A101089 -id:A101089 - cp's OEIS frontend

A213554 Principal diagonal of the convolution array A213553.

1, 43, 334, 1406, 4271, 10577, 22764, 44220, 79437, 134167, 215578, 332410, 495131, 716093, 1009688, 1392504, 1883481, 2504067, 3278374, 4233334, 5398855, 6807977, 8497028, 10505780, 12877605, 15659631, 18902898, 22662514

Offset: 1

Clark Kimberling, Jun 17 2012

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A213500, A213553.

List([1..30], n-> n*(39*n^4 +15*n^3 -25*n^2 +1)/30); # G. C. Greubel, Jul 31 2019

[n*(39*n^4 +15*n^3 -25*n^2 +1)/30: n in [1..30]]; // G. C. Greubel, Jul 31 2019

(* First program *)
b[n_]:= n; c[n_]:= n^3;
T[n_, k_]:= Sum[b[k-i] c[n+i], {i, 0, k-1}]
TableForm[Table[T[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[T[n-k+1, k], {n, 12}, {k, n, 1, -1}]]
r[n_]:= Table[T[n, k], {k, 1, 60}]  (* A213553 *)
d = Table[T[n, n], {n, 1, 40}] (* A213554 *)
s[n_]:= Sum[T[i, n+1-i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A101089 *)
(* Second program *)
Table[(39n^5+15n^4-25n^3+n)/30,{n,30}] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{1,43,334,1406,4271,10577},30] (* Harvey P. Dale, Jan 15 2013 *)

vector(30, n, n*(39*n^4 +15*n^3 -25*n^2 +1)/30) \\ G. C. Greubel, Jul 31 2019

[n*(39*n^4 +15*n^3 -25*n^2 +1)/30 for n in (1..30)] # G. C. Greubel, Jul 31 2019

a(n) = n*(39*n^4 + 15*n^3 - 25*n^2 + 1)/30.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
G.f.: x*(1 + 37*x  + 91*x^2 + 27*x^3)/(1 - x)^6.

A302352 a(n) = Sum_{k=0..n} k^4binomial(2n-k,n).

Original entry on oeis.org

0, 1, 19, 155, 936, 4884, 23465, 107107, 472600, 2036838, 8631206, 36119798, 149724940, 616104450, 2520629685, 10265200035, 41650094640, 168481778790, 679847488650, 2737640364810, 11005139655744, 44176226269728, 177114113623194, 709364594864910, 2838599638596176, 11350436081373340

Offset: 0

Ilya Gutkovskiy, Apr 06 2018

nonn

Main diagonal of iterated partial sums array of fourth powers (starting with the first partial sums). For nonnegative integers see A002054, for squares see A265612, for cubes see A293550.

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, Biquadratic Number

Cf. A000538, A000583, A002054, A101089, A101090, A101091, A265612, A293550, A302353.

Table[Sum[k^4 Binomial[2 n - k, n], {k, 0, n}], {n, 0, 25}]
Table[SeriesCoefficient[x (1 + 11 x + 11 x^2 + x^3)/(1 - x)^(n + 6), {x, 0, n}], {n, 0, 25}]
Table[2^(2 n + 1) n (75 n^3 + 52 n^2 - 3 n - 4) Gamma[n + 3/2]/(Sqrt[Pi] Gamma[n + 6]), {n, 0, 25}]
CoefficientList[Series[(24 - 180 x + 410 x^2 - 285 x^3 + 31 x^4 + Sqrt[1 - 4 x] (-24 + 132 x - 194 x^2 + 65 x^3 - x^4))/(2 Sqrt[1 - 4 x] x^5), {x, 0, 25}], x]
CoefficientList[Series[E^(2 x) (-576 + 360 x - 244 x^2 + 75 x^3) BesselI[0, 2 x]/x^3 + E^(2 x) (576 - 360 x + 532 x^2 - 255 x^3 + 75 x^4) BesselI[1, 2 x]/x^4, {x, 0, 25}], x]* Range[0, 25]!

a(n) = sum(k=0, n, k^4*binomial(2*n-k,n)); \\ Michel Marcus, Apr 07 2018

a(n) = [x^n] x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^(n+6).
a(n) = 2^(2*n+1)*n*(75*n^3 + 52*n^2 - 3*n - 4)*Gamma(n+3/2)/(sqrt(Pi)*Gamma(n+6)).
a(n) ~ 75*2^(2*n+1)/sqrt(Pi*n).

A257449 a(n) = 75(2^n - 1) - 4n^3 - 18n^2 - 52n.

Original entry on oeis.org

1, 17, 99, 373, 1115, 2901, 6907, 15509, 33483, 70405, 145451, 296997, 601819, 1213493, 2439195, 4893301, 9804587, 19630629, 39286603, 78602885, 157240251, 314520277, 629086139, 1258224213, 2516507275, 5033080901, 10066236267, 20132555749, 40265204123

Offset: 1

Luciano Ancora, Apr 23 2015

See the first comment of A257448.

			This sequence provides the antidiagonal sums of the array:
1, 16,  81, 256,  625,  1296, ...   A000583
1, 17,  98, 354,  979,  2275, ...   A000538
1, 18, 116, 470, 1449,  3724, ...   A101089
1, 19, 135, 605, 2054,  5778, ...   A101090
1, 20, 155, 760, 2814,  8592, ...   A101091
1, 21, 176, 936, 3750, 12342, ...   A254681
...
See also A254681 (Example field).

Index entries for linear recurrences with constant coefficients, signature (6,-14,16,-9,2).

Cf. A000225, A000670, A050488, A208744, A257448, A257450.

[75*(2^n-1)-4*n^3-18*n^2-52*n: n in [1..30]]; // Vincenzo Librandi, Apr 24 2015

Table[75 (2^n - 1) - 4 n^3 - 18 n^2 - 52 n, {n, 30}]

G.f.: -x*(1 + x)*(1 + 10*x + x^2)/((-1 + x)^4*(-1 + 2*x)).

a(n) = 6*a(n-1) -14*a(n-2) +16*a(n-3) -9*a(n-4) +2*a(n-5) for n>5.

A302353 a(n) = Sum_{k=0..n} k^nbinomial(2n-k,n).

Original entry on oeis.org

1, 1, 7, 69, 936, 16290, 345857, 8666413, 250355800, 8191830942, 299452606190, 12095028921250, 534924268768540, 25710497506696860, 1334410348734174285, 74379234152676275325, 4431350132232658244400, 281020603194039519937590, 18900157831016574533520330, 1343698678390575915132318870

Offset: 0

Ilya Gutkovskiy, Apr 06 2018

nonn

a(n) is the n-th term of the main diagonal of iterated partial sums array of n-th powers (starting with the first partial sums).

			For n = 4 we have:
------------------------
0   1    2    3    [4]
------------------------
0,  1,  17,   98,  354,  ... A000538 (partial sums of fourth powers)
0,  1,  18,  116,  470,  ... A101089 (partial sums of A000538)
0,  1,  19,  135,  605,  ... A101090 (partial sums of A101089)
0,  1,  20,  155,  760,  ... A101091 (partial sums of A101090)
0,  1,  21,  176, [936], ... A254681 (partial sums of A101091)
------------------------
therefore a(4) = 936.

Cf. A002054, A031971, A265612, A293550, A293574, A302352.

Join[{1}, Table[Sum[k^n Binomial[2 n - k, n], {k, 0, n}], {n, 19}]]
Table[SeriesCoefficient[HurwitzLerchPhi[x, -n, 0]/(1 - x)^(n + 1), {x, 0, n}], {n, 0, 19}]

a(n) ~ c * (r * (2-r)^(2-r) / (1-r)^(1-r))^n * n^n, where r = 0.69176629470097668698335106516328398961170464277337300459988208658267146... is the root of the equation (2-r) = (1-r) * exp(1/r) and c = 0.96374921279011282619632879505754646526289414675402231447188230355850496... - Vaclav Kotesovec, Apr 08 2018

cp's OEIS Frontend

A213554 Principal diagonal of the convolution array A213553.

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A302352 a(n) = Sum_{k=0..n} k^4*binomial(2*n-k,n).

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PARI

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A257449 a(n) = 75*(2^n - 1) - 4*n^3 - 18*n^2 - 52*n.

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Author

Keywords

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Links

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Programs

Magma

Mathematica

Formula

A302353 a(n) = Sum_{k=0..n} k^n*binomial(2*n-k,n).

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Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A302352 a(n) = Sum_{k=0..n} k^4binomial(2n-k,n).

A257449 a(n) = 75(2^n - 1) - 4n^3 - 18n^2 - 52n.

A302353 a(n) = Sum_{k=0..n} k^nbinomial(2n-k,n).