A084901 -id:A084901 - cp's OEIS frontend

A077616 Binomial transform of n^2*2^n/2.

1, 10, 63, 324, 1485, 6318, 25515, 99144, 373977, 1377810, 4979799, 17714700, 62178597, 215765046, 741360195, 2525407632, 8537599665, 28669116186, 95692860783, 317684800980, 1049522104701, 3451916556990, 11307641812443

Offset: 1

Karol A. Penson, Nov 12 2002

With a leading zero, this is second binomial transform of the hexagonal numbers A000384 (with leading zero). - Paul Barry, Jun 09 2003

Coefficients in the hypergeometric series identity 1 - 10*x/(x + 9) + 63*x*(x - 1)/((x + 9)*(x + 12)) - 324*x*(x - 1)*(x - 2)/((x + 9)*(x + 12)*(x + 15)) + ... = 0, valid in the half-plane Re(x) > 0. Cf. A276289 and A084901. - Peter Bala, May 30 2019

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (9,-27,27).

Cf. A084901, A276289.

List([1..30], n-> 3^(n-2)*n*(1+2*n)) # G. C. Greubel, Jun 03 2019

[3^(n-2)*n*(1+2*n): n in [1..30]]; // G. C. Greubel, Jun 03 2019

LinearRecurrence[{9, -27, 27}, {1, 10, 63}, 30] (* Jean-François Alcover, May 23 2016 *)

a(n)=n*(2*n+1)*3^(n-2) \\ Charles R Greathouse IV, Mar 19 2017

[3^(n-2)*n*(1+2*n) for n in (1..30)] # G. C. Greubel, Jun 03 2019

E.g.f: x*(1+2*x)*exp(3*x).
O.g.f: ((1/3)*x^(3/4)*3^(3/4)/(-(3*x+1)/(3*x-1)+1)^(1/4))*(-(3*x+1)/(3*x-1)-1)^(1/4)*hypergeom([ -1, 2], [3/2], 3*x/(3*x-1))/(3*x-1)^2, which can also be represented as associated Legendre function: 1/6*x^(3/4)*Pi^(1/2)*3^(3/4)*LegendreP(1, -1/2, (3*x+1)/(1-3*x))/(3*x-1)^2.
G.f.: x*(1+x)/(1-3*x)^3. - Paul Barry, Jun 09 2003
a(n) = n*(2*n+1)*3^(n-2). - Paul Barry, Jul 24 2003

A084902 a(n) = 5^(n-1)n(n+1)/2.

Original entry on oeis.org

0, 1, 15, 150, 1250, 9375, 65625, 437500, 2812500, 17578125, 107421875, 644531250, 3808593750, 22216796875, 128173828125, 732421875000, 4150390625000, 23345947265625, 130462646484375, 724792480468750, 4005432128906250

Offset: 0

Paul Barry, Jun 10 2003

Binomial transform of A084901. Fourth binomial transform of heptagonal numbers A000566. Fifth binomial transform of (0,1,5,0,0,0,...).
Number of n-permutations of 6 objects u, v, w, z, x, y with repetition allowed, containing exactly two u's. Example: a(2)=15 because we have uuw, uuv, uuz, uux, uuy, uwu, uvu, uzu, uxu, uyu, wuu, vuu, zuu, xuu, yuu. - Zerinvary Lajos, Dec 30 2007
A shifted version of A081135. - R. J. Mathar, Sep 17 2008

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (15,-75,125).

Cf. A000566, A038243, A081135, A084901.

[5^(n-1)*Binomial(n+1,2): n in [0..30]]; // G. C. Greubel, May 17 2021

Table[5^(n-1)n(n+1)/2,{n,0,30}] (* or *) LinearRecurrence[{15,-75,125},{0,1,15},30] (* Harvey P. Dale, Sep 18 2018 *)

a(n)=5^(n-1)*n*(n+1)/2 \\ Charles R Greathouse IV, Oct 07 2015

[5^(n-1)*binomial(n+1,2) for n in (0..30)] # G. C. Greubel, May 17 2021

G.f.: x/(1 - 5*x)^3.
E.g.f.: (x/2)*(2 + 5*x)*exp(5*x). - G. C. Greubel, May 17 2021
a(n) = 15*a(n-1) - 75*a(n-2) + 125*a(n-3). - Wesley Ivan Hurt, May 17 2021

A276289 Expansion of x(1 + x)/(1 - 2x)^3.

Original entry on oeis.org

0, 1, 7, 30, 104, 320, 912, 2464, 6400, 16128, 39680, 95744, 227328, 532480, 1232896, 2826240, 6422528, 14483456, 32440320, 72220672, 159907840, 352321536, 772800512, 1688207360, 3674210304, 7969177600, 17230200832, 37144756224, 79859548160, 171261820928, 366414397440

Offset: 0

Ilya Gutkovskiy, Aug 27 2016

Binomial transform of pentagonal numbers (A000326).
More generally, the binomial transform of k-gonal numbers is n*Hypergeometric2F1(k/(k-2),1-n;2/(k-2);-1), where Hypergeometric2F1(a,b;c;x) is the hypergeometric function.
Coefficients in the hypergeometric series identity 1 - 7*x/(x + 6) + 30*x*(x - 1)/((x + 6)*(x + 8)) - 104*x*(x - 1)*(x - 2)/((x + 6)*(x + 8)*(x + 10)) + ... = 0, valid in the half-plane Re(x) > 0. Cf. A077616 and A084901. - Peter Bala, May 30 2019

G. C. Greubel, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Pentagonal Number
Index entries for linear recurrences with constant coefficients, signature (6,-12,8).

Cf. A000326, A084901.

Cf. A001793 (binomial transform of triangular numbers), A001788 (binomial transform of squares), A084899 (binomial transform of heptagonal numbers).

List([0..40], n-> 2^(n-3)*n*(3*n+1)); # G. C. Greubel, Jun 02 2019

[2^(n-3)*n*(3*n+1): n in [0..40]]; // G. C. Greubel, Jun 02 2019

a:=series(x*(1+x)/(1-2*x)^3,x=0,31): seq(coeff(a,x,n),n=0..40); # Paolo P. Lava, Mar 27 2019

LinearRecurrence[{6, -12, 8}, {0, 1, 7}, 40]
Table[2^(n - 3) n (3 n + 1), {n, 0, 40}]

concat(0, Vec(x*(1+x)/(1-2*x)^3 + O(x^40))) \\ Altug Alkan, Aug 27 2016

[2^(n-3)*n*(3*n+1) for n in (0..40)] # G. C. Greubel, Jun 02 2019

O.g.f.: x*(1 + x)/(1 - 2*x)^3.
E.g.f.: x*(2 + 3*x)*exp(2*x)/2.
a(n) = 6*a(n-1) - 12*a(n-2) + 8*a(n-3).
a(n) = Sum_{k = 0..n} binomial(n,k)*k*(3*k - 1)/2.
a(n) = 2^(n-3)*n*(3*n + 1).
Sum_{n>=1} 1/a(n) = 8*(-3*2^(1/3)*Hypergeometric2F1(1/3,1/3;4/3;-1) + 3 + log(2)) = 1.1906948190529335181687...

A084900 a(n) = 3^(n-2)n(5*n+1)/2.

Original entry on oeis.org

0, 1, 11, 72, 378, 1755, 7533, 30618, 119556, 452709, 1673055, 6062364, 21611934, 75996063, 264126177, 908764110, 3099363912, 10489051017, 35255264499, 117775828656, 391294693890, 1293597012771, 4257363753621, 13954111172802

Offset: 0

Paul Barry, Jun 10 2003

Binomial transform of A084899. Second binomial transform of heptagonal numbers A000566. Third binomial transform of (0,1,5,0,0,0,...).

Index entries for linear recurrences with constant coefficients, signature (9,-27,27).

Cf. A000566, A084899, A084901.

Table[(3^(n-2) n(5n+1))/2,{n,0,30}] (* or *) LinearRecurrence[{9,-27,27},{0,1,11},30] (* Harvey P. Dale, Jul 21 2016 *)

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

E.g.f.: exp(3*x)*x*(2 + 5*x)/2. - Stefano Spezia, Oct 28 2023

cp's OEIS Frontend

A077616 Binomial transform of n^2*2^n/2.

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A084902 a(n) = 5^(n-1)*n*(n+1)/2.

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A276289 Expansion of x*(1 + x)/(1 - 2*x)^3.

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GAP

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A084900 a(n) = 3^(n-2)*n*(5*n+1)/2.

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Mathematica

Formula

A084902 a(n) = 5^(n-1)n(n+1)/2.

A276289 Expansion of x(1 + x)/(1 - 2x)^3.

A084900 a(n) = 3^(n-2)n(5*n+1)/2.