))^(2n-1) + 2 ).
a(n) = ceiling( (1/6)*(2 + sqrt(3))^(2n-1) ).
a(n) = (1/6)*( (tan(5*Pi/12))^(2n-1) + (tan(Pi/12))^(2n-1) + 2 ).
a(n) = ceiling ( (1/6)*(tan(5*Pi/12))^(2n-1) ).
G.f.: x*(1-6*x+x^2) / ((1-x)*(1-14*x+x^2)). (End)
a(n) = A006253(2n-2). - Andrey Goder, Oct 17 2021

A342709 12-gonal (dodecagonal) square numbers.

Original entry on oeis.org

1, 64, 3025, 142129, 6677056, 313679521, 14736260449, 692290561600, 32522920134769, 1527884955772561, 71778070001175616, 3372041405099481409, 158414167969674450625, 7442093853169599697984, 349619996931001511354641, 16424697761903901433970161

Offset: 1

Bernard Schott, Mar 19 2021

The 12-gonal square numbers k correspond to the nonnegative integer solutions of the Diophantine equation k = d*(5*d-4) = c^2, equivalent to (5*d-2)^2 - 5*c^2 = 4. Substituting x = 5*d-2 and y = c gives the Pell-Fermat's equation x^2 - 5*y^2 = 4.
The solutions x are in A342710, while corresponding solutions y that are also the indices c of the squares which are 12-gonal are in A033890.
The indices d of the corresponding 12-gonal which are squares are in A081068.

			142129 = 169*(5*169-4) = 377^2, so 142129 is the 169th 12-gonal number and the 377th square, hence 142129 is a term.

Index entries for linear recurrences with constant coefficients, signature (48,-48,1).

Intersection of A000290 (squares) and A051624 (12-gonal numbers).

Similar for n-gonal squares: A001110 (triangular), A036353 (pentagonal), A046177 (hexagonal), A036354 (heptagonal), A036428 (octagonal), A036411 (9-gonal), A188896 (there are no 10-gonal squares > 1), A333641 (11-gonal), this sequence (12-gonal).

with(combinat):
seq(fibonacci(4*n-2)^2, n=1..16);

Table[Fibonacci[4*n - 2]^2, {n, 1, 16}] (* Amiram Eldar, Mar 19 2021 *)

a(n) = fibonacci(4*n-2)^2; \\ Michel Marcus, Mar 21 2021

G.f.: x*(1 + 16*x + x^2)/((1 - x)*(1 - 47*x + x^2)). - Stefano Spezia, Mar 20 2021
a(n) = 48*a(n-1) - 48*a(n-2) + a(n-3). - Kevin Ryde, Mar 20 2021
a(n) = 9*A161582(n) + 1. - Hugo Pfoertner, Mar 19 2021
a(n) = A033890(n-1)^2.

A114618 Numbers k such that the k-th octagonal number is 4-almost prime.

Original entry on oeis.org

4, 9, 27, 39, 49, 57, 59, 69, 75, 85, 87, 105, 109, 117, 119, 121, 125, 143, 147, 153, 161, 169, 175, 177, 185, 187, 199, 207, 217, 219, 231, 235, 239, 245, 249, 265, 267, 269, 275, 283, 285, 289, 291, 299, 301, 305, 311, 319, 321, 327, 329, 333, 335, 345, 349, 357, 359, 361, 363, 371, 381, 385

Offset: 1

Jonathan Vos Post, Feb 17 2006

It is necessary but not sufficient that k must be prime (A000040), semiprime (A001358), or 3-almost prime (A014612).

			a(1) = 4 because OctagonalNumber(4) = Oct(4) = 4*(3*4-2) = 40 = 2^3 * 5 has exactly 4 prime factors (3 are all equally 2; factors need not be distinct).
a(2) = 9 because Oct(9) = 9*(3*9-2) = 225 = 3^2 * 5^2, a 4-almost prime [225 is also a square, hence a square octagonal number A036428, as is Oct(121)].
a(3) = 27 because Oct(27) = 27*(3*27-2) = 2133 = 3^3 * 79.
a(4) = 39 because Oct(39) = 39*(3*39-2) = 4485 = 3 * 5 * 13 * 23 has exactly 4 prime factors, in this case distinct.
a(26) = 187 because Oct(187) = 187*(3*187-2) = 104533 = 11 * 13 * 17 * 43 [a 4-brilliant number, that is with 4 prime factors that are each the same number of digits in length].

Amiram Eldar, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Almost Prime.
Eric Weisstein's World of Mathematics, Octagonal Number.

Cf. A000040, A000567, A001222, A001358, A014612, A014613, A036428.

Select[Range[400],PrimeOmega[#(3#-2)]==4&] (* Harvey P. Dale, Sep 07 2011 *)

Numbers k such that k*(3*k-2) has exactly four prime factors (with multiplicity).
Numbers k such that A000567(k) is a term of A014613.
Numbers k such that A001222(A000567(k)) = 4.
Numbers k such that A001222(k) + A001222(3*k-2) = 4.
Numbers k such that [(3*k-2)*(3*k-1)*(3*k)]/[(3*k-2)+(3*k-1)+(3*k)] is a term of A014613.

265 inserted by R. J. Mathar, Dec 22 2010

A333641 11-gonal (or hendecagonal) square numbers.

Original entry on oeis.org

0, 1, 196, 29241, 1755625, 261468900, 38941102225, 2337990844401, 348201795147556, 51858411008887561, 3113535139359330841, 463705205422871375236, 69060571958250748760481, 4146338334574433921200225, 617522713934165528806340100, 91968930524758079223806760025

Offset: 1

Bernard Schott, Mar 31 2020

The 11-gonal square numbers correspond to the nonnegative integer solutions of the Diophantine equation k*(9*k-7)/2 = m^2, equivalent to (18*k-7)^2 - 72*m^2 = 49. Substituting x = 18*k-7 and y = m gives the Pell equation x^2-72*y^2 = 49. The integer solutions (x,y) = (-7,0), (11,1), (119,14), (1451,171), (11243,1325), ... correspond to the following solutions (k,m) = (0,0), (1,1), (7,14), (81,171), (625,1325), ...

			1755625 is a term because 625*(9*625-7)/2 = 1325^2 = 1755625; that means that 1755625 is the 625th 11-gonal number and the square of 1325.

Index entries for linear recurrences with constant coefficients, signature (1,0,1331714,-1331714,0,-1,1).

Intersection of A000290 (squares) and A051682 (11-gonals).
Cf. A106525.
Cf. A001110 (square triangulars), A036353 (square pentagonals), A046177 (square hexagonals), A036354 (square heptagonals), A036428 (square octagonals), A036411 (square 9-gonals), A188896 (only {0,1} are square 10-gonals), this sequence (square 11-gonals), A342709 (square 12-gonals).

for k from 0 to 8000000 do
d:= k*(9*k-7)/2;
if issqr(d) then print(k,sqrt(d),d); else fi; od:

Last /@ Solve[(18*x - 7)^2 - 72*y^2 == 49 && x >= 0 && y >= 0 && y < 10^16, {x, y}, Integers] /. Rule -> (#2^2 &) (* Amiram Eldar, Mar 31 2020 *)

concat(0, Vec(-x*(1 + 195*x + 29045*x^2 + 394670*x^3 + 29045*x^4 + 195*x^5 + x^6)/(-1 + x + 1331714*x^3 - 1331714*x^4 - x^6 + x^7) + O(x^20))) \\ Jinyuan Wang, Mar 31 2020

a(n) = k*(9*k-7)/2 for n > 1, where k = (A106525(4*n-6) + 7)/18. - Jinyuan Wang, Mar 31 2020

More terms from Amiram Eldar, Mar 31 2020

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A028230 Bisection of A001353. Indices of square numbers which are also octagonal.

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A046184 Indices of octagonal numbers which are also squares.

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A342709 12-gonal (dodecagonal) square numbers.

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A114618 Numbers k such that the k-th octagonal number is 4-almost prime.

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A333641 11-gonal (or hendecagonal) square numbers.

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