A142983 -id:A142983 - cp's OEIS frontend

A142986 a(1) = 1, a(2) = 8, a(n+2) = 8a(n+1) + (n + 1)(n + 2)*a(n).

1, 8, 70, 656, 6648, 72864, 862128, 10977408, 149892480, 2187106560, 33985025280, 560578268160, 9786290088960, 180315565516800, 3497645442816000, 71256899266560000, 1521414754578432000, 33975929212194816000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 4 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. A014820, A142983, A142984, A142985, A142987.

Cf. A002378.

a142986 n = a142986_list !! (n-1)
a142986_list = 1 : 8 : zipWith (+)
                       (map (* 8) $ tail a142986_list)
                       (zipWith (*) (drop 2 a002378_list) a142986_list)
-- Reinhard Zumkeller, Jul 17 2015

p := n -> (n^4+2*n^2)/3: 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]==8,a[n]==8a[n-1]+n(n-1)a[n-2]},a,{n,20}] (* Harvey P. Dale, Apr 08 2015 *)

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (n^4 + 2*n^2)/3 = A014820(n).
Recurrence: a(1) = 1, a(2) = 8, a(n+2) = 8*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 8 and b(2) = 66.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(8 + 1*2/(8 + 2*3/(8 + 3*4/(8 + ... + n*(n - 1)/8)))), 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/(8 + 1*2/(8 + 2*3/(8 + 3*4/(8 + ... + n*(n - 1)/(8 + ...))))) = 17/3 - 8*log(2), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).

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

Original entry on oeis.org

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)]).

A142989 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1)+(n+1)(n+3)*a(n).

Original entry on oeis.org

1, 5, 33, 240, 1992, 18360, 187416, 2093760, 25462080, 334592640, 4728412800, 71488811520, 1151817408000, 19699405286400, 356504125824000, 6805868977152000, 136702533123072000, 2881808345235456000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 1 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1)+(n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142990 (m=2) and A142991 (m=3).

Harvey P. Dale, Table of n, a(n) for n = 1..447

Cf. A142979, A142983, A142988, A142990, A142991.

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

RecurrenceTable[{a[1]==1,a[2]==5,a[n]==5a[n-1]+(n-1)(n+1)a[n-2]},a,{n,20}] (* Harvey P. Dale, Jun 17 2013 *)

a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (2*n-1)/3. Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 5, b(2) = 28. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5+1*3/(5+2*4/(5+3*5/(5+...+(n-1)*(n+1)/5)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(5+1*3/(5+2*4/(5+3*5/(5+...+(n-1)*(n+1)/(5+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 17/2-12*log(2).

A142990 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1)+(n+1)(n+3)*a(n).

Original entry on oeis.org

1, 7, 57, 504, 4896, 51912, 598392, 7459200, 100085760, 1439061120, 22083719040, 360371773440, 6232667212800, 113901166310400, 2193425619840000, 44398776748032000, 942498015750144000, 20938290999865344000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 2 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1)+(n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142989 (m=1) and A142991 (m=3).

Seiichi Manyama, Table of n, a(n) for n = 1..446

Cf. A142979, A142983, A142988, A142989, A142991.

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

a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (n^2-n+1)/3. Recurrence: a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 7, b(2) = 52. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7+1*3/(7+2*4/(7+3*5/(7+...+(n-1)*(n+1)/7)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(7+1*3/(7+2*4/(7+3*5/(7+...+(n-1)*(n+1)/(7+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 24*log(2)-33/2.

A142991 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1)+(n+1)(n+3)*a(n).

Original entry on oeis.org

1, 9, 89, 936, 10560, 127800, 1657080, 22965120, 339252480, 5326819200, 88651670400, 1559600179200, 28929882240000, 564490975104000, 11560712397696000, 247991610230784000, 5561409662613504000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 3 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1) + (n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142989 (m=1) and A142990 (m=2).

Seiichi Manyama, Table of n, a(n) for n = 1..445

Cf. A142979, A142983, A142988, A142989, A142990.

p := n -> (2*n^3-3*n^2+7*n-3)/15: a := n -> (n+2)!*p(n+2)*sum ((-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), k = 1..n): seq(a(n), n = 1..20);

RecurrenceTable[{a[1]==1,a[2]==9,a[n+2]==9a[n+1]+(n+1)(n+3)a[n]},a,{n,20}] (* Harvey P. Dale, Jul 18 2020 *)

a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (2*n^3-3*n^2+7*n-3)/15. Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 9, b(2) = 84. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/9)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/(9+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 167/6 - 40*log(2).

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

A156136 A triangle of polynomial coefficients related to Mittag-Leffler polynomials: p(x,n)=Sum[Binomial[n, k]Binomial[n - 1, n - k]2^kx^k, {k, 0, n}]/(2x).

Original entry on oeis.org

1, 2, 2, 3, 12, 4, 4, 36, 48, 8, 5, 80, 240, 160, 16, 6, 150, 800, 1200, 480, 32, 7, 252, 2100, 5600, 5040, 1344, 64, 8, 392, 4704, 19600, 31360, 18816, 3584, 128, 9, 576, 9408, 56448, 141120, 150528, 64512, 9216, 256, 10, 810, 17280, 141120, 508032

Offset: 0

Roger L. Bagula, Feb 04 2009

			1;
2, 2;
3, 12, 4;
4, 36, 48, 8;
5, 80, 240, 160, 16;
6, 150, 800, 1200, 480, 32;
7, 252, 2100, 5600, 5040, 1344, 64;
8, 392, 4704, 19600, 31360, 18816, 3584, 128;
9, 576, 9408, 56448, 141120, 150528, 64512, 9216, 256;
10, 810, 17280, 141120, 508032, 846720, 645120, 207360, 23040, 512;

Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 75-76

A142983, A142978, A047781 (row sums).

Clear[t0, p, x, n, m];
p[x_, n_] = Sum[Binomial[n, k]*Binomial[n - 1, n - k]*2^k*x^k, {k, 0, n}]/(2*x);
Table[FullSimplify[ExpandAll[p[x, n]]], {n, 1, 10}];
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 1, 10}];
Flatten[%]

p(x,n)=Sum[Binomial[n, k]*Binomial[n - 1, n - k]*2^k*x^k, {k, 0, n}]/(2*x);
p(x,n)=n Hypergeometric2F1[1 - n, 1 - n, 2, 2 x];
t(n,m)=coefficiemts(p(x,n))
T(n,m) = 2^m*A103371(n,m). - R. J. Mathar, Dec 05 2017

cp's OEIS Frontend

A142986 a(1) = 1, a(2) = 8, a(n+2) = 8*a(n+1) + (n + 1)*(n + 2)*a(n).

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

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A142989 a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1)+(n+1)*(n+3)*a(n).

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A142990 a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1)+(n+1)*(n+3)*a(n).

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A142991 a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*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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A156136 A triangle of polynomial coefficients related to Mittag-Leffler polynomials: p(x,n)=Sum[Binomial[n, k]*Binomial[n - 1, n - k]*2^k*x^k, {k, 0, n}]/(2*x).

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A142986 a(1) = 1, a(2) = 8, a(n+2) = 8a(n+1) + (n + 1)(n + 2)*a(n).

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

A142989 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1)+(n+1)(n+3)*a(n).

A142990 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1)+(n+1)(n+3)*a(n).

A142991 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1)+(n+1)(n+3)*a(n).

A156136 A triangle of polynomial coefficients related to Mittag-Leffler polynomials: p(x,n)=Sum[Binomial[n, k]Binomial[n - 1, n - k]2^kx^k, {k, 0, n}]/(2x).