A003215 -id:A003215 - cp's OEIS frontend

A113530 Semiprimes in A003215.

91, 169, 217, 469, 721, 817, 1027, 1141, 1261, 1387, 2611, 2977, 3781, 3997, 4681, 5677, 5941, 6487, 6769, 7651, 7957, 8587, 9577, 10981, 11347, 12481, 12871, 14077, 14491, 15769, 16207, 17557, 18019, 18961, 20419, 20917, 21421, 22969, 24031

Offset: 1

Jonathan Vos Post, Jan 12 2006

Intersection of A003215 and A001358.

			a(1) = 91 because A003215(5) = (5+1)^3 - 5^3 = 91 = 7 * 13 is semiprime.
a(7) = 121 because A003215(7) = (7+1)^3 - 7^3 = 169 = 13^2 is semiprime; the two prime factors need not be distinct.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A001358, A003215.

Cf. A113519, A113524, A113525, A113527, A113528.

Select[Array[3 #^2 + 3 # + 1 &, 90], PrimeOmega[#] == 2 &] (* Michael De Vlieger, Mar 17 2021 *)

A001922 Numbers k such that 3k^2 - 3k + 1 is both a square (A000290) and a centered hexagonal number (A003215).

Original entry on oeis.org

1, 8, 105, 1456, 20273, 282360, 3932761, 54776288, 762935265, 10626317416, 148005508553, 2061450802320, 28712305723921, 399910829332568, 5570039304932025, 77580639439715776, 1080558912851088833, 15050244140475527880, 209622859053806301481

Offset: 0

N. J. A. Sloane

Also larger of two consecutive integers whose cubes differ by a square. Defined by a(n)^3 - (a(n) - 1)^3 = square.
Let m be the n-th ratio 2/1, 7/4, 26/15, 97/56, 362/209, ... Then a(n) = m*(2-m)/(m^2-3). The numerators 2, 7, 26, ... of m are A001075. The denominators 1, 4, 15, ... of m are A001353.
From Colin Barker, Jan 06 2015: (Start)
Also indices of centered triangular numbers (A005448) which are also centered square numbers (A001844).
Also indices of centered hexagonal numbers (A003215) which are also centered octagonal numbers (A016754).
Also positive integers x in the solutions to 3*x^2 - 4*y^2 - 3*x + 4*y = 0, the corresponding values of y being A156712.
(End)

			8 is in the sequence because 3*8^2 - 3*8 + 1 = 169 is a square and also a centered hexagonal number. - _Colin Barker_, Jan 07 2015

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).

Vincenzo Librandi, Table of n, a(n) for n = 0..800
J. Brenner and E. P. Starke, Problem E702, Amer. Math. Monthly, 53 (1946), 465.
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
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001075, A001353, A001570, A001844, A001921, A003215.

Cf. A005448, A006051, A016754, A076139, A156712.

I:=[1, 8, 105]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Apr 16 2012

seq(simplify((1 +ChebyshevU(n,7) +ChebyshevU(n-1,7))/2), n=0..30); # G. C. Greubel, Oct 07 2022

With[{s1=3+2Sqrt[3],s2=3-2Sqrt[3],t1=7+4Sqrt[3],t2=7-4Sqrt[3]}, Simplify[ Table[(s1 t1^n+s2 t2^n+6)/12,{n,0,20}]]] (* or *) LinearRecurrence[ {15,-15,1},{1,8,105},21] (* Harvey P. Dale, Aug 14 2011 *)
CoefficientList[Series[(1-7*x)/(1-15*x+15*x^2-x^3),{x,0,30}],x] (* Vincenzo Librandi, Apr 16 2012 *)

Vec((1-7*x)/(1-15*x+15*x^2-x^3) + O(x^100)) \\ Colin Barker, Jan 06 2015

[(1+chebyshev_U(n,7) +chebyshev_U(n-1,7))/2 for n in range(30)] # G. C. Greubel, Oct 07 2022

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = (s1*t1^n + s2*t2^n + 6)/12 where s1 = 3 + 2*sqrt(3), s2 = 3 - 2*sqrt(3), t1 = 7 + 4*sqrt(3), t2 = 7 - 4*sqrt(3).
a(n) = A001075(n)*A001353(n+1).
G.f.: (1-7*x)/((1-x)*(1-14*x+x^2)). - Simon Plouffe (in his 1992 dissertation) and Colin Barker, Jan 01 2012
a(n) = A076139(n+1) - 7*A076139(n). - R. J. Mathar, Jul 14 2015
a(n) = (1/2)*(1 + ChebyshevU(n, 7) + ChebyshevU(n-1, 7)). G. C. Greubel, Oct 07 2022
a(n) = 1 - a(-1-n) = 1 + A001921(n) for all integers n. - Michael Somos, Jul 10 2025

Additional comments from James R. Buddenhagen, Mar 04 2001
Name improved by Colin Barker, Jan 07 2015
Edited by Robert Israel, Feb 20 2017

A253175 Indices of hexagonal numbers (A000384) which are also centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 7, 67, 661, 6541, 64747, 640927, 6344521, 62804281, 621698287, 6154178587, 60920087581, 603046697221, 5969546884627, 59092422149047, 584954674605841, 5790454323909361, 57319588564487767, 567405431320968307, 5616734724645195301, 55599941815130984701

Offset: 1

Colin Barker, Jan 08 2015

Also positive integers x in the solutions to 4*x^2-6*y^2-2*x+6*y-2 = 0, the corresponding values of y being A253475.

			7 is in the sequence because the 7th hexagonal number is 91, which is also the 6th centered hexagonal number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A000384, A003215, A253475, A006244, A350923.

LinearRecurrence[{11, -11, 1}, {1, 7, 67}, 25] (* Paolo Xausa, May 30 2025 *)

Vec(-x*(x^2-4*x+1)/((x-1)*(x^2-10*x+1)) + O(x^100))

a(n) = 11*a(n-1)-11*a(n-2)+a(n-3).
G.f.: -x*(x^2-4*x+1) / ((x-1)*(x^2-10*x+1)).
a(n) = (2+(5-2*sqrt(6))^n*(3+sqrt(6))-(-3+sqrt(6))*(5+2*sqrt(6))^n)/8. - Colin Barker, Mar 05 2016
4*a(n) = 1+3*A072256(n). - R. J. Mathar, Feb 07 2022
a(n) = A350923(n)/2. - Paolo Xausa, May 30 2025

A253475 Indices of centered square numbers (A001844) which are also centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 6, 55, 540, 5341, 52866, 523315, 5180280, 51279481, 507614526, 5024865775, 49741043220, 492385566421, 4874114620986, 48248760643435, 477613491813360, 4727886157490161, 46801248083088246, 463284594673392295, 4586044698650834700, 45397162391834954701

Offset: 1

Colin Barker, Jan 02 2015

Also positive integers x in the solutions to 4*x^2 - 6*y^2 - 4*x + 6*y = 0, the corresponding values of y being A054318.
Also indices of centered hexagonal numbers (A003215) which are also hexagonal numbers (A000384).
Also indices of terms in sequence A193218 which are the square root of a sum of 5th powers (A000539). - Daniel Poveda Parrilla, Jun 10 2017

			6 is in the sequence because the 6th centered square number is 61, which is also the 5th centered hexagonal number.

Colin Barker, Table of n, a(n) for n = 1..1000
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).
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A000384, A000539, A001844, A003215, A054318, A193218, A253175.

LinearRecurrence[{11, -11, 1}, {1, 6, 55}, 25] (* Paolo Xausa, May 30 2025 *)

Vec(x*(5*x-1)/((x-1)*(x^2-10*x+1)) + O(x^100))

a(n) = 11*a(n-1)-11*a(n-2)+a(n-3).
G.f.: x*(5*x-1) / ((x-1)*(x^2-10*x+1)).
a(n) = sqrt((-2-(5-2*sqrt(6))^n-(5+2*sqrt(6))^n)*(2-(5-2*sqrt(6))^(1+n)-(5+2*sqrt(6))^(1+n)))/(4*sqrt(2)). - Gerry Martens, Jun 04 2015
2*a(n) = 1+A054320(n-1). - R. J. Mathar, Feb 07 2022

A350090 a(n) is the number of indices i in the range 0 <= i <= n-1 such that A003215(n) - A003215(i) is an oblong number (A002378), where A003215 are the hex numbers.

Original entry on oeis.org

0, 1, 1, 1, 1, 3, 1, 2, 3, 1, 1, 1, 3, 1, 1, 3, 3, 1, 3, 3, 3, 3, 5, 1, 1, 1, 5, 1, 1, 3, 1, 3, 1, 7, 1, 3, 3, 1, 1, 3, 7, 1, 1, 3, 3, 1, 3, 3, 1, 1, 3, 3, 1, 3, 7, 1, 3, 7, 1, 7, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 7, 5, 3, 3, 1, 5, 3, 3, 7, 3, 1, 1, 3, 3, 3, 7, 1, 3, 1, 3, 1

Offset: 0

Klaus Purath and Michel Marcus, Dec 14 2021

nonn

There are very few even terms in the data (3 up to 10000). They are obtained for indices coming from A001921. For odd terms see A350120.

a(n) = 1 for n in A111251.

			For n=5, the 5 numbers hex(5)-hex(i), for i=0 to 4, are (90, 84, 72, 54, 30) out of which 90, 72 and 30 are oblong, so a(5) = 3.

Michel Marcus, Table of n, a(n) for n = 0..10000

Cf. A002378, A003215.

Cf. also A001921, A111251, A350120.

obQ[n_] := IntegerQ @ Sqrt[4*n + 1]; hex[n_] := 3*n*(n + 1) + 1; a[n_] := Module[{h = hex[n]}, Count[Range[0, n - 1], ?(obQ[h - hex[#]] &)]]; Array[a, 100, 0] (* _Amiram Eldar, Dec 14 2021 *)

hex(n) = 3*n*(n+1)+1; \\ A003215
isob(n) = my(m=sqrtint(n)); m*(m+1)==n; \\ A002378
a(n) = my(h=hex(n)); sum(k=0, n-1, isob(h - hex(k)));

a(n) = numdiv(3*n*n + 3*n + 1) - 1; \\ Jinyuan Wang, Dec 19 2021

a(n) = A000005(A003215(n)) - 1. - Jinyuan Wang, Dec 19 2021

Edited by N. J. A. Sloane, Dec 25 2021

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

A133141 Numbers which are both centered pentagonal (A005891) and centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 331, 159391, 76825981, 37029963301, 17848365484951, 8602875133782931, 4146567966117887641, 1998637156793688059881, 963338963006591526974851, 464327381532020322313818151, 223804834559470788763733373781

Offset: 1

Richard Choulet, Sep 21 2007

The problem is to find p and r such that 6*(2*p-1)^2 = 5*(2*r+1)^2 + 1 equivalent to 3*p^2 - 3*p + 1 = (5*r^2 + 5*r + 2)/2. The Diophantine equation (6*X)^2 = 30*Y^2 + 6 is such that
X is given by 1, 21, 461, 10121, ... with a(n+2) = 22*a(n+1) - a(n) and also a(n+1) = 11*a(n) + sqrt(120*a(n)^2 - 20);
Y is given by 1, 23, 805, 11087, ... with a(n+2) = 22*a(n+1) - a(n) and also a(n+1) = 11*a(n) + sqrt(120*a(n)^2+24);
r is given by 0, 11, 252, 5543, 121704, ... with a(n+2) = 22*a(n+1) - a(n) + 10 and also a(n+1) = 11*a(n) + 5 + sqrt(120*a(n)^2 + 120*a(n) + 36);
p is given by 1, 11, 231, 5061, ... with a(n+2) = 22*a(n+1) - a(n) - 10 and also a(n+1) = 11*a(n) - 5 + sqrt(120*a(n)^2 - 120*a(n) + 25).

Colin Barker, Table of n, a(n) for n = 1..373
Index entries for linear recurrences with constant coefficients, signature (483,-483,1).

Cf. A003215, A005891, A254782, A133285.

LinearRecurrence[{483,-483,1},{1,331,159391},20] (* Paolo Xausa, Jan 07 2024 *)

Vec(-x*(x^2-152*x+1)/((x-1)*(x^2-482*x+1)) + O(x^100)) \\ Colin Barker, Feb 07 2015

a(n+2) = 482*a(n+1) - a(n) - 150.
a(n+1) = 241*a(n) - 75 + 11*sqrt(480*a(n)^2 - 300*a(n) + 45).
G.f.: z*(1-152*z+z^2)/((1-z)*(1-482*z+z^2)).

More terms from Paolo P. Lava, Sep 26 2008

A253457 Indices of centered hexagonal numbers (A003215) which are also centered heptagonal numbers (A069099).

Original entry on oeis.org

1, 14, 351, 9100, 236237, 6133050, 159223051, 4133666264, 107316099801, 2786084928550, 72330892042487, 1877817108176100, 48750913920536101, 1265645944825762514, 32858043651549289251, 853043488995455758000, 22146272670230300418737, 574950045936992355129150

Offset: 1

Colin Barker, Jan 01 2015

Also positive integers x in the solutions to 6*x^2 - 7*y^2 - 6*x + 7*y = 0, the corresponding values of y being A253458.

			14 is in the sequence because the 14th centered hexagonal number is 547, which is also the 13th centered heptagonal number.

Colin Barker, Table of n, a(n) for n = 1..708
Index entries for linear recurrences with constant coefficients, signature (27,-27,1).

Cf. A003215, A069099, A253458, A253546.

Vec(x*(13*x-1)/((x-1)*(x^2-26*x+1)) + O(x^100))

a(n) = 27*a(n-1)-27*a(n-2)+a(n-3).
G.f.: x*(13*x-1) / ((x-1)*(x^2-26*x+1)).
a(n) = sqrt((-2-(13-2*sqrt(42))^n-(13+2*sqrt(42))^n)*(2-(13-2*sqrt(42))^(1+n)-(13+2*sqrt(42))^(1+n)))/(4*sqrt(6)). - Gerry Martens, Jun 04 2015

A253458 Indices of centered heptagonal numbers (A069099) which are also centered hexagonal numbers (A003215).

Original entry on oeis.org

1, 13, 325, 8425, 218713, 5678101, 147411901, 3827031313, 99355402225, 2579413426525, 66965393687413, 1738520822446201, 45134575989913801, 1171760454915312613, 30420637251808214125, 789764808092098254625, 20503464373142746406113, 532300308893619308304301

Offset: 1

Colin Barker, Jan 01 2015

Also positive integers y in the solutions to 6*x^2 - 7*y^2 - 6*x + 7*y = 0, the corresponding values of x being A253457.

			13 is in the sequence because the 13th centered heptagonal number is 547, which is also the 14th centered hexagonal number.

Colin Barker, Table of n, a(n) for n = 1..708
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for linear recurrences with constant coefficients, signature (27,-27,1).

Cf. A003215, A069099, A253457, A253546.

LinearRecurrence[{27,-27,1},{1,13,325},20] (* Harvey P. Dale, Oct 13 2022 *)

Vec(-x*(x^2-14*x+1)/((x-1)*(x^2-26*x+1)) + O(x^100))

a(n) = 27*a(n-1)-27*a(n-2)+a(n-3).
G.f.: -x*(x^2-14*x+1) / ((x-1)*(x^2-26*x+1)).
a(n) = 1/2+(13+2*sqrt(42))^(-n)*(7+sqrt(42)-(-7+sqrt(42))*(13+2*sqrt(42))^(2*n))/28. - Colin Barker, Mar 03 2016

A253546 Centered hexagonal numbers (A003215) which are also centered heptagonal numbers (A069099).

Original entry on oeis.org

1, 547, 368551, 248402701, 167423051797, 112842888508351, 76055939431576651, 51261590333994154297, 34550235829172628419401, 23286807687272017560521851, 15695273830985510663163308047, 10578591275276546914954509101701, 7129954824262561635168675971238301

Offset: 1

Colin Barker, Jan 03 2015

			547 is in the sequence because it is the 14th centered hexagonal number and the 13th centered heptagonal number.

Colin Barker, Table of n, a(n) for n = 1..354
Index entries for linear recurrences with constant coefficients, signature (675,-675,1).

Cf. A003215, A069099, A253457, A253458.

LinearRecurrence[{675,-675,1},{1,547,368551},20] (* Harvey P. Dale, Aug 03 2021 *)

Vec(-x*(x^2-128*x+1) / ((x-1)*(x^2-674*x+1)) + O(x^100))

a(n) = 675*a(n-1)-675*a(n-2)+a(n-3).

G.f.: -x*(x^2-128*x+1) / ((x-1)*(x^2-674*x+1)).

cp's OEIS Frontend

A113530 Semiprimes in A003215.

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A001922 Numbers k such that 3*k^2 - 3*k + 1 is both a square (A000290) and a centered hexagonal number (A003215).

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A253175 Indices of hexagonal numbers (A000384) which are also centered hexagonal numbers (A003215).

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A253475 Indices of centered square numbers (A001844) which are also centered hexagonal numbers (A003215).

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A350090 a(n) is the number of indices i in the range 0 <= i <= n-1 such that A003215(n) - A003215(i) is an oblong number (A002378), where A003215 are the hex numbers.

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

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A001922 Numbers k such that 3k^2 - 3k + 1 is both a square (A000290) and a centered hexagonal number (A003215).