A007667 -id:A007667 - cp's OEIS frontend

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, 102558961, 1015229051, 10049731549, 99482086439, 984771132841, 9748229241971, 96497521286869, 955226983626719, 9455772314980321, 93602496166176491, 926569189346784589, 9172089397301669399, 90794324783669909401

Offset: 0

Ignacio Larrosa Cañestro, Feb 27 2000

Chebyshev's even-indexed U-polynomials evaluated at sqrt(3).
a(n)^2 is a star number (A003154).
Any k in the sequence has the successor 5*k + 2*sqrt(3(2*k^2 + 1)). - Lekraj Beedassy, Jul 08 2002
{a(n)} give the values of x solving: 3*y^2 - 2*x^2 = 1. Corresponding values of y are given by A072256(n+1). x + y = A001078(n+1). - Richard R. Forberg, Nov 21 2013
The aerated sequence (b(n))n>=1 = [1, 0, 11, 0, 109, 0, 1079, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

			a(1)^2 = 121 is the 5th star number (A003154).

G. C. Greubel, Table of n, a(n) for n = 0..1000
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.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Star Number
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (10,-1).
Index entries for sequences related to Chebyshev polynomials.

A member of the family A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, which are the expansions of (1+x) / (1-kx+x^2) with k = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, May 04 2004
Cf. A003154, A006061, A031138, A007667, A004189.
Cf. A138281. Cf. A100047.
Cf. A142238.

a:=[1,11];; for n in [3..30] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jul 22 2019

I:=[1,11]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

CoefficientList[Series[(1+x)/(1-10x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Mar 22 2015 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Numerator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A142238 *)
a[3/2, 20] (* Gerry Martens, Jun 07 2015 *)

a(n)=subst(poltchebi(n+1)-poltchebi(n),x,5)/4;

(a(n)-1)^2 + a(n)^2 + (a(n)+1)^2 = b(n)^2 + (b(n)+1)^2 = c(n), where b(n) is A031138 and c(n) is A007667.
a(n) = 10*a(n-1) - a(n-2).
a(n) = (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1).
a(n) = U(2*(n-1), sqrt(3)) = S(n-1, 10) + S(n-2, 10) with Chebyshev's U(n, x) and S(n, x) := U(n, x/2) polynomials and S(-1, x) := 0. S(n, 10) = A004189(n+1), n >= 0.
6*a(n)^2 + 3 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then (-1)^n*q(n, -12) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-10)*(-1)^n, where L is defined as in A108299; see also A072256 for L(n,+10). - Reinhard Zumkeller, Jun 01 2005
From Reinhard Zumkeller, Mar 12 2008: (Start)
(sqrt(2) + sqrt(3))^(2*n+1) = a(n)*sqrt(2) + A138288(n)*sqrt(3);
a(n) = A138288(n) + A001078(n).
a(n) = A001079(n) + 3*A001078(n). (End)
a(n) = A142238(2n) = A041006(2n)/2 = A041038(2n)/4. - M. F. Hasler, Feb 14 2009
a(n) = sqrt(A006061(n)). - Zak Seidov, Oct 22 2012
a(n) = sqrt((3*A072256(n)^2 - 1)/2). - T. D. Noe, Oct 23 2012
(sqrt(3) + sqrt(2))^(2*n+1) - (sqrt(3) - sqrt(2))^(2*n+1) = a(n)*sqrt(8). - Bruno Berselli, Oct 29 2019
a(n) = A004189(n)+A004189(n+1). - R. J. Mathar, Oct 01 2021
E.g.f.: exp(5*x)*(2*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/2. - Stefano Spezia, May 16 2023
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 5), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 10*a(n)*a(n+1) + a(n+1)^2 = 12.
More generally, for arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 12 with a(n) := (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1) as given above.
a(n+1/2) = sqrt(3) * A001078(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6)*sqrt(sqrt(3) + 1) * A001078(n+1).
a(n+3/4) - a(n+1/4) = sqrt(sqrt(3) - 1) * A001079(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/12 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A004291(n) + 1/A004291(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3/2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3/2 * (1 - 1/A171640(k+2))). (End)

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A031138 Numbers k such that 1^5 + 2^5 + ... + k^5 is a square.

Original entry on oeis.org

1, 13, 133, 1321, 13081, 129493, 1281853, 12689041, 125608561, 1243396573, 12308357173, 121840175161, 1206093394441, 11939093769253, 118184844298093, 1169909349211681, 11580908647818721, 114639177128975533, 1134810862641936613, 11233469449290390601

Offset: 1

Ignacio Larrosa Cañestro, entry revised Feb 27 2000

Partial sums of A004291 or convolution of A040000 with A054320. - R. J. Mathar, Oct 26 2009
This is a 6th-degree Diophantine equation 12*m^2 = n^2*(n+1)^2*(2*n^2 + 2*n - 1) which reduces to the generalized Pell equation 6*q^2 = (2*n + 1)^2 - 3 where q = 3*m/(n*(n+1)), so there is no surprise that the solutions satisfy a linear recurrent equation. - Charles R Greathouse IV, Max Alekseyev, Oct 22 2012
Also k such that k^2 + (k+1)^2 is equal to the sum of three consecutive squares, for example 13^2 + 14^2 = 10^2 + 11^2 + 12^2. - Colin Barker, Sep 06 2015

			a(2) = 13 because 1^5+2^5+...13^5 = 1001^2; a(1) = 1 because 1^5 = 1^2.

Colin Barker, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (11, -11, 1).

Cf. A000539, A006061, A054320, A007667.

[Round(-1/2 + ((3 - Sqrt(6))/4)*(5 + 2*Sqrt(6))^n + ((3 + Sqrt(6) )/4)*(5 - 2*Sqrt(6))^n): n in [0..50]]; // G. C. Greubel, Nov 04 2017

LinearRecurrence[{11,-11,1},{1,13,133},20 ] (* Harvey P. Dale, Oct 23 2012 *)

isok(n) = issquare(sum(i=1, n, i^5)); \\ Michel Marcus, Dec 28 2013

Vec(x*(1+x)^2/((1-x)*(x^2-10*x+1)) + O(x^40)) \\ Colin Barker, Sep 06 2015

a(n) = 11*(a(n-1) - a(n-2)) + a(n-3).
a(n) = -1/2 + ((3 - sqrt(6))/4)*(5 + 2*sqrt(6))^n + ((3 + sqrt(6))/4)*(5 - 2*sqrt(6))^n.
a(n)^2 + (a(n) + 1)^2 = (b(n) - 1)^2 + b(n)^2 + (b(n) + 1)^2 = c(n) = 3*d(n) + 2; where b(n) is A054320, c(n) is A007667 and d(n) is A006061.
a(n) = 10*a(n-1) - a(n-2) + 4; a(0) = a(1) = 1. Also sum of first a(n) fifth powers is a square m^2, where m has factors A000217{a(n)} and A054320(n). - Lekraj Beedassy, Jul 08 2002
contfrac(sqrt(6)/A054320(n))[4]/2 - Thomas Baruchel, Dec 02 2003
G.f.: x*(1+x)^2/((1-x)*(x^2-10*x+1)). - R. J. Mathar, Oct 26 2009

A006061 Star numbers (A003154) that are squares.

Original entry on oeis.org

1, 121, 11881, 1164241, 114083761, 11179044361, 1095432263641, 107341182792481, 10518340481399521, 1030690025994360601, 100997104206965939401, 9896685522256667700721, 969774184076946468731281

Offset: 1

N. J. A. Sloane

			a(2)=121 because this is the 2nd star number (A003154) that is a square.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 121, p. 42, Ellipses, Paris 2008.
M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 22.
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..500
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, Star Number
Index entries for linear recurrences with constant coefficients, signature (99, -99, 1).

A007667 is 3*a(n)+2, sqrt(a(n)) is A054320.

Cf. A003154.

a:=[1,121,11881];; for n in [4..20] do a[n]:=99*a[n-1]-99*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Jul 23 2019

R:=PowerSeriesRing(Integers(), 20); Coefficients(R!( (1+22*x+x^2)/((1-x)*(1-98*x+x^2)) )); // G. C. Greubel, Jul 23 2019

Digits := 1000:q := seq(floor(evalf(( (5+2*sqrt(6))^n*(sqrt(6)-2)-(5-2*sqrt(6))^n*(sqrt(6)+2))^2/16)),n=1..100);
A006061:=-(1+22*z+z**2)/(z-1)/(z**2-98*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation

CoefficientList[Series[(1+22*x+x^2)/((1-x)*(1-98*x+x^2)), {x,0,20}], x] (* or *) LinearRecurrence[{99,-99,1}, {1,121,11881}, 20] (* G. C. Greubel, Jul 23 2019 *)

my(x='x+O('x^20)); Vec((1+22*x+x^2)/((1-x)*(1-98*x+x^2))) \\ G. C. Greubel, Jul 23 2019

((1+22*x+x^2)/((1-x)*(1-98*x+x^2))).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Jul 23 2019

A007667 = 3*square star numbers (A006061) + 2.
a(n) = denominator of kappa(sqrt(6)/A054320(n)) where kappa(x) is the sum of successive remainders by computing the Euclidean algorithm for (1, x). - Thomas Baruchel, Nov 29 2003
From Ignacio Larrosa Cañestro, Feb 27 2000: (Start)
a(n) = 99*(a(n-1) - a(n-2)) + a(n-3).
a(n) = (5 - 2*sqrt(6))/8*(sqrt(3) + sqrt(2))^(4*n) + (5 + 2*sqrt(6))/8*(sqrt(3) - sqrt(2))^(4*n) - 1/4. (End)
a(n) = 98*a(n-1) - a(n-2) + 24. - Lekraj Beedassy, Jul 14 2008

More terms from Eric W. Weisstein and Sascha Kurz, Mar 24 2002

A129863 Sums of three consecutive pentagonal numbers.

Original entry on oeis.org

6, 18, 39, 69, 108, 156, 213, 279, 354, 438, 531, 633, 744, 864, 993, 1131, 1278, 1434, 1599, 1773, 1956, 2148, 2349, 2559, 2778, 3006, 3243, 3489, 3744, 4008, 4281, 4563, 4854, 5154, 5463, 5781, 6108, 6444, 6789, 7143, 7506, 7878, 8259, 8649, 9048, 9456, 9873

Offset: 0

Jonathan Vos Post, May 23 2007, May 24 2007

Arises in pentagonal number analog to A129803, Triangular numbers that are the sum of three consecutive triangular numbers. What are the pentagonal numbers which are the sum of three consecutive pentagonal numbers?

			a(0) = 6 = A000326(0) + A000326(1) + A000326(2) = 0 + 1 + 5.
a(1) = 18 = A000326(1) + A000326(2) + A000326(3) = 1 + 5 + 12.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000326, A007667, A034961, A129803.

[(9/2)*(n^2)+(15/2)*n+6: n in [0..50]]; // Vincenzo Librandi, Aug 16 2017

Table[(3/2)*(4 + 5*n + 3*n^2), {n, 0, 100}] (* Stefan Steinerberger, May 27 2007 *)
CoefficientList[Series[3 (2 + x^2) / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Aug 16 2017 *)
Total/@Partition[PolygonalNumber[5,Range[0,50]],3,1] (* Requires Mathematica version 10 or later *) (* or *) LinearRecurrence[{3,-3,1},{6,18,39},50] (* Harvey P. Dale, Nov 22 2018 *)

a(n)=n*(9*n+15)/2+6 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = P(n) + P(n+1) + P(n+2) where P(n) = A000326(n) = n*(3*n-1)/2.
a(n) = (9/2)*(n^2) + (15/2)*n + 6.
a(n) = (3*n^2 + 5*n + 4)*(3/2). - Stefan Steinerberger, May 27 2007
G.f.: 3*(2+x^2)/(1-x)^3. - Colin Barker, Feb 13 2012
From Elmo R. Oliveira, Nov 16 2024: (Start)
E.g.f.: 3*exp(x)*(3*x^2 + 8*x + 4)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

Offset corrected by Eric Rowland, Aug 15 2017

A006244 Hexagonal numbers (A000384) which are also centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 91, 8911, 873181, 85562821, 8384283271, 821574197731, 80505887094361, 7888755361049641, 773017519495770451, 75747828155224454551, 7422514141692500775541, 727330638057709851548461, 71270980015513872950973631, 6983828710882301839343867371, 684343942686450066382748028721

Offset: 1

N. J. A. Sloane, Jeffrey Shallit

Equivalently, triangular hex numbers.

			a(1)=91 because 91 is the sixth centered hexagonal number and the seventh hexagonal number.

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).

Jon E. Schoenfield, Table of n, a(n) for n = 1..500
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
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, Hex Number
Index entries for linear recurrences with constant coefficients, signature (99,-99,1).

Cf. A001570, A001921, A000384, A003215, A253175.

CP := n -> 1+1/2*6*(n^2-n): N:=10: u:=5: v:=1: x:=6: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+24*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
A006244:=-(1-8*z+z**2)/(z-1)/(z**2-98*z+1); # Conjectured (correctly) by Simon Plouffe in his 1992 dissertation.
a := n -> (Matrix([[91,1,1]]). Matrix([[99,1,0],[ -99,0,1],[1,0,0]])^n)[1,3]; seq (a(n), n=1..20); # Alois P. Heinz, Aug 14 2008

CoefficientList[Series[(1 - 8*x + x^2)/(1 - 99*x + 99*x^2 - x^3), {x, 0, 20}], x] (* Jean-François Alcover, Feb 26 2015 *)

Vec(-x*(x^2-8*x+1)/((x-1)*(x^2-98*x+1)) + O(x^100)) \\ Colin Barker, Jan 08 2015

From Richard Choulet, Sep 19 2007: (Start)
We must solve 2*r^2-r=3*p^2-3*p+1, which gives X^2=6*Y^2+3 with X=4*r-1 and Y=2*p-1. We obtain at the same time the following sequences:
X is given by 3, 27, 267, ... sequence for which a(n+2)=10*a(n+1)-a(n) and a(n+1)=5*a(n)+2*(6a(n)^2-18)^0.5
Y is given by 1, 11, 109, ... sequence for which a(n+2)=10*a(n+1)-a(n) and a(n+1)=5*a(n)+2*(6a(n)^2+3)^0.5
p is given by 1, 6, 55, 540, ... sequence for which a(n+2)=10*a(n+1)-a(n)-4 and a(n+1)=5*a(n)-2+(24*a(n)^2-24*a(n)+9)^0.5
r is given by 1, 7, 67, 661, ... sequence for which a(n+2)=10*a(n+1)-a(n)-2 and a(n+1)=5*a(n)-1+(24*a(n)^2-12*a(n)-3)^0.5
a(n+2) = 98*a(n+1)-a(n)-6, a(n+1)=49*a(n)-3+5*(96*a(n)^2-12*a(n)-3)^0.5.
G.f.: z*(1-8*z+z^2)/((1-z)*(1-98*z+z^2)). (End)
Define x(n) + y(n)*sqrt(24) = (6+sqrt(24))*(5+sqrt(24))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+6*(s(n)^2-s(n))). - Steven Schlicker, Apr 24 2007
a(n) = (A007667(n+1)-1)/4. - Ralf Stephan, Mar 03 2004
a(n) = 99*a(n-1)-99*a(n-2)+a(n-3). - Colin Barker, Jan 08 2015

Edited by N. J. A. Sloane, Sep 25 2007
More terms from Alois P. Heinz, Aug 14 2008
More terms from Jon E. Schoenfield, Dec 26 2008

A129109 Sums of three consecutive hexagonal numbers.

Original entry on oeis.org

7, 22, 49, 88, 139, 202, 277, 364, 463, 574, 697, 832, 979, 1138, 1309, 1492, 1687, 1894, 2113, 2344, 2587, 2842, 3109, 3388, 3679, 3982, 4297, 4624, 4963, 5314, 5677, 6052, 6439, 6838, 7249, 7672, 8107, 8554, 9013, 9484, 9967, 10462, 10969, 11488, 12019, 12562

Offset: 0

Jonathan Vos Post, May 24 2007

Arises in hexagonal number analog to A129803 Triangular numbers which are the sum of three consecutive triangular numbers. What are the hexagonal numbers which are the sum of three consecutive hexagonal numbers? Prime for a(0) = 7, a(4) = 139, a(6) = 277, a(8) = 463, a(18) = 2113, a(22) = 3109, a(26) = 4297, a(38) = 9013, a(40) = 9967.

			a(0) = H(0) + H(1) + H(2) = 0 + 1 + 6 = 7 = 6*0^2 + 9*0 + 7.
a(1) = H(1) + H(2) + H(3) = 1 + 6 + 15 = 22 = 6*1^2 + 9*1 + 7.
a(2) = H(2) + H(3) + H(4) = 6 + 15 + 28 = 49 = 6*2^2 + 9*2 + 7.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000384, A007667, A034961, A129803, A129863.

I:=[7,22,49]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 20 2012

LinearRecurrence[{3,-3,1},{7,22,49},50] (* Vincenzo Librandi, Feb 20 2012 *)
Total/@Partition[PolygonalNumber[6,Range[0,50]],3,1] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Mar 14 2020 *)

a(n)=6*n^2+9*n+7 \\ Charles R Greathouse IV, Feb 20 2012

a(n) = H(n) + H(n+1) + H(n+2) where H(n) = A000384(n) = n*(2*n-1).
a(n) = 6*n^2 + 9*n + 7.
From Colin Barker, Feb 20 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: (7 + x + 4*x^2)/(1-x)^3. (End)
E.g.f.: (7 + 15*x + 6*x^2)*exp(x). - Elmo R. Oliveira, Nov 16 2024

A129111 Sums of three consecutive heptagonal numbers.

Original entry on oeis.org

8, 26, 59, 107, 170, 248, 341, 449, 572, 710, 863, 1031, 1214, 1412, 1625, 1853, 2096, 2354, 2627, 2915, 3218, 3536, 3869, 4217, 4580, 4958, 5351, 5759, 6182, 6620, 7073, 7541, 8024, 8522, 9035, 9563, 10106, 10664, 11237, 11825, 12428, 13046, 13679, 14327, 14990

Offset: 0

Jonathan Vos Post, May 24 2007

Arises in heptagonal number analog to A129803 (Triangular numbers which are the sum of three consecutive triangular numbers).
What are the heptagonal numbers which are the sum of three consecutive heptagonal numbers?
Prime for a(2) = 59, a(3) = 107, a(7) = 449, a(10) = 863, a(11) = 1031, a(23) = 4217, a(26) = 5351, a(31) = 7541, a(42) = 13679, a(43) = 14327, a(46) = 16361, a(51) = 20051.

			a(0) = Hep(0) + Hep(1) + Hep(2) = 0 + 1 + 7 = 8 = (15/2)*0^2 + (21/2)*0 + 8.
a(1) = Hep(1) + Hep(2) + Hep(3) = 1 + 7 + 18 = 26 = (15/2)*1^2 + (21/2)*1 + 8.
a(2) = Hep(2) + Hep(3) + Hep(4) = 7 + 18 + 34 = 59 = (15/2)*2^2 + (21/2)*2 + 8.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000566, A007667, A034961, A129803, A129863.

I:=[8,26,59]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 20 2012

LinearRecurrence[{3,-3,1},{8,26,59},50] (* Vincenzo Librandi, Feb 12 2012 *)

a(n)=3*n*(5*n+7)/2+8 \\ Charles R Greathouse IV, Jun 17 2017

def a(n): return 3*n*(5*n+7)//2 + 8
print([a(n) for n in range(44)]) # Michael S. Branicky, Aug 26 2021

a(n) = Hep(n) + Hep(n+1) + Hep(n+2) where Hep(n) = A000566(n) = n*(5*n-3)/2.
a(n) = (15/2)*n^2 + (21/2)*n + 8.
From Colin Barker, Feb 20 2012: (Start)
G.f.: (8 + 2*x + 5*x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
E.g.f.: exp(x)*(16 + 36*x + 15*x^2)/2. - Elmo R. Oliveira, Nov 16 2024

cp's OEIS Frontend

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

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A031138 Numbers k such that 1^5 + 2^5 + ... + k^5 is a square.

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A006061 Star numbers (A003154) that are squares.

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A129863 Sums of three consecutive pentagonal numbers.

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A006244 Hexagonal numbers (A000384) which are also centered hexagonal numbers (A003215).

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A129109 Sums of three consecutive hexagonal numbers.

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