A069038 -id:A069038 - cp's OEIS frontend

A142987 a(1) = 1, a(2) = 10, a(n+2) = 10a(n+1) + (n + 1)(n + 2)*a(n).

1, 10, 106, 1180, 13920, 174600, 2330640, 33084000, 498646080, 7964020800, 134491276800, 2396163513600, 44942274316800, 885524502643200, 18293122632960000, 395457106963968000, 8930300425804800000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 5 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n + 1)*(n + 2)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Cf. A069038, A142983, A142984, A142985, A142986.

Cf. A002378.

a142987 n = a142987_list !! (n-1)
a142987_list = 1 : 10 : zipWith (+)
                        (map (* 10) $ tail a142987_list)
                        (zipWith (*) (drop 2 a002378_list) a142987_list)
-- Reinhard Zumkeller, Jul 17 2015

p := n -> (2*n^5+10*n^3+3*n)/15: a := n -> n!*p(n+1)*sum ((-1)^(k+1)/(p(k)*p(k+1)), k = 1..n): seq(a(n), n = 1..20);

RecurrenceTable[{a[1]==1,a[2]==10,a[n+2]==10a[n+1]+(n+1)(n+2)a[n]},a,{n,20}] (* Harvey P. Dale, Mar 23 2021 *)

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (2*n^5 + 10*n^3 + 3*n)/15 = A069038(n).
Recurrence: a(1) = 1, a(2) = 10, a(n+2) = 10*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 10 and b(2) = 102.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(10 + 1*2/(10 + 2*3/(10 + 3*4/(10 + .. +n*(n - 1)/(10))))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(10 + 1*2/(10 + 2*3/(10 + 3*4/(10 + ... + n*(n - 1)/(10 + ...))))) = 10*log(2) - 41/6, where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).

A142977 Table of coefficients in the expansion of the rational function 1/{(1-x)^2 - y*(1+x)^2}.

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 10, 19, 4, 1, 14, 51, 44, 5, 1, 18, 99, 180, 85, 6, 1, 22, 163, 476, 501, 146, 7, 1, 26, 243, 996, 1765, 1182, 231, 8, 1, 30, 339, 1804, 4645, 5418, 2471, 344, 9, 1, 34, 451, 2964, 10165, 17718, 14407, 4712, 489, 10

Offset: 0

Peter Bala, Jul 15 2008

The row entries are the figurate numbers of the odd dimensional cross polytopes. See A142978 for the complete table of figurate numbers of n-dimensional cross polytopes. The rows are the partial sums of the even-numbered rows of the square array of Delannoy numbers A008288.

			The square array begins
 n\k| 0...1....2.....3.....4.......5
------------------------------------
 .0.| 1...2....3.....4......5......6 ... A000027
 .1.| 1...6...19....44.....85....146 ... A005900
 .2.| 1..10...51...180....501...1182 ... A069038
 .3.| 1..14...99...476...1765...5418 ... A099193
 .4.| 1..18..163...996...4645..17718 ... A099196
 .5.| 1..22..243..1804..10165..46530 ... A300624
 ...

Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2003), 65-75.

Cf. A005900 (row 1), A008288, A069038 (row 2), A099193 (row 3), A099196 (row 4), A300624 (row 5), A142978, A142983.

with(combinat): T:=(n,k) -> add(binomial(2n,k-j)*binomial(2n+j+1,j), j = 0..k): for n from 0 to 9 do seq(T(n,k), k = 0..9) end do;

T(n,k) = Sum_{j = 0..k} C(2*n, k-j)*C(2*n+j+1, j).
O.g.f.: 1/{(1 - x)^2 - y*(1 + x)^2} = Sum_{n, k >= 0} T(n,k)*x^k*y^n = 1/(1 - y) * Sum_{m >= 0} U(m, (1 + y)/(1 - y))*x^m, where U(m, y) denotes the m-th Chebyshev polynomial of the second kind.
O.g.f. row n: (1 + x)^(2*n)/(1 - x)^(2*n+2).
O.g.f. column k: 1/(1 - y)*U(k, (1 + y)/(1 - y)).
The entries in the n-th row appear in the series acceleration formula for the constant log(2): Sum_{k >= 1} (-1)^(k+1)/(T(n,k)*T(n,k+1)) = 1 + (4*n + 2)*( log(2) - (1 - 1/2 + 1/3 - ... + 1/(2*n + 1)) ).
For example, n = 1 gives log(2) = 4/6 + (1/6)*( 1/(1*6) - 1/(6*19) + 1/(19*44) - 1/(44*85) + ... ). See A142983 for further details.

Restored missing program. - Peter Bala, Oct 02 2008

A300624 Figurate numbers based on the 11-dimensional regular convex polytope called the 11-dimensional cross-polytope, or 11-dimensional hyperoctahedron.

Original entry on oeis.org

0, 1, 22, 243, 1804, 10165, 46530, 180775, 614680, 1871145, 5188590, 13286043, 31760676, 71513949, 152784282, 311603535, 609802800, 1150082385, 2098144710, 3714481475, 6399123260, 10753517061, 17664712562, 28418229623, 44847366984, 69528316025, 106032285086

Offset: 0

Alejandro J. Becerra Jr., Aug 14 2018

The 11-dimensional cross-polytope is represented by the Schlaefli symbol {3, 3, 3, 3, 3, 3, 3, 3, 3, 4}. It is the dual of the 11-dimensional hypercube.

Georg Fischer, Table of n, a(n) for n = 0..100 (first 61 terms from Alejandro J. Becerra Jr.)
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193 (m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).

[(n*(14175 + 83754*n^2 + 50270*n^4 + 7392*n^6 + 330*n^8 + 4*n^10)) / 155925 : n in [0..40]]; // Wesley Ivan Hurt, Jul 17 2020

concat(0, Vec(x*(1 + x)^10 / (1 - x)^12 + O(x^40))) \\ Colin Barker, Aug 15 2018

a(n) = (n*(14175 + 83754*n^2 + 50270*n^4 + 7392*n^6 + 330*n^8 + 4*n^10)) / 155925 \\ Colin Barker, Aug 15 2018

a(n) = 11-crosspolytope(n).
From Colin Barker, Aug 15 2018: (Start)
G.f.: x*(1 + x)^10 / (1 - x)^12.
a(n) = (n*(14175 + 83754*n^2 + 50270*n^4 + 7392*n^6 + 330*n^8 + 4*n^10)) / 155925.
(End)

A191596 Expansion of (1+x)^4/(1-x)^7.

Original entry on oeis.org

1, 11, 62, 242, 743, 1925, 4396, 9108, 17469, 31471, 53834, 88166, 139139, 212681, 316184, 458728, 651321, 907155, 1241878, 1673882, 2224607, 2918861, 3785156, 4856060, 6168565, 7764471, 9690786, 12000142, 14751227, 18009233, 21846320

Offset: 0

Bruno Berselli, Jun 08 2011

The first, second and third differences are in A069038, A001846 and A008412, respectively.
Inverse binomial transform of this sequence: 1, 10, 41, 88, 104, 64, 16, 0, 0 (0 continued).
Also (by Superseeker), the n-th coefficient of the expansion of ((1+x)^4/(1-x)^7)*(1+x)^n is A006976(n-1).

Bruno Berselli, Table of n, a(n) for n = 0..1000
M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550, 2013
M. Janjic, B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A008415, A001848, A069039, A008412, A001846, A069038, A061927 (for type of g.f.).

[(2*n^6+18*n^5+80*n^4+210*n^3+323*n^2+267*n+90)/90: n in [0..30]]; // Vincenzo Librandi, Jun 08 2011

A191596:=n->(n+1)*(n+2)*(2*n^4+12*n^3+40*n^2+66*n+45)/90: seq(A191596(n), n=0..40); # Wesley Ivan Hurt, Nov 20 2014

CoefficientList[Series[(1 + x)^4/(1 - x)^7, {x, 0, 30}], x] (* Wesley Ivan Hurt, Nov 20 2014 *)

makelist(coeff(taylor((1+x)^4/(1-x)^7, x, 0, n), x, n), n, 0, 30);

a(n)=(((((n+n+18)*n+80)*n+210)*n+323)*n+267)/90*n+1 \\ Charles R Greathouse IV, Jun 08 2011

G.f.: (1+x)^4/(1-x)^7.
a(n) = (n+1)*(n+2)*(2*n^4+12*n^3+40*n^2+66*n+45)/90.
a(n) = a(-n-3) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+21*a(n-5)-7*a(n-6)+a(n-7).
By Superseeker:
a(n)+a(n+1) = A069039(n+2),
a(n+2)-a(n) = A001847(n+2),
a(n+2)+2*a(n+1)+a(n) = A001848(n+2).

A364429 a(0) = 1, a(n) = (2n^5 + 20n^3 + 23n) 2/15 for n>=1.

Original entry on oeis.org

1, 6, 36, 146, 456, 1182, 2668, 5418, 10128, 17718, 29364, 46530, 71000, 104910, 150780, 211546, 290592, 391782, 519492, 678642, 874728, 1113854, 1402764, 1748874, 2160304, 2645910, 3215316, 3878946, 4648056, 5534766, 6552092, 7713978, 9035328, 10532038, 12221028

Offset: 0

Steven Lu, Jul 24 2023

a(n) is the 6th n-orthoplex (n-dimensional cross-polytope) number.

			a(3) = 146 since the 6th octahedral number is 146; A005900(6) = 146.
a(4) = 456 since the 6th 16-cell number is 456; A014820(5) = 456.

OEIS Wiki, Orthoplex numbers.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A005900, A014820, A069038, A069039, A099193, A099195, A099196, A099197, A058331.

Cf. A142978 (column 6 with an initial 1).

Prepend[Table[2/15 (2 x^5 + 20 x^3 + 23 x), {x, 100}], 1]

print([1]+[(2*i**5+20*i**3+23*i)*2//15 for i in range(1,101)])

a(0) = 1, a(n) = (2*n^5 + 20*n^3 + 23*n) * 2/15 for n>=1.

G.f.: (1 + 15*x^2 + 15*x^4 + x^6)/(1 - x)^6. - Stefano Spezia, Jul 24 2023

cp's OEIS Frontend

A142987 a(1) = 1, a(2) = 10, a(n+2) = 10*a(n+1) + (n + 1)*(n + 2)*a(n).

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A142977 Table of coefficients in the expansion of the rational function 1/{(1-x)^2 - y*(1+x)^2}.

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A300624 Figurate numbers based on the 11-dimensional regular convex polytope called the 11-dimensional cross-polytope, or 11-dimensional hyperoctahedron.

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A191596 Expansion of (1+x)^4/(1-x)^7.

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A364429 a(0) = 1, a(n) = (2*n^5 + 20*n^3 + 23*n) * 2/15 for n>=1.

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A142987 a(1) = 1, a(2) = 10, a(n+2) = 10a(n+1) + (n + 1)(n + 2)*a(n).

A364429 a(0) = 1, a(n) = (2n^5 + 20n^3 + 23n) 2/15 for n>=1.