A155988 -id:A155988 - cp's OEIS frontend

A171220 a(n) = (2n + 1)*5^n.

1, 15, 125, 875, 5625, 34375, 203125, 1171875, 6640625, 37109375, 205078125, 1123046875, 6103515625, 32958984375, 177001953125, 946044921875, 5035400390625, 26702880859375, 141143798828125, 743865966796875, 3910064697265625, 20503997802734375, 107288360595703125

Offset: 0

Jaume Oliver Lafont, Dec 05 2009

Inserting x=1/sqrt(b) into the power series expansion of arctanh(x) yields the general BBP-type formula log((sqrt(b)+1)/(sqrt(b)-1))*sqrt(b)/2 = Sum_{k>=0} 1/((2k+1)b^k).
This sequence illustrates case b=5, with
Sum_{k>=0} 1/a(k) = sqrt(5)*log((1+sqrt(5))/2).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
David H. Bailey, Compendium of BBP formulas for mathematical constants. See formula 86 at p. 26.
Index entries for linear recurrences with constant coefficients, signature (10,-25).

Cf. A014480 ((2n+1)*2^n), A124647 ((2n+1)*3^n), A058962 ((2n+1)*4^n), A155988 ((2n+1)*9^n), A165283 ((2n+1)*16^n), A166725 ((2n+1)*25^n).

[(2*n+1)*5^n: n in [0..25]]; // Vincenzo Librandi, Jun 08 2011

PARI
```
a(n)=(2*n+1)*5^n
```

a(n) = 10*a(n-1) - 25*a(n-2).
O.g.f: (1+5*x)/(1-5*x)^2.
Sum_{n>=0} (-1)^n/a(n) = sqrt(5)*arctan(1/sqrt(5)). - Amiram Eldar, Feb 26 2022
E.g.f.: exp(5*x)*(1 + 10*x). - Stefano Spezia, May 09 2023

A199299 a(n) = (2n + 1)6^n.

Original entry on oeis.org

1, 18, 180, 1512, 11664, 85536, 606528, 4199040, 28553472, 191476224, 1269789696, 8344332288, 54419558400, 352638738432, 2272560758784, 14575734521856, 93096626946048, 592433080565760, 3757718396731392, 23765029860409344, 149902496042582016, 943288877536247808

Offset: 0

Philippe Deléham, Nov 04 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (12,-36).

Cf. A014480, A058962, A124647, A155988, A171220, A199300, A199301.

[(2*n+1)*6^n: n in [0..30]]; // Vincenzo Librandi, Nov 05 2011

a[n_] := (2*n + 1)*6^n; Array[a, 25, 0] (* Amiram Eldar, Dec 10 2022 *)

a(n) = (2*n+1)*6^n \\ Amiram Eldar, Dec 10 2022

a(n) = 12*a(n-1) - 36*a(n-2).
G.f.: (1+6*x)/(1-6*x)^2.
a(n) = 6*a(n-1) + 2*6^n. - Vincenzo Librandi, Nov 05 2011
From Amiram Eldar, Dec 10 2022: (Start)
Sum_{n>=0} 1/a(n) = sqrt(6)*arccoth(sqrt(6)).
Sum_{n>=0} (-1)^n/a(n) = sqrt(6)*arccot(sqrt(6)). (End)
E.g.f.: exp(6*x)*(1 + 12*x). - Stefano Spezia, May 07 2023

A060851 a(n) = (2n-1) * 3^(2n-1).

Original entry on oeis.org

3, 81, 1215, 15309, 177147, 1948617, 20726199, 215233605, 2195382771, 22082967873, 219667417263, 2165293113021, 21182215236075, 205891132094649, 1990280943581607, 19147875284802357, 183448998696332259, 1751104078464989745, 16660504517966902431

Offset: 1

Frank Ellermann, May 03 2001

Denominators of odd terms in expansion of arctanh(s/3); numerators are all 1. - Gerry Martens, Jul 26 2015

Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 28-40.

Harry J. Smith, Table of n, a(n) for n = 1..200
Xavier Gourdon and Pascal Sebah, Riemann's zeta function.
Pablo A. Panzone, Formulas for the Euler-Mascheroni constant, Rev. Un. Mat. Argentina, Vol 50, No. 1 (2009), pp. 161-164.
Simon Plouffe, Other interesting computations at numberworld.org.
Index entries for linear recurrences with constant coefficients, signature (18,-81).

Cf. A002162 (log(2)), A001620 (Euler's constant).

[ (2*n-1) * (3^(2*n-1)): n in [1..100]]; // Vincenzo Librandi, Apr 20 2011

A060851:=n->(2*n-1)*3^(2*n-1); seq(A060851(n), n=1..20); # Wesley Ivan Hurt, Dec 02 2013

Table[(2*n - 1)*3^(2*n - 1), {n, 20}] (* Wesley Ivan Hurt, Dec 02 2013 *)
a[n_] := 1/SeriesCoefficient[ArcTanh[s/3],{s,0,n}]
Table[a[n], {n, 1, 40, 2}] (* Gerry Martens, Jul 26 2015 *)

a(n) = (2*n - 1)*(3^(2*n - 1)) \\ Harry J. Smith, Jul 13 2009

Sum_{n>=1} 2/a(n) = log(2).
Sum_{n>=1} (2/a(n) - zeta(2n+1)/(2^(2n)*(2n+1))) = gamma (Euler's constant).
Sum_{n>=1} ((4n+2)/a(n) - zeta(2n+1)/2^(2n))/(2n+1) = gamma (Euler's constant).
Sum_{n>=1} ((4n+2)/a(n) - zeta(2n+1)/2^(2n)) = 7/4.
Sum_{n>=1} ((2n+1)/a(n) - zeta(2n+1)/2^(2n+1)) = 7/8.
From R. J. Mathar, May 07 2013: (Start)
G.f.: 3*x*(1+9*x) / (9*x-1)^2.
a(n+1) = 3*A155988(n). (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = arctan(1/3). - Amiram Eldar, Feb 26 2022
E.g.f.: (1 + exp(9*x)*(18*x - 1))/3. - Stefano Spezia, Dec 26 2024

More terms from Larry Reeves (larryr(AT)acm.org), May 07 2001

A199300 a(n) = (2n + 1)7^n.

Original entry on oeis.org

1, 21, 245, 2401, 21609, 184877, 1529437, 12353145, 98001617, 766718533, 5931980229, 45478515089, 346032180025, 2616003280989, 19668469112621, 147174406808233, 1096686708796833, 8142067989552245, 60251303122686613, 444556912229552577, 3271482918202092041

Offset: 0

Philippe Deléham, Nov 04 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (14,-49).

Cf. A014480, A058962, A124647, A155988, A171220, A199299, A199301.

[(2*n+1)*7^n: n in [0..30]]; // Vincenzo Librandi, Nov 05 2011

a[n_] := (2*n + 1)*7^n; Array[a, 25, 0] (* Amiram Eldar, Dec 10 2022 *)
LinearRecurrence[{14,-49},{1,21},30] (* Harvey P. Dale, Mar 26 2025 *)

a(n) = (2*n+1)*7^n \\ Amiram Eldar, Dec 10 2022

a(n) = 14*a(n-1) - 49*a(n-2).
G.f.: (1+7*x)/(1-7*x)^2.
a(n) = 7*a(n-1) + 2*7^n. - Vincenzo Librandi, Nov 05 2011
From Amiram Eldar, Dec 10 2022: (Start)
Sum_{n>=0} 1/a(n) = sqrt(7)*arccoth(sqrt(7)).
Sum_{n>=0} (-1)^n/a(n) = sqrt(7)*arccot(sqrt(7)). (End)
E.g.f.: exp(7*x)*(1 + 14*x). - Stefano Spezia, May 09 2023

a(15) corrected by Vincenzo Librandi, Nov 05 2011

A199301 a(n) = (2n+1)*8^n.

Original entry on oeis.org

1, 24, 320, 3584, 36864, 360448, 3407872, 31457280, 285212672, 2550136832, 22548578304, 197568495616, 1717986918400, 14843406974976, 127543348822016, 1090715534753792, 9288674231451648, 78812993478983680, 666532744850833408, 5620492334958379008, 47269781688880726016

Offset: 0

Philippe Deléham, Nov 04 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (16,-64).

Cf. A001018 (Powers of 8), A005408 (2n+1).

Cf. A014480, A058962, A124647, A155988, A171220, A199299, A199300.

[(2*n+1)*8^n: n in [0..30]]; // Vincenzo Librandi, Nov 05 2011

A199301:=n->(2*n+1)*8^n: seq(A199301(n), n=0..20); # Wesley Ivan Hurt, Oct 30 2014

Table[(2 n + 1)*8^n, {n, 0, 20}] (* Wesley Ivan Hurt, Oct 30 2014 *)

a(n) = (2*n+1)*8^n \\ Amiram Eldar, Dec 10 2022

a(n) = 16*a(n-1)-64*a(n-2).
G.f.: (1+8*x)/(1-8*x)^2.
a(n) = 8*(a(n-1)+2^(3*n-2)). - Vincenzo Librandi, Nov 05 2011
a(n) = A005408(n) * A001018(n). - Wesley Ivan Hurt, Oct 30 2014
From Amiram Eldar, Dec 10 2022: (Start)
Sum_{n>=0} 1/a(n) = sqrt(8)*arccoth(sqrt(8)).
Sum_{n>=0} (-1)^n/a(n) = sqrt(8)*arccot(sqrt(8)). (End)
E.g.f.: exp(8*x)*(1 + 16*x). - Stefano Spezia, May 09 2023

a(18) corrected by Vincenzo Librandi, Nov 05 2011

A157327 Egyptian fraction expansion for Pi/4 = arctan(1/2) + arctan(1/3) (Hutton 1776).

Original entry on oeis.org

2, 3, -24, -81, 160, 1215, -896, -15309, 4608, 177147, -22528, -1948617, 106496, 20726199, -491520, -215233605, 2228224, 2195382771, -9961472, -22082967873, 44040192, 219667417263, -192937984, -2165293113021, 838860800

Offset: 0

Jaume Oliver Lafont, Feb 27 2009

Sum_{n>=0} 1/a(n) = Pi/4.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
X. Gourdon and P. Sebah, The constant Pi. The classic period

Cf. A157142, A155988, A058962.

CoefficientList[Series[2 (1 - 4 x^2)/(1 + 4 x^2)^2 + 3 x (1 - 9 x^2)/(1 + 9 x^2)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 12 2012 *)

G.f.: 2*(1-4*x^2)/(1+4*x^2)^2 + 3*x*(1-9*x^2)/(1+9*x^2)^2.

A165283 a(n) = (2n + 1)16^n.

Original entry on oeis.org

1, 48, 1280, 28672, 589824, 11534336, 218103808, 4026531840, 73014444032, 1305670057984, 23089744183296, 404620279021568, 7036874417766400, 121597189939003392, 2089670227099910144, 35740566642812256256, 608742554432415203328, 10330176681277348904960

Offset: 0

Jaume Oliver Lafont, Sep 13 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (32,-256).

Cf. A058962 ((2n+1)4^n), A155988 ((2n+1)9^n).

[(2*n+1)*16^n: n in [0..20]]; // Vincenzo Librandi, Jun 08 2011

PARI
```
a(n)=(2*n+1)*16^n
```

G.f.: (1+16*x)/(1-16*x)^2.
Sum_{n>=0} 1/a(n) = 2*log(5/3).
Sum_{n>=0} (-1)^n/a(n) = 4 * arctan(1/4). - Amiram Eldar, Jul 12 2020
E.g.f.: exp(16*x)*(1 + 32*x). - Stefano Spezia, May 09 2023

A166725 a(n) = (2n+1)25^n.

Original entry on oeis.org

1, 75, 3125, 109375, 3515625, 107421875, 3173828125, 91552734375, 2593994140625, 72479248046875, 2002716064453125, 54836273193359375, 1490116119384765625, 40233135223388671875, 1080334186553955078125, 28870999813079833984375, 768341124057769775390625

Offset: 0

Jaume Oliver Lafont, Oct 20 2009

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (50,-625).

Cf. A058962 ((2n+1)*4^n), A155988 ((2n+1)*9^n), A016578 (log(3/2)).

[(2*n+1)*25^n: n in [0..20]]; // Vincenzo Librandi, Jun 08 2011

Table[5^(2*n) *(2*n + 1), {n,0,10}] (* G. C. Greubel, May 24 2016 *)
LinearRecurrence[{50,-625},{1,75},30] (* Harvey P. Dale, Mar 02 2018 *)

PARI
```
a(n)=(2*n+1)*25^n
```

G.f.: (1+25*x)/(1-25*x)^2.
Sum_{k>=0} 1/a(k) = (5/2)*log(3/2).
E.g.f.: (50*x + 1)*exp(25*x). - G. C. Greubel, May 24 2016
Sum_{n>=0} (-1)^n/a(n) = 5*arctan(1/5). - Amiram Eldar, Feb 26 2022

A362885 Array read by ascending antidiagonals: A(n, k) = (1 + 2n)k^n.

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 5, 6, 1, 0, 7, 20, 9, 1, 0, 9, 56, 45, 12, 1, 0, 11, 144, 189, 80, 15, 1, 0, 13, 352, 729, 448, 125, 18, 1, 0, 15, 832, 2673, 2304, 875, 180, 21, 1, 0, 17, 1920, 9477, 11264, 5625, 1512, 245, 24, 1, 0, 19, 4352, 32805, 53248, 34375, 11664, 2401, 320, 27, 1

Offset: 0

Stefano Spezia, May 08 2023

			The array begins:
    1,  1,   1,    1,     1,     1, ...
    0,  3,   6,    9,    12,    15, ...
    0,  5,  20,   45,    80,   125, ...
    0,  7,  56,  189,   448,   875, ...
    0,  9, 144,  729,  2304,  5625, ...
    0, 11, 352, 2673, 11264, 34375, ...
    ...

Cf. A000007 (k=0), A000012 (n=0), A004248, A005408 (k=1), A008585 (n=1), A014480 (k=2), A033429 (n=2), A058962 (k=4), A124647 (k=3), A155988 (k=9), A171220 (k=5), A176043, A199299 (k=6), A199300 (k=7), A199301 (k=8), A244727 (n=3), A362886 (antidiagonal sums).

A[n_,k_]:=(1+2n)k^n; Join[{1}, Table[A[n-k,k],{n,10},{k,0,n}]]//Flatten (* or *)
A[n_,k_]:=SeriesCoefficient[(1+k*x)/(1-k*x)^2,{x,0,n}]; Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten (* or *)
A[n_,k_]:=n!SeriesCoefficient[Exp[k*x](1+2k*x),{x,0,n}]; Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten

A(n, k) = A005408(n)*A004248(n, k).
O.g.f. of column k: (1 + k*x)/(1 - k*x)^2.
E.g.f. of column k: exp(k*x)*(1 + 2*k*x).
A(n, n) = A176043(n+1).

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A171220 a(n) = (2n + 1)*5^n.

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A199299 a(n) = (2*n + 1)*6^n.

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A060851 a(n) = (2n-1) * 3^(2n-1).

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A199300 a(n) = (2*n + 1)*7^n.

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A199301 a(n) = (2n+1)*8^n.

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A157327 Egyptian fraction expansion for Pi/4 = arctan(1/2) + arctan(1/3) (Hutton 1776).

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Mathematica

Formula

A165283 a(n) = (2*n + 1)*16^n.

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A199299 a(n) = (2n + 1)6^n.

A199300 a(n) = (2n + 1)7^n.

A165283 a(n) = (2n + 1)16^n.

A166725 a(n) = (2n+1)25^n.

A362885 Array read by ascending antidiagonals: A(n, k) = (1 + 2n)k^n.