A005891 -id:A005891 - cp's OEIS frontend

A129556 Numbers k such that the k-th centered pentagonal number A005891(k) = (5k^2 + 5k + 2)/2 is a square.

0, 2, 21, 95, 816, 3626, 31005, 137711, 1177392, 5229410, 44709909, 198579887, 1697799168, 7540806314, 64471658493, 286352060063, 2448225223584, 10873837476098, 92968086837717, 412919472031679, 3530339074609680, 15680066099727722, 134059916748330141

Offset: 1

Alexander Adamchuk, Apr 20 2007

Corresponding numbers m > 0 such that m^2 is a centered pentagonal number are listed in A129557 = {1, 4, 34, 151, 1291, 5734, 49024, ...}.
From Andrea Pinos, Nov 02 2022: (Start)
By definition: 5*T(a(n)) = A129557(n)^2 - 1 where triangular number T(j) = j*(j+1)/2. This implies:
Every odd prime factor of a(n) and d(n)=a(n)+1 is present in b(n)=A129557(n)+1 or in c(n)=A129557(n)-1. (End)
From the law of cosines the non-Pythagorean triple {a(n), a(n)+1=A254332(n), A129557(n+1)} forms a near-isosceles triangle whose angle between the consecutive integer sides is equal to the central angle of the regular pentachoron polytope (4-simplex) (see A140244 and A140245). This implies that the terms {a(n)} are also those numbers k such that 1 + 5*A000217(k) is a square. - Federico Provvedi, Apr 04 2023

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Centered Pentagonal Number
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. A005891 (centered pentagonal numbers), A129557 (numbers k>0 such that k^2 is a centered pentagonal number), A221874.

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; A233450 for k=3; A001652 for k=4; this sequence for k=5; A001921 for k=6. - Bruno Berselli, Dec 16 2013

A005891 := proc(n) (5*n^2+5*n+2)/2 ; end: n := 0 : while true do if issqr(A005891(n)) then print(n) ; fi ; n := n+1 ; od : # R. J. Mathar, Jun 06 2007

Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[n] ], {n,1,40000} ]
LinearRecurrence[{1,38,-38,-1,1},{0,2,21,95,816},30] (* Harvey P. Dale, Nov 09 2017 *)
Table[(((x^(n+2))+(((-1)^n*(x^(2*n+1)+1)-x)/(x^n)))/(x^2+1)-1)/2/.x->3+Sqrt[10],{n,0,50}]//Round (* Federico Provvedi, Apr 04 2023 *)

a(n)=([0,1,0,0,0; 0,0,1,0,0; 0,0,0,1,0; 0,0,0,0,1; 1,-1,-38,38,1]^(n-1)*[0;2;21;95;816])[1,1] \\ Charles R Greathouse IV, Feb 11 2019

For n >= 5, a(n) = 38*a(n-2) - a(n-4) + 18. - Max Alekseyev, May 08 2009
G.f.: x^2*(x^3+2*x^2-19*x-2) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)). - Colin Barker, Feb 21 2013
a(n) = (A221874(n) - 1) / 2. - Bruno Berselli, Feb 21 2013
From Andrea Pinos, Oct 24 2022: (Start)
The ratios of successive terms converge to two different limits:
lower: D = lim_{n->oo} a(2n)/a(2n-1) = (7+2*sqrt(10))/3;
upper: E = lim_{n->oo} a(2n+1)/a(2n) = (13+4*sqrt(10))/3.
So lim_{n->oo} a(n+2)/a(n) = D*E = 19 + 6*sqrt(10).  (End)
a(n) = (x^(2*(n+1)) + (-1)^n*(x^(2*n+1)+1) - x) / (2*x^n*(x^2 + 1)) - (1/2), with x=3+sqrt(10). - Federico Provvedi, Apr 04 2023

More terms from R. J. Mathar, Jun 06 2007
Further terms from Max Alekseyev, May 08 2009
a(22)-a(23) from Colin Barker, Feb 21 2013

A254627 Indices of centered pentagonal numbers (A005891) that are also triangular numbers (A000217).

Original entry on oeis.org

1, 2, 11, 28, 189, 494, 3383, 8856, 60697, 158906, 1089155, 2851444, 19544085, 51167078, 350704367, 918155952, 6293134513, 16475640050, 112925716859, 295643364940, 2026369768941, 5305104928862, 36361730124071, 95196245354568, 652484772464329

Offset: 1

Colin Barker, Feb 03 2015

Also positive integers y in the solutions to x^2 - 5*y^2 + x + 5*y - 2 = 0, the corresponding values of x being A254626.

Also indices of centered pentagonal numbers (A005891) that are also hexagonal numbers (A000384). - Colin Barker, Feb 11 2015

			2 is in the sequence because the 2nd centered pentagonal number is 6, which is also the 3rd triangular number.

Colin Barker, Table of n, a(n) for n = 1..1000
Hermann Stamm-Wilbrandt, 6 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,18,-18,-1,1).

Cf. A000217, A005891, A254626, A254628.

Cf. A000384, A254962.

[(2 +(1+2*(-1)^n)*Fibonacci(3*n) -(-1)^n*Lucas(3*n))/4 : n in [1..30]]; // G. C. Greubel, Apr 19 2019

CoefficientList[Series[x (x^3 + 9 x^2 - x - 1)/((x - 1) (x^2 - 4 x - 1) (x^2 + 4 x - 1)), {x, 0, 25}], x] (* Michael De Vlieger, Jun 06 2016 *)
LinearRecurrence[{1,18,-18,-1,1},{1,2,11,28,189},30] (* Harvey P. Dale, Apr 23 2017 *)

Vec(x*(x^3+9*x^2-x-1)/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)) + O(x^30))

{a(n) = (2 +(1+3*(-1)^n)*fibonacci(3*n) - 2*(-1)^n*fibonacci(3*n+1))/4}; \\ G. C. Greubel, Apr 19 2019

[(2 +(1+3*(-1)^n)*fibonacci(3*n) -2*(-1)^n*fibonacci(3*n+1))/4 for n in (1..30)] # G. C. Greubel, Apr 19 2019

a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(1+x-9*x^2-x^3)/((1-x)*(1+4*x-x^2)*(1-4*x-x^2)).
a(n) = (10 - sqrt(5)*(2-sqrt(5))^n - 5*(-2+sqrt(5))^n - 2*sqrt(5)*(-2+sqrt(5))^n + sqrt(5)*(2+sqrt(5))^n + (-2-sqrt(5))^n*(-5+2*sqrt(5)))/20. - Colin Barker, Jun 06 2016
a(2*n+2) = A232970(2*n+1); a(2*n+1) = A110679(2*n). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019
a(n) = (2 +(1+2*(-1)^n)*Fibonacci(3*n) -(-1)^n*Lucas(3*n))/4. - G. C. Greubel, Apr 19 2019

A322638 Numbers that are sums of consecutive centered pentagonal numbers (A005891).

Original entry on oeis.org

0, 1, 6, 7, 16, 22, 23, 31, 47, 51, 53, 54, 76, 82, 98, 104, 105, 106, 127, 141, 158, 174, 180, 181, 182, 226, 233, 247, 264, 276, 280, 286, 287, 322, 323, 331, 374, 391, 405, 407, 421, 427, 428, 456, 502, 504, 526, 548, 555, 586, 601, 602, 607, 608, 609, 654, 681, 683, 722

Offset: 1

Ilya Gutkovskiy, Dec 21 2018

nonn

Robert Israel, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Centered Pentagonal Number

Cf. A004068, A005891, A152043, A319184, A322610, A322611, A322640.

L:= [seq((5*n^2+5*n+2)/2,n=0..30)]: N:= L[-1]:
S:=[0,op(ListTools:-PartialSums(L))]:
R:=select(`<=`,{0,seq(seq(S[n]-S[m],m=1..n-1),n=1..nops(S))},N):
sort(convert(R,list)); # Robert Israel, Mar 19 2023

terms = 59;
nmax = 16; kmax = 9; (* empirical *)
T = Table[(5n^2 + 5n + 2)/2, {n, 0, nmax}];
Union[{0}, T, Table[k MovingAverage[T, k], {k, 2, kmax}] // Flatten][[1 ;; terms]] (* Jean-François Alcover, Dec 26 2018 *)

A129557 Numbers k > 0 such that k^2 is a centered pentagonal number (A005891).

Original entry on oeis.org

1, 4, 34, 151, 1291, 5734, 49024, 217741, 1861621, 8268424, 70692574, 313982371, 2684456191, 11923061674, 101938642684, 452762361241, 3870983965801, 17193046665484, 146995452057754, 652883010927151, 5581956194228851, 24792361368566254

Offset: 1

Alexander Adamchuk, Apr 20 2007

Corresponding numbers m such that centered pentagonal number A005891(m) = (5*m^2 + 5*m + 2)/2 is a perfect square are listed in A129556 = {0, 2, 21, 95, 816, 3626, 31005, ...}.

Also positive integers x in the solutions to 2*x^2 - 5*y^2 + 5*y - 2 = 0, the corresponding values of y being A254332. - Colin Barker, Jan 28 2015

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Centered Pentagonal Number.
Index entries for linear recurrences with constant coefficients, signature (0,38,0,-1).

Cf. A005891 (centered pentagonal numbers).
Cf. A129556 (k such that A005891(k) is a perfect square).
Cf. A000290, A005667, A254332, A254333.

Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[ Sqrt[f] ] ], {n,1,40000} ]
CoefficientList[Series[(1-x)*(1+5*x+x^2)/((1+6*x-x^2)*(1-6*x-x^2)),{x,0,30}],x] (* Vincenzo Librandi, Apr 11 2012 *)

A129557()={ for(n=1,1000000000, f=(5*n^2+5*n+2)/2 ; if(issquare(f), print1(sqrtint(f), ", ") ; ); ) ; } \\ R. J. Mathar, Oct 11 2007

Vec(x*(1-x)*(1+5*x+x^2)/((1+6*x-x^2)*(1-6*x-x^2)) + O(x^100)) \\ Colin Barker, Jan 28 2015

a(n) = sqrt( (5*A129556(n)^2 + 5*A129556(n) + 2)/2 ).
For n >= 5, a(n) = 38*a(n-2) - a(n-4). - Max Alekseyev, May 08 2009
G.f.: x*(1-x)*(1 + 5*x + x^2)/((1 + 6*x - x^2)*(1 - 6*x - x^2)). - Colin Barker, Apr 11 2012
From Andrea Pinos, Oct 07 2022: (Start)
The ratios of successive terms converge to two different limits:
  lower: D = lim_{n->oo} a(2n)/a(2n-1) = (7 + 2*sqrt(10))/3;
  upper: E = lim_{n->oo} a(2n+1)/a(2n) = (13 + 4*sqrt(10))/3.
So lim_{n->oo} a(n+2)/a(n) = D*E = 19 + 6*sqrt(10).
a(n) = (A005667(n) - (-1)^n*A005667(n-1))/4. (End)

More terms from R. J. Mathar, Oct 11 2007

More terms from Max Alekseyev, May 08 2009

A145838 Primes in A005891 = Centered pentagonal numbers: (5n^2 + 5n + 2)/2.

Original entry on oeis.org

31, 181, 331, 601, 1051, 1381, 3331, 4951, 5641, 5881, 9151, 11731, 12781, 14251, 17431, 17851, 19141, 21391, 31081, 33931, 41281, 43891, 51481, 52201, 61231, 63601, 67651, 70141, 70981, 84181, 92641, 100501, 104551, 107641, 116101, 126001

Offset: 1

Alexander Adamchuk, Oct 21 2008

nonn

Eric Weisstein's World of Mathematics, Centerd Pentagonal Number

Cf. A005891 = Centered pentagonal numbers: (5n^2 + 5n + 2)/2.

A005891 := proc(n) (5*n*(n+1)+2)/2 ; end: for n from 1 to 1000 do c := A005891(n) ; if isprime(c) then printf("%d,",c) ; fi; od: # R. J. Mathar, Nov 02 2008

More terms from R. J. Mathar, Nov 02 2008

A253410 Indices of centered pentagonal numbers (A005891) which are also centered octagonal numbers (A016754).

Original entry on oeis.org

1, 96, 817, 137712, 1177393, 198579888, 1697799169, 286352060064, 2448225223585, 412919472031680, 3530339074609681, 595429592317621776, 5090746497361935697, 858609059202538568592, 7340852918856836664673, 1238113667940468298287168, 10585504818245061108522049

Offset: 1

Colin Barker, Dec 31 2014

Also positive integers x in the solutions to 5*x^2 - 8*y^2 - 5*x + 8*y = 0, the corresponding values of y being A253411.

			96 is in the sequence because the 96th centered pentagonal number is 22801, which is also the 76th centered octagonal number.

Colin Barker, Table of n, a(n) for n = 1..633
Index entries for linear recurrences with constant coefficients, signature (1,1442,-1442,-1,1).

Cf. A005891, A016754, A253411, A253579.

LinearRecurrence[{1,1442,-1442,-1,1},{1,96,817,137712,1177393},20] (* Harvey P. Dale, Jul 12 2021 *)

Vec(x*(95*x^3+721*x^2-95*x-1)/((x-1)*(x^2-38*x+1)*(x^2+38*x+1)) + O(x^100))

a(n) = a(n-1) + 1442*a(n-2) - 1442*a(n-3) - a(n-4) + a(n-5).

G.f.: x*(95*x^3 + 721*x^2 - 95*x - 1) / ((x-1)*(x^2 - 38*x + 1)*(x^2 + 38*x + 1)).

A254332 Indices of centered pentagonal numbers (A005891) which are also squares (A000290).

Original entry on oeis.org

1, 3, 22, 96, 817, 3627, 31006, 137712, 1177393, 5229411, 44709910, 198579888, 1697799169, 7540806315, 64471658494, 286352060064, 2448225223585, 10873837476099, 92968086837718, 412919472031680, 3530339074609681, 15680066099727723, 134059916748330142

Offset: 1

Colin Barker, Jan 28 2015

Also positive integers y in the solutions to 2*x^2 - 5*y^2 + 5*y - 2 = 0, the corresponding values of x being A129557.

			3 is in the sequence because the 3rd centered pentagonal number is 16, which is also the 4th square number.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,38,-38,-1,1).

Cf. A000290, A005891, A129557, A254333.

LinearRecurrence[{1,38,-38,-1,1},{1,3,22,96,817},30] (* Harvey P. Dale, Mar 27 2017 *)

Vec(x*(2*x^3+19*x^2-2*x-1) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)) + O(x^100))

a(n) = a(n-1) + 38*a(n-2) - 38*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(2*x^3 + 19*x^2 - 2*x - 1) / ((x-1)*(x^2 - 6*x - 1)*(x^2 + 6*x - 1)).
a(n) = (1/40)*(20 - b^n*(19 + 3*b) + (3 + b)*c^n - (b^n*(3 + b) + (1 - 3*b)*c^n)*(-1)^n) with b = sqrt(10) - 3 and c = sqrt(10) + 3. - Alan Michael Gómez Calderón, Jul 02 2024

A254333 Squares (A000290) which are also centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 16, 1156, 22801, 1666681, 32878756, 2403352576, 47411143081, 3465632747641, 68366835443776, 4997440018745476, 98584929298781641, 7206305041398228481, 142159399682007682276, 10391486872256226723856, 204993755756525779060081, 14984516863488437537571601

Offset: 1

Colin Barker, Jan 28 2015

			16 is in the sequence because it is the 4th square number and the 3rd centered pentagonal number.

Colin Barker, Table of n, a(n) for n = 1..634
Index entries for linear recurrences with constant coefficients, signature (1,1442,-1442,-1,1).

Cf. A000290, A005891, A129557, A254332.

LinearRecurrence[{1,1442,-1442,-1,1},{1,16,1156,22801,1666681},20] (* Harvey P. Dale, Jul 26 2015 *)

Vec(-x*(x^4+15*x^3-302*x^2+15*x+1) / ((x-1)*(x^2-38*x+1)*(x^2+38*x+1)) + O(x^100))

a(n) = a(n-1)+1442*a(n-2)-1442*a(n-3)-a(n-4)+a(n-5).

G.f.: -x*(x^4+15*x^3-302*x^2+15*x+1) / ((x-1)*(x^2-38*x+1)*(x^2+38*x+1)).

A254628 Triangular numbers (A000217) that are also centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 6, 276, 1891, 88831, 608856, 28603266, 196049701, 9210162781, 63127394826, 2965643812176, 20326825084231, 954928097357851, 6545174549727516, 307483881705415806, 2107525878187175881, 99008854981046531641, 678616787601720906126, 31880543820015277772556

Offset: 1

Colin Barker, Feb 03 2015

Also hexagonal numbers (A000384) that are also centered pentagonal numbers (A005891). - Colin Barker, Feb 11 2015

			6 is in the sequence because it is the 3rd triangular number and the 2nd centered pentagonal number.

Colin Barker, Table of n, a(n) for n = 1..798
Index entries for linear recurrences with constant coefficients, signature (1,322,-322,-1,1).

Cf. A000217, A005891, A254626, A254627.

Cf. A000384, A254962.

Vec(-x*(x^4+5*x^3-52*x^2+5*x+1)/((x-1)*(x^2-18*x+1)*(x^2+18*x+1)) + O(x^100))

a(n) = a(n-1)+322*a(n-2)-322*a(n-3)-a(n-4)+a(n-5).

G.f.: -x*(x^4+5*x^3-52*x^2+5*x+1) / ((x-1)*(x^2-18*x+1)*(x^2+18*x+1)).

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

cp's OEIS Frontend

A129556 Numbers k such that the k-th centered pentagonal number A005891(k) = (5k^2 + 5k + 2)/2 is a square.

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A254627 Indices of centered pentagonal numbers (A005891) that are also triangular numbers (A000217).

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A322638 Numbers that are sums of consecutive centered pentagonal numbers (A005891).

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A129557 Numbers k > 0 such that k^2 is a centered pentagonal number (A005891).

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A145838 Primes in A005891 = Centered pentagonal numbers: (5n^2 + 5n + 2)/2.

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A253410 Indices of centered pentagonal numbers (A005891) which are also centered octagonal numbers (A016754).

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A254332 Indices of centered pentagonal numbers (A005891) which are also squares (A000290).

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