A272129 -id:A272129 - cp's OEIS frontend

A272126 a(n) = 120n^3 + 60n^2 + 2*n + 1.

1, 183, 1205, 3787, 8649, 16511, 28093, 44115, 65297, 92359, 126021, 167003, 216025, 273807, 341069, 418531, 506913, 606935, 719317, 844779, 984041, 1137823, 1306845, 1491827, 1693489, 1912551, 2149733, 2405755, 2681337, 2977199, 3294061, 3632643, 3993665

Offset: 0

Vincenzo Librandi, Apr 25 2016

This is the polynomial Qbar(3,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials. - Peter Bala, Jan 22 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A014641, A158673, A160485, A245244, A272127, A272129.

[120*n^3 + 60*n^2 + 2*n + 1: n in [0..50]];

Table[120 n^3 + 60 n^2 + 2 n + 1, {n, 0, 40}]
LinearRecurrence[{4,-6,4,-1},{1,183,1205,3787},40] (* Harvey P. Dale, Nov 08 2020 *)

a(n) = 120*n^3 + 60*n^2 + 2*n + 1; \\ Altug Alkan, Apr 30 2016

O.g.f.: (1 + 179*x + 479*x^2 + 61*x^3)/(1-x)^4.
E.g.f.: (1 + 182*x + 420*x^2 + 120*x^3)*exp(x).
a(n) = (2*n+1)*(60*n^2+1).
a(n) = (2*n+1) * A158673(n).
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3.
See page 7 in Brent's paper:
a(n) = (2*n+1)^2*A014641(n) - 2*n*(2*n+1)*A014641(n-1).
A272127(n) = (2*n+1)^2*a(n) - 2*n*(2*n+1)*a(n-1).
From Peter Bala, Jan 22 2019: (Start)
a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^6 * binomial(2*n + 1, n - k).
a(n-1) = 2/4^n * binomial(2*n,n) * ( 1 + 3^6*(n - 1)/(n + 1) + 5^6*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^6*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)

A272127 a(n) = 1680n^4 - 168n^2 + 128*n + 1.

Original entry on oeis.org

1, 1641, 26465, 134953, 427905, 1046441, 2172001, 4026345, 6871553, 11010025, 16784481, 24577961, 34813825, 47955753, 64507745, 85014121, 110059521, 140268905, 176307553, 218881065, 268735361, 326656681, 393471585, 470046953, 557289985, 656148201, 767609441

Offset: 0

Vincenzo Librandi, Apr 25 2016

This is the polynomial Qbar(4,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials. - Peter Bala, Jan 21 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A014641, A272126, A160485, A245244, A272129.

[1680*n^4-168*n^2+128*n+1: n in [0..50]];

Table[1680 n^4 - 168 n^2 + 128 n + 1, {n, 0, 30}]
LinearRecurrence[{5,-10,10,-5,1},{1,1641,26465,134953,427905},30] (* Harvey P. Dale, Nov 27 2017 *)

a(n) = 1680*n^4 - 168*n^2 + 128*n + 1; \\ Altug Alkan, Apr 30 2016

O.g.f.: (1+1636*x+18270*x^2+19028*x^3+1385*x^4)/(1-x)^5.
E.g.f.: (1+1640*x+11592*x^2+10080*x^3+1680*x^4)*exp(x).
a(n) = (2*n+1)*(840*n^3-420*n^2+126*n+1).
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5), for n>4
See page 7 in Brent's paper:
a(n) = (2*n+1)^2*A272126(n) - 2*n*(2*n+1)*A272126(n-1).
A272128(n) = (2*n+1)^2*a(n) - 2*n*(2*n+1)*a(n-1).
From Peter Bala, Jan 22 2019: (Start)
a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^8 * binomial(2*n + 1, n - k).
a(n-1) = 2/4^n * binomial(2*n,n) * ( 1 + 3^8*(n - 1)/(n + 1) + 5^8*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^8*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)

A272131 a(n) = 384n^3 - 1184n^2 + 1228*n - 427.

Original entry on oeis.org

-427, 1, 365, 2969, 10117, 24113, 47261, 81865, 130229, 194657, 277453, 380921, 507365, 659089, 838397, 1047593, 1288981, 1564865, 1877549, 2229337, 2622533, 3059441, 3542365, 4073609, 4655477, 5290273, 5980301, 6727865, 7535269, 8404817, 9338813, 10339561

Offset: 0

Vincenzo Librandi, Apr 26 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014 (page 16).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A272129, A272132, A245244.

[384*n^3 - 1184*n^2 + 1228*n - 427: n in [0..50]];

[384*n^3-1184*n^2+1228*n-427$n=0..35]; # Muniru A Asiru, Jan 28 2019

Table[384 n^3 - 1184 n^2 + 1228 n - 427, {n, 0, 40}]
LinearRecurrence[{4,-6,4,-1},{-427,1,365,2969},40] (* Harvey P. Dale, Aug 24 2024 *)

lista(nn) = for(n=0, nn, print1(384*n^3-1184*n^2+1228*n-427, ", ")); \\ Altug Alkan, Apr 26 2016

O.g.f.: (-427 + 1709*x - 2201*x^2 + 3223*x^3)/(1-x)^4.
E.g.f.: (-427 + 428*x - 32*x^2 + 384*x^3)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3.
See page 7 in Brent's paper:
a(n) = (2*n-1)^2*A272129(n) - 4*(n-1)^2*A272129(n-1).
A272132(n) =  (2*n-1)^2*a(n) - 4*(n-1)^2*a(n-1).
n*a(n) = 1 + 3^7*(n-1)/(n+1) + 5^7*((n-1)*(n-2))/((n+1)*(n+2)) + ... for n >= 1. See A245244. - Peter Bala, Jan 19 2019

cp's OEIS Frontend

A272126 a(n) = 120n^3 + 60n^2 + 2*n + 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A272127 a(n) = 1680n^4 - 168n^2 + 128*n + 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A272131 a(n) = 384n^3 - 1184n^2 + 1228*n - 427.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

cp's OEIS Frontend

A272126 a(n) = 120*n^3 + 60*n^2 + 2*n + 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A272127 a(n) = 1680*n^4 - 168*n^2 + 128*n + 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A272131 a(n) = 384*n^3 - 1184*n^2 + 1228*n - 427.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A272126 a(n) = 120n^3 + 60n^2 + 2*n + 1.

A272127 a(n) = 1680n^4 - 168n^2 + 128*n + 1.

A272131 a(n) = 384n^3 - 1184n^2 + 1228*n - 427.