A046175 -id:A046175 - cp's OEIS frontend

A175410 a(n) = (b(m)-1)*b(m) = Sum_{n=b(m)+1,...,c(m)}n, b=A046174, c=A046175, m=n+1.

132, 27060, 5269320, 1022496552, 198362899020, 38481433358940, 7465200453136272, 1448210416843244880, 280945355811546307860, 54501950819034511436292, 10573097513564898455783640

Offset: 1

Manuel Valdivia, May 05 2010

nonn

Solution to (b-1)*b = (c^2+2bc+c)/2.

			A046174(2) = 12, then 11*12 = 13+14+15+16+17+18+19+20 = 132, is a term. A046174(3) = 165, then 164*165 = 166+167+,....,+285 = 27060, is a term.

A046174, A046175.

lst={};k=1;j=0;Do[b=14*k-j-2;AppendTo[lst,(b-1)*b];j=k;k=b,{n,1,16}];lst

A045899 Numbers k such that k+1 and 3*k+1 are perfect squares.

Original entry on oeis.org

0, 8, 120, 1680, 23408, 326040, 4541160, 63250208, 880961760, 12270214440, 170902040408, 2380358351280, 33154114877520, 461777249934008, 6431727384198600, 89582406128846400, 1247721958419651008, 17378525011746267720, 242051628206028097080, 3371344269872647091408

Offset: 1

Andrej Dujella (duje(AT)math.hr)

Essentially the same as A051047.
It appears that a(n) = A046175(n)-A046174(n), that is, the triangular index of the n-th pentagonal triangular number minus its pentagonal index. - Jonathan Vos Post, Feb 28 2011
Sequence lists the nonnegative x solutions when (x + 1)*(3*x + 1) is a square. Positive x solutions when (x - 1)*(3*x - 1) is a square are in A011922. - Bruno Berselli, Feb 20 2018

G. C. Greubel, Table of n, a(n) for n = 1..870
A. Baker and H. Davenport, The Equations 3x^2-2=y^2 and 8x^2-7=z^2, Quart. J. Math. Oxford 20 (1969).
A. Dujella and A. Pethoe, A generalization of a theorem of Baker and Davenport, Quart. J. Math. Oxford Ser. (2) 49 (1998), 291-306.
A. Dujella, The Problem of Diophantus and Davenport, References
A. Dujella, Publications of Andrej Dujella
Z. Franusic, On the Extension of the Diophantine Pair {1,3} in Z[surd d], J. Int. Seq. 13 (2010) # 10.9.6
P. Gibbs, 1,3,8,120 ... A Diophantine Problem
P. Gibbs, Diophantine quadruples and Cayley's hyperdeterminant, arXiv:math/0107203 [math.NT], 2001.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A051047, A067900. A076139, A076140, A123480, A217855.

Cf. A046184, A245031.

f[n_] := FullSimplify[((Sqrt[3] + 2)*(7 + 4*Sqrt[3])^n - (Sqrt[3] - 2) (7 - 4 Sqrt[3])^n - 4)/6]; Array[f, 18, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006 *)
Rest[CoefficientList[Series[-8*x^2/((x - 1)*(x^2 - 14*x + 1)), {x,0,50}], x]] (* G. C. Greubel, Jun 07 2017 *)
LinearRecurrence[{15,-15,1},{0,8,120},20] (* Harvey P. Dale, Jul 14 2024 *)

x='x+O('x^50); concat([0], Vec(-8*x^2/((x - 1)*(x^2 - 14*x + 1)))) \\ G. C. Greubel, Jun 07 2017

a(n) = A046184(n+1) - 1.
a(n) = 14*a(n-1) - a(n-2) + 8.
a(n) = ((2 + sqrt(3))*(7 + 4*sqrt(3))^n + (2 - sqrt(3))*(7 - 4*sqrt(3))^n - 4)/6. - Joseph Biberstine (jrbibers(AT)indiana.edu), Apr 23 2006
a(n) = 8*A076139(n-1) = 4*A217855(n-1) = 2*A123480(n-1) = 8/3*A076140(n-1). - Peter Bala, Dec 31 2012
From Colin Barker, Jul 30 2013: (Start)
G.f.: -8*x^2 / ((x - 1)*(x^2 - 14*x + 1)).
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). (End)
E.g.f.: (-4*exp(x) + (2 + sqrt(3))*exp((7-4*sqrt(3))*x) + (2 - sqrt(3))*exp((7+4*sqrt(3))*x))/6. - Ilya Gutkovskiy, Apr 28 2016

A046174 Indices of pentagonal numbers which are also triangular.

Original entry on oeis.org

0, 1, 12, 165, 2296, 31977, 445380, 6203341, 86401392, 1203416145, 16761424636, 233456528757, 3251629977960, 45289363162681, 630799454299572, 8785902997031325, 122371842504138976, 1704419892060914337

Offset: 0

Eric W. Weisstein

Vincenzo Librandi, Table of n, a(n) for n = 0..500
W. Sierpiński, Sur les nombres pentagonaux, Bull. Soc. Roy. Sci. Liège 33 (1964) 513-517.
W. Sierpiński, Sur les nombres pentagonaux, Bull. Soc. Roy. Sci. Liège 33 (1964) 513-517.
Eric Weisstein's World of Mathematics, Pentagonal Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1)

Cf. A014979, A046175, A001834.

[ n eq 1 select 0 else n eq 2 select 1 else 14*Self(n-1)-Self(n-2)-2: n in [1..20] ]; // Vincenzo Librandi, Aug 23 2011

LinearRecurrence[{15,-15,1},{0,1,12},20] (* Harvey P. Dale, Aug 22 2011 *)

From Warut Roonguthai, Jan 05 2001: (Start)
a(n) = 14*a(n-1) - a(n-2) - 2.
G.f.: x*(1-3*x)/((1-x)*(1-14*x+x^2)). (End)
a(n+1) = 7*a(n) - 1 + 2*sqrt(12*a(n)^2 - 4*a(n) + 1). - Richard Choulet, Sep 19 2007
a(n+1) = 15*a(n) - 15*a(n-1) + a(n-2), a(1)=1, a(2)=12, a(3)=165. - Sture Sjöstedt, May 29 2009
a(n) = (1/12)*(2 - (7 - 4*sqrt(3))^n*(1 + sqrt(3)) + (-1 + sqrt(3))*(7 + 4*sqrt(3))^n). - Alan Michael Gómez Calderón, Jun 30 2024

A046173 Indices of square numbers that are also pentagonal.

Original entry on oeis.org

1, 99, 9701, 950599, 93149001, 9127651499, 894416697901, 87643708742799, 8588189040096401, 841554882220704499, 82463790268588944501, 8080609891439495856599, 791817305570802005002201, 77590015336047156994359099, 7603029685627050583442189501

Offset: 1

Eric W. Weisstein

As n increases, this sequence is approximately geometric with common ratio r = lim_{n->oo} a(n)/a(n-1) = (sqrt(2) + sqrt(3))^4 = 49 + 20 * sqrt(6). - Ant King, Nov 07 2011
a(n)^2 is of the form (2*m-1)*(3*m-2), and the corresponding values of m are 1, 41, 3961, 388081, 38027921, 3726348121, 365144087881, ..., with closed form ((5-2*sqrt(6))^(2n-1)+(5+2*sqrt(6))^(2n-1)+14)/24 (for n>0). - Bruno Berselli, Dec 12 2013
The terms of this sequence satisfy the Diophantine equation m^2 = k * (3k-1)/2, which is equivalent to (6k-1)^2 - 6*(2*m)^2 = 1. Now, with x=6k-1 and y=2*m, we get the Pell-Fermat equation x^2 - 6*y^2 = 1. The solutions (x,y) of this equation are respectively in A046174 and A046175. The indices m=y/2 of the square numbers which are also pentagonal are the terms of this sequence, the indices k=(x+1)/6 of the pentagonal numbers which are also square are in A046172, and the pentagonal square numbers are in A036353. - Bernard Schott, Mar 10 2019
Also, this sequence is related to A302330 by (sqrt(2) + sqrt(3))^(4*n-2) = A302330(n-1)*5 + a(n)*sqrt(24). - Bruno Berselli, Oct 29 2019

			G.f. = x + 99*x^2 + 9701*x^3 + 950599*x^4 + 93149001*x^5 + ...
99 is a term because 99^2 = 9801 = (1/2) * 81 * (3*81 - 1), so 9801 is the 99th square number, also the 81st pentagonal number, and the second pentagonal square number after 1. - _Bernard Schott_, Mar 10 2019

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 35.

Colin Barker, Table of n, a(n) for n = 1..503
M. A. Asiru, All square chiliagonal numbers, Int J Math Edu Sci Technol, 47:7(2016), 1123-1134.
L. Euler, De solutione problematum diophanteorum per numeros integros, par. 21
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Pentagonal Square Number
Index entries for linear recurrences with constant coefficients, signature (98,-1).

Cf. A036353 (pentagonal square numbers), A046172 (indices of pentagonal numbers that are also square).
Cf. A046174, A046175 (solutions of x^2 - 6*y^2 = 1).
Cf. A302330.

CoefficientList[Series[(1 + x)/(1 - 98* x + x^2), {x, 0, 30}], x] (* T. D. Noe, Aug 01 2011 *)
LinearRecurrence[{98, -1}, {1, 99}, 30] (* Harvey P. Dale, Jul 31 2017 *)

{a(n) = subst( poltchebi(n) - poltchebi(n-1), 'x, 49) / 48}; /* Michael Somos, Sep 05 2006 */

Vec(x*(x+1)/(x^2-98*x+1) + O(x^30)) \\ Colin Barker, Jun 23 2015

a(n) = 98*a(n-1) - a(n-2); g.f.: (1+x)/(1-98*x+x^2). - Warut Roonguthai, Jan 05 2001
a(1-n) = -a(n) for all n in Z. - Michael Somos, Sep 05 2006
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(f(a(n-1),5),5). - Marcos Carreira, Dec 27 2006
a(n) = ((12+5*sqrt(6))/24)*(5+2*sqrt(6))^(2*n)+((12-5*sqrt(6))/24)*(5-2*sqrt(6))^(2*n) for n>=0. - Richard Choulet, Apr 29 2009
a(n+1) = 49*a(n) + 10*sqrt(24*a(n)^2+1) for n > =0 with a(0)=1. - Richard Choulet, Apr 29 2009
a(n) = b such that (-1)^n*Integral_{x=-Pi/2..Pi/2} (cos(2*n-1)*x)/(5-sin(x)) dx = c + b*(log(2)-log(3)). - Francesco Daddi, Aug 01 2011
a(n) = floor((1/24) * sqrt(6) * (sqrt(2) + sqrt(3))^(4n-2)). - Ant King, Nov 07 2011
a(n) = A138288(n)*A054320(n). - Gerry Martens, May 13 2024

A014979 Numbers that are both triangular and pentagonal.

Original entry on oeis.org

0, 1, 210, 40755, 7906276, 1533776805, 297544793910, 57722156241751, 11197800766105800, 2172315626468283465, 421418033734080886426, 81752926228785223683195, 15859646270350599313653420

Offset: 1

Glenn Johnston (glennj(AT)sonic.net)

			G.f. = x^2 + 210*x^3 + 40755*x^4 + 7906276*x^5 + 1533776805*x^6 + ...
a(4) = 40755 which is 285*(285-1)/2 = 165*(3*165-1)/2.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 210, p. 61, Ellipses, Paris 2008.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 22.

C. Gill, solution to question no. 8, Mathematical Miscellany, 1 (1836), pp. 220-225, at p. 223.
J. C. Su, On some properties of two simultaneous polygonal sequences, JIS 10 (2007) 07.10.4, example 4.2.
Eric Weisstein's World of Mathematics, Pentagonal Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (195,-195,1).

Cf. A046174, A046175, A076139, A108281.

a[ n_] := ChebyshevU[ 2 n - 3, 7] / 14 + ChebyshevT[ 2 n - 3, 7] / 84 - 1/12; (* Michael Somos, Feb 24 2015 *)
LinearRecurrence[{195,-195,1},{0,1,210},20] (* Harvey P. Dale, May 19 2017 *)

{a(n) = polchebyshev( 2*n - 3, 2, 7) / 14 + polchebyshev( 2*n - 3, 1, 7) / 84 - 1 / 12}; /* Michael Somos, Jun 16 2011 */

a(n) = 194 * a(n-1) - a(n-2) + 16.
G.f.: x^2 * (1 + 15*x) / ((1 - x) * (1 - 194*x + x^2)).
a(n)=((((1+sqrt(3))^(4*n-1)-(1-sqrt(3))^(4*n-1))/(2^(2*n+1)*sqrt(3)))^2)/2-1/8. - John Sillcox (johnsillcox(AT)hotmail.com), Sep 01 2003
a(n+1) = 97*a(n)+8+7*(192*a(n)^2+32*a(n)+1)^(1/2) - Richard Choulet, Sep 19 2007
a(n) = A076139(2*n - 3) = A108281(2 - n). for all n in Z. - Michael Somos, Jun 16 2011

Corrected and extended by Warut Roonguthai

Edited by N. J. A. Sloane, Jul 24 2006

A276915 Indices of triangular numbers in A276914 which are also pentagonal.

Original entry on oeis.org

0, 1, 10, 143, 1988, 27693, 385710, 5372251, 74825800, 1042188953, 14515819538, 202179284583, 2815994164620, 39221739020101, 546288352116790, 7608815190614963, 105977124316492688, 1476070925240282673, 20559015829047464730, 286350150681424223551

Offset: 0

Daniel Poveda Parrilla, Sep 22 2016

A276914(a(n)) = A014979(n + 1). All numbers which are both triangular and pentagonal can be found in sequence A276914.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (13,13,-1).

Cf. A014979, A046175, A276914.

RecurrenceTable[{a[n] == 14 a[n - 1] - a[n - 2] - 4 (-1)^n, a[0] == 0, a[1] == 1}, a, {n, 19}] (* Michael De Vlieger, Sep 23 2016 *)

concat(0, Vec(x*(1-3*x)/((1+x)*(1-14*x+x^2)) + O(x^30))) \\ Colin Barker, Sep 23 2016

a(n) = 14*a(n-1) - a(n-2) - 4*(-1)^n for n>1, a(0)=0, a(1)=1.
a(n) = (A046175(n) + (A046175(n) mod 2))/2.
From Colin Barker, Sep 23 2016: (Start)
G.f.: x*(1 - 3*x) / ((1 + x)*(1 - 14*x + x^2)).
a(n) = 13*a(n-1) + 13*a(n-2) - a(n-3) for n>2.
a(n) = ( -6*(-1)^n + (3+sqrt(3))*(7-4*sqrt(3))^n - (-3+sqrt(3))*(7+4*sqrt(3))^n )/24. (End)

A081065 Numbers n such that n^2 = (1/3)(n+floor(sqrt(3)nfloor(sqrt(3)n))).

Original entry on oeis.org

2, 24, 330, 4592, 63954, 890760, 12406682, 172802784, 2406832290, 33522849272, 466913057514, 6503259955920, 90578726325362, 1261598908599144, 17571805994062650, 244743685008277952, 3408839784121828674, 47479013292697323480, 661297346313640700042

Offset: 1

Benoit Cloitre, Apr 15 2003

a(n)/2 gives indices of pentagonal numbers which are also triangular.

a(n) itself gives x-values solving the Diophantine equation 2*x^2 + (x-1)^2 = y^2.

Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A046090, A046174, A046175.

LinearRecurrence[{15,-15,1},{2,24,330},20] (* Harvey P. Dale, Mar 14 2016 *)

Vec(2*(1-3*x)/((1-x)*(1-14*x+x^2)) + O(x^40)) \\ Michel Marcus, Nov 17 2014

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = 14*a(n-1) - a(n-2) - 4. [Sture Sjöstedt, May 02 2011]
G.f.: 2*(1-3*x)/((1-x)*(1-14*x+x^2)). - Bruno Berselli, Nov 11 2011

A189356 a(n) gives y-values solving the Diophantine equation 2*x^2 + (x-1)^2 = y^2 for positive x.

Original entry on oeis.org

3, 41, 571, 7953, 110771, 1542841, 21489003, 299303201, 4168755811, 58063278153, 808717138331, 11263976658481, 156886956080403, 2185153408467161, 30435260762459851, 423908497265970753, 5904283700961130691, 82236063316189858921, 1145400602725696894203

Offset: 1

Sture Sjöstedt, May 02 2011

nonn

(a(n)-1)/2 gives indices of triangular numbers which are also pentagonal (A046175).

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Cf. A081065, A046174, A046175.

[n le 2 select 38*n-35 else 14*Self(n-1)-Self(n-2): n in [1..19]]; // Bruno Berselli, May 03 2011

LinearRecurrence[{14,-1}, {3, 41}, 19]  (* Bruno Berselli, Nov 11 2011 *)

a(n) = 14*a(n-1) - a(n-2).

G.f.: x*(3-x)/(1-14*x+x^2). - Bruno Berselli, May 03 2011

Extended by T. D. Noe, May 02 2011

cp's OEIS Frontend

A175410 a(n) = (b(m)-1)*b(m) = Sum_{n=b(m)+1,...,c(m)}n, b=A046174, c=A046175, m=n+1.

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A045899 Numbers k such that k+1 and 3*k+1 are perfect squares.

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A046174 Indices of pentagonal numbers which are also triangular.

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A046173 Indices of square numbers that are also pentagonal.

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A014979 Numbers that are both triangular and pentagonal.

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Extensions

A276915 Indices of triangular numbers in A276914 which are also pentagonal.

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A081065 Numbers n such that n^2 = (1/3)*(n+floor(sqrt(3)*n*floor(sqrt(3)*n))).

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A081065 Numbers n such that n^2 = (1/3)(n+floor(sqrt(3)nfloor(sqrt(3)n))).