A157142 -id:A157142 - cp's OEIS frontend

A128139 Triangle read by rows: matrix product A004736 * A128132.

1, 1, 2, 1, 3, 3, 1, 4, 5, 4, 1, 5, 7, 7, 5, 1, 6, 9, 10, 9, 6, 1, 7, 11, 13, 13, 11, 7, 1, 8, 13, 16, 17, 16, 13, 8, 1, 9, 15, 19, 21, 21, 19, 15, 9, 1, 10, 17, 22, 25, 26, 25, 22, 17, 10

Offset: 0

Gary W. Adamson, Feb 16 2007

A077028 with the final term in each row omitted.
Interchanging the factors in the matrix product leads to A128140 = A128132 * A004736.
From Gary W. Adamson, Jul 01 2012: (Start)
Alternatively, antidiagonals of an array A(n,k) of sequences with arithmetic progressions as follows:
  1, 2, 3,  4,  5,  6, ...
  1, 3, 5,  7,  9, 11, ...
  1, 4, 7, 10, 13, 16, ...
  1, 5, 9, 13, 17, 21, ...
  ... (End)
From Gary W. Adamson, Jul 02 2012: (Start)
A summation generalization for Sum_{k>=1} 1/(A(n,k)*A(n,k+1)) (formulas copied from A002378, A000466, A085001, A003185):
  1 = 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4)  + ...
  1 = 2/(1)*(3) + 2/(3)*(5) + 2/(5)*(7)  + ...
  1 = 3/(1)*(4) + 3/(4)*(7) + 3/(7)*(10) + ...
  1 = 4/(1)*(5) + 4/(5)*(9) + 4/(9)*(13) + ...
  ...
As a summation of terms equating to a definite integral:
  Integral_{0..1} dx/(1+x) = ... 1 - 1/2 + 1/3 - 1/4 + ... = log(2).
  Integral_{0..1} dx/(1+x^2) = 1 - 1/3 + 1/5 - 1/7  + ... = Pi/4 (see A157142)
  Integral_{0..1} dx/(1+x^3) = 1 - 1/4 + 1/7 - 1/10 + ... (see A016777)
  Integral_{0..1} dx/(1+x^4) = 1 - 1/5 + 1/9 - 1/13 + ... (see A016813). (End)

			First few rows of the triangle:
  1;
  1,  2;
  1,  3,  3;
  1,  4,  5,  4;
  1,  5,  7,  7,  5;
  1,  6,  9, 10,  9,  6;
  1,  7, 11, 13, 13, 11,  7;
  1,  8, 13, 16, 17, 16, 13,  8;
  1,  9, 15, 19, 21, 21, 19, 15,  9;
  1, 10, 17, 22, 25, 26, 25, 22, 17, 10;
  ...

Cf. A004736, A128132, A128140, A004006 (row sums).

A004736 * A128132 as infinite lower triangular matrices.

T(n,k) = k*(1+n-k)+1 = 1 + A094053(n+1,1+n-k). - R. J. Mathar, Jul 09 2012

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.

A164991 Number of triangular involutions of n. A triangular involution is a square involution with at most three faces.

Original entry on oeis.org

1, 1, 3, 6, 13, 26, 54, 108, 221, 442, 898, 1796, 3634, 7268, 14668, 29336, 59101, 118202, 237834, 475668, 956198, 1912396, 3841588, 7683176, 15425138, 30850276, 61908564, 123817128, 248377156, 496754312

Offset: 1

Simone Rinaldi (rinaldi(AT)unisi.it), Sep 04 2009

The sequence 2^(n+1) - binomial(n, floor(n/2)), which begins 1,3,6,... has Hankel transform (-1)^n*(2*n+1) (A157142). - Paul Barry, Nov 03 2010

For n >= 2 also row sums of A258445. - Wolfdieter Lang, Jun 27 2015

F. Disanto, A. Frosini, S. Rinaldi, Square Involutions, Proceedings of Permutation Patterns, July, 13-17 2009, Florence.
T. Mansour, S. Severini, Grid polygons from permutations and their enumeration by the kernel method, 19th Conference on Formal Power Series and Algebraic Combinatorics, Tianjin, China, July 2-6, 2007.

F. Disanto, A. Frosini, S. Rinaldi, Square involutions, J. Int. Seq. 14 (2011) # 11.3.5.
T. Mansour, S. Severini, Grid polygons from permutations and their enumeration by the kernel method, arXiv:math/0603225 [math.CO], 2006.

Cf. A128652, A128650, A258445.

Join[{1},Table[2^(n-1)-Binomial[n-2,Floor[(n-2)/2]],{n,2,30}]] (* Harvey P. Dale, Dec 26 2015 *)

a(n) =  2^(n-1) - binomial(n-2, (n-2)\2) \\ Michel Marcus, May 27 2013

a(n) = 2^(n-1) - binomial(n-2, floor((n-2)/2)) for n>1, a(1)=1.
From Wolfdieter Lang, Jun 27 2015: (Start)
a(n) = Sum_{k = 1..2*n-3} A258445(n-1, k), n >= 2.
a(2*k+1) = 4*Sum_{j = 0..(k-2)} binomial(2*k-1,j) + 3*binomial(2*k-1,k-1), k >= 1.
a(2*k) = 4*Sum_{j = 0..(k-2)} binomial(2*(k-1),j) + binomial(2*(k-1),k-1), k >= 1. (End)
(-n+1)*a(n) + 2*(n-1)*a(n-1) + 4*(n-4)*a(n-2) + 8*(-n+4)*a(n-3) = 0. - R. J. Mathar, Aug 09 2017

A164916 Denominators of a BBP series for Pi/4.

Original entry on oeis.org

1, -8, -20, -24, 144, -384, -832, -896, 4352, -10240, -21504, -22528, 102400, -229376, -475136, -491520, 2162688, -4718592, -9699328, -9961472, 42991616, -92274688, -188743680, -192937984, 822083584, -1744830464, -3556769792

Offset: 0

Jaume Oliver Lafont, Aug 31 2009

From the BBP formula for Pi, the following expression for Pi/4 in unit numerators is obtained
Pi/4 = Sum((1/(8k+1)+1/(-2*(8k+4))+1/(-4*(8k+5))+1/(-4*(8k+6)))/16^k, k>=0)
Therefore a(n) such that
a(4*n) = (8*n+1)*16^n.
a(4*n+1) = -2*(8*n+4)*16^n.
a(4*n+2) = -4*(8*n+5)*16^n.
a(4*n+3) = -4*(8*n+6)*16^n.
has
Sum_{n >= 0} (1/a(n)) = Pi/4.
Using PARI/GP suminf(n=0,1/(2^(n-2)*(2*(-1+(-1)^n+(1-I)*(-I)^n+(1+I)*I^n)+(-3+3*(-1)^n+(4-I)*(-I)^n+(4+I)*I^n)*n)))= 0.7853981633974483096156608454...=Pi/4. - Alexander R. Povolotsky, Sep 01 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A048581, A066968, A154925, A154962, A156269, A157142.

CoefficientList[Series[(1 - 8*x - 20*x^2 - 24*x^3 + 112*x^4 - 128*x^5 - 192*x^6 - 128*x^7)/(1 - 16*x^4)^2, {x,0,50}], x] (* G. C. Greubel, Feb 25 2017 *)

x='x + O('x^50); Vec((1 - 8*x - 20*x^2 - 24*x^3 + 112*x^4 - 128*x^5 - 192*x^6 - 128*x^7)/(1 - 16*x^4)^2) \\ G. C. Greubel, Feb 25 2017

G.f.: (1-8*x-20*x^2-24*x^3+112*x^4-128*x^5-192*x^6-128*x^7)/(1-16*x^4)^2.

a(n)= 2^(n-2)*(2*(-1+(-1)^n+(1-I)*(-I)^n+(1+I)*I^n)+(-3+3*(-1)^n+(4-I)*(-I)^n+(4+I)*I^n)*n). - Alexander R. Povolotsky, Sep 01 2009

Comment section corrected by Jaume Oliver Lafont, Sep 03 2009

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A128139 Triangle read by rows: matrix product A004736 * A128132.

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

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A164991 Number of triangular involutions of n. A triangular involution is a square involution with at most three faces.

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A164916 Denominators of a BBP series for Pi/4.

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