A048907 -id:A048907 - cp's OEIS frontend

A001080 a(n) = 16*a(n-1) - a(n-2) with a(0) = 0, a(1) = 3.

0, 3, 48, 765, 12192, 194307, 3096720, 49353213, 786554688, 12535521795, 199781794032, 3183973182717, 50743789129440, 808716652888323, 12888722657083728, 205410845860451325, 3273684811110137472, 52173546131901748227, 831503053299317834160

Offset: 0

N. J. A. Sloane

Also 7*x^2 + 1 is a square; n=7 in PARI script below. - Cino Hilliard, Mar 08 2003
That is, the terms are solutions y of the Pell-Fermat equation x^2 - 7 * y^2 = 1. The corresponding values of x are in A001081. (x,y) = (1,0), (8,3), (127,48), ... - Bernard Schott, Feb 23 2019
The first solution to the equation x^2 - 7*y^2 = 1 is (X(0); Y(0)) = (1; 0) and the other solutions are defined by: (X(n); Y(n))= (8*X(n-1) + 21*Y(n-1); 3*X(n-1) + 8*Y(n-1)), with n >= 1. - Mohamed Bouhamida, Jan 16 2020

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).
V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 281.

T. D. Noe, Table of n, a(n) for n = 0..200
H. Brocard, Notes élémentaires sur le problème de Peel [sic], Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
M. Davis, One equation to rule them all, Trans. New York Acad. Sci. Ser. II, 30 (1968), 766-773.
Tanya Khovanova, Recursive Sequences
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.
Mihai Prunescu, On other two representations of the C-recursive integer sequences by terms in modular arithmetic, arXiv:2406.06436 [math.NT], 2024. See p. 17.
Mihai Prunescu and Lorenzo Sauras-Altuzarra, On the representation of C-recursive integer sequences by arithmetic terms, arXiv:2405.04083 [math.LO], 2024. See p. 15.
Mihai Prunescu and Joseph M. Shunia, On modular representations of C-recursive integer sequences, arXiv:2502.16928 [math.NT], 2025. See p. 5.
Index entries for linear recurrences with constant coefficients, signature (16,-1).

Equals 3 * A077412. Bisection of A084069.
Cf. A048907.
Cf. A001081, A010727. - Vincenzo Librandi, Feb 16 2009

a:=[0,3];; for n in [3..30] do a[n]:=16*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Feb 23 2019

I:=[0,3]; [n le 2 select I[n] else 16*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 20 2017

A001080:=3*z/(1-16*z+z**2); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation

LinearRecurrence[{16,-1},{0,3},30] (* Harvey P. Dale, Nov 01 2011 *)
CoefficientList[Series[3*x/(1-16*x+x^2), {x, 0, 30}], x] (* G. C. Greubel, Dec 20 2017 *)

nxsqp1(m,n) = { for(x=1,m, y = n*x*x+1; if(issquare(y),print1(x" ")) ) }

x='x+O('x^30); concat([0], Vec(3*x/(1-16*x+x^2))) \\ G. C. Greubel, Dec 20 2017

(3*x/(1-16*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 23 2019

G.f.: 3*x/(1-16*x+x^2).
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 15*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 17*(a(n-1) - a(n-2)) + a(n-3). (End)
a(n) = 16*a(n-1) - a(n-2) with a(1)=0 and a(2)=3. - Sture Sjöstedt, Nov 18 2011
E.g.f.: exp(8*x)*sinh(3*sqrt(7)*x)/sqrt(7). - G. C. Greubel, Feb 23 2019

A276123 a(0) = a(1) = a(2) = 1; for n > 2, a(n) = (a(n-1) + 1)*(a(n-2) + 1) / a(n-3).

Original entry on oeis.org

1, 1, 1, 4, 10, 55, 154, 868, 2449, 13825, 39025, 220324, 621946, 3511351, 9912106, 55961284, 157971745, 891869185, 2517635809, 14213945668, 40124201194, 226531261495, 639469583290, 3610286238244, 10191389131441, 57538048550401, 162422756519761

Offset: 0

Bruno Langlois, Aug 21 2016

Colin Barker, Table of n, a(n) for n = 0..1000
Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences", PhD Dissertation, Mathematics Department, Rutgers University, May 2016.
Index entries for linear recurrences with constant coefficients, signature (0,17,0,-17,0,1).

Cf. A072881, A076839, A276175.

I:=[1,1,1,4,10,55]; [n le 6 select I[n] else 17*Self(n-2)-17*Self(n-4)+Self(n-6): n in [1..30]]; // Vincenzo Librandi, Aug 27 2016

LinearRecurrence[{0, 17, 0, -17, 0, 1}, {1, 1, 1, 4, 10, 55}, 40] (* Vincenzo Librandi, Aug 27 2016 *)
nxt[{a_,b_,c_}]:={b,c,((c+1)(b+1))/a}; NestList[nxt,{1,1,1},30][[All,1]] (* Harvey P. Dale, Oct 01 2021 *)

Vec((1+x-16*x^2-13*x^3+10*x^4+4*x^5)/((1-x)*(1+x)*(1-16*x^2+x^4)) + O(x^30)) \\ Colin Barker, Aug 21 2016

a(n) = (9-3*(-1)^n)/2*a(n-1) - a(n-2) - 1.
From Colin Barker, Aug 21 2016: (Start)
a(n) = 17*a(n-2) - 17*a(n-4) + a(n-6) for n > 5.
G.f.: (1 + x - 16*x^2 - 13*x^3 + 10*x^4 + 4*x^5) / ((1-x)*(1+x)*(1 - 16*x^2 + x^4)). (End)
a(2n+1) = A073352(n). a(2n) = A048907(n). - R. J. Mathar, Jul 04 2024

More terms from Colin Barker, Aug 21 2016

A048909 9-gonal (or nonagonal) triangular numbers.

Original entry on oeis.org

1, 325, 82621, 20985481, 5330229625, 1353857339341, 343874433963061, 87342752369278225, 22184715227362706161, 5634830324997758086741, 1431224717834203191326125, 363525443499562612838749081, 92334031424171069457850940521, 23452480456295952079681300143325

Offset: 1

Eric W. Weisstein

We want solutions to m(7m-5)/2 = n(n+1)/2, or equivalently (14m-5)^2 = 7(2n+1)^2 + 18. This is the Pell-type equation x^2 - 7y^2 = 18.
This equation has unit solutions (x,y) = (5,1), (9, 3) and (19, 7), which lead to the family of solutions (5, 1), (9, 3), (19, 7), (61, 23), (135, 51), (299, 113), (971, 367), .... The corresponding integer solutions are (m,n) = (1,1), (10, 25), (154, 406), (2449, 6478), ... (A048907 and A048908), giving the nonagonal triangular numbers 1, 325, 82621, 20985481, ... shown here.
Also, numbers simultaneously 9-gonal and centered 9-gonal, the intersection of A001106 and A060544. - Steven Schlicker, Apr 24 2007

Colin Barker, Table of n, a(n) for n = 1..416
S. C. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5, December 2011, pp. 339-350.
Eric Weisstein's World of Mathematics, Nonagonal Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (255,-255,1).

Cf. A001106, A060544, A048907, A048908.

CP := n -> 1+1/2*9*(n^2-n): N:=10: u:=8: v:=1: x:=9: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+63*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp; # Steven Schlicker, Apr 24 2007

LinearRecurrence[{255, -255, 1}, {1, 325, 82621}, 12]; (* Ant King, Nov 03 2011 *)

Vec(-x*(x^2+70*x+1)/((x-1)*(x^2-254*x+1)) + O(x^20)) \\ Colin Barker, Jun 22 2015

Define x(n) + y(n)*sqrt(63) = (9+sqrt(63))*(8+sqrt(63))^n, s(n) = (y(n)+1)/2; then a(n) = (2+9*(s(n)^2-s(n)))/2. - Steven Schlicker, Apr 24 2007
a(n+1) = 254*a(n+1)-a(n)+72. - Richard Choulet, Sep 22 2007
a(n+1) = 127*a(n+1)+36+6*(448*a(n)^2+256*a(n)+25)^0.5. - Richard Choulet, Sep 22 2007
G.f.: z*(1+70*z+z^2)/((1-z)*(1-254*z+z^2)). - Richard Choulet, Sep 22 2007
From Ant King, Nov 03 2011: (Start)
a(n) = 255*a(n-1) - 255*a(n-2) + a(n-3).
a(n) = 1/112*(9*(8 + 3*sqrt(7))^(2n-1) + 9*(8-3* sqrt(7))^(2n-1) - 32).
a(n) = floor(9/112*(8 + 3*sqrt(7))^(2n-1)).
Limit_{n -> oo} a(n)/a(n-1) = (8 + 3*sqrt(7))^2. (End)

Edited by N. J. A. Sloane at the suggestion of Richard Choulet, Sep 22 2007

A048908 Indices of triangular numbers which are also 9-gonal.

Original entry on oeis.org

1, 25, 406, 6478, 103249, 1645513, 26224966, 417953950, 6661038241, 106158657913, 1691877488374, 26963881156078, 429730221008881, 6848719654986025, 109149784258767526, 1739547828485294398, 27723615471505942849, 441838299715609791193, 7041689179978250716246

Offset: 1

Eric W. Weisstein

lim( n -> Infinity , a(n)/a(n-1)) = 8 + 3*sqrt(7). - Ant King, Nov 03 2011

Colin Barker, Table of n, a(n) for n = 1..832
Eric Weisstein's World of Mathematics, Nonagonal Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (17,-17,1).

Cf. A048907, A048909.

LinearRecurrence[{17, -17, 1}, {1, 25, 406}, 16]; (* Ant King, Nov 03 2011 *)

Vec(x*(2*x^2-8*x-1)/((x-1)*(x^2-16*x+1)) + O(x^50)) \\ Colin Barker, Jun 22 2015

a(n+2) = 16*a(n+1)-a(n)+7, a(n+1) = 8*a(n)+3.5+1.5*(28*a(n)^2+28*a(n)+25)^0.5 - Richard Choulet, Sep 22 2007
G.f.: f(z) = a(1)*z+a(2)*z^2+... = (z+8z^2-2*z^3)/((1-z)*(1-16*z+z^2)) - Richard Choulet, Oct 09 2007
From Ant King, Nov 03 2011: (Start)
a(n) = 17*a(n-1) - 17*a(n-2) + a(n-3).
a(n) = floor(3/28*sqrt(7)*(3 - sqrt(7))*(8 + 3* sqrt(7))^n).
(End)

cp's OEIS Frontend

A001080 a(n) = 16*a(n-1) - a(n-2) with a(0) = 0, a(1) = 3.

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A276123 a(0) = a(1) = a(2) = 1; for n > 2, a(n) = (a(n-1) + 1)*(a(n-2) + 1) / a(n-3).

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A048909 9-gonal (or nonagonal) triangular numbers.

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A048908 Indices of triangular numbers which are also 9-gonal.

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