A245244 -id:A245244 - cp's OEIS frontend

A014641 Odd octagonal numbers: (2n+1)*(6n+1).

1, 21, 65, 133, 225, 341, 481, 645, 833, 1045, 1281, 1541, 1825, 2133, 2465, 2821, 3201, 3605, 4033, 4485, 4961, 5461, 5985, 6533, 7105, 7701, 8321, 8965, 9633, 10325, 11041, 11781, 12545, 13333, 14145, 14981, 15841, 16725, 17633, 18565, 19521, 20501, 21505

Offset: 0

Mohammad K. Azarian, Dec 11 1999

Sequence found by reading the line from 1, in the direction 1, 21, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Jul 18 2012

Muniru A Asiru, Table of n, a(n) for n = 0..5000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014.
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, Vol. 18 (2015), Article 15.3.2.
Leo Tavares, Illustration: Square Block Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A001082, A014642, A014793, A014794, A243201, A289873.
Cf. A160485, A245244.
Cf. A016754, A014105.

List([0..50],n->(2*n+1)*(6*n+1)); # Muniru A Asiru, Feb 05 2019

[ (2*n+1)*(6*n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 08 2014

A014641:=n->(2*n+1)*(6*n+1); seq(A014641(n), n=0..50); # Wesley Ivan Hurt, Jun 08 2014

Table[(2n + 1)(6n + 1), {n, 0, 49}]  (* Harvey P. Dale, Mar 24 2011 *)

a(n)=(2*n+1)*(6*n+1) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = a(n-1) + 24*n - 4, with n > 0, a(0)=1. - Vincenzo Librandi, Dec 28 2010
G.f.: (1 + 18*x + 5*x^2)/(1 - 3*x + 3*x^2 - x^3). - Colin Barker, Jan 06 2012
a(n) = A289873(6*n+2). - Hugo Pfoertner, Jul 15 2017
From Peter Bala, Jan 22 2018: (Start)
This is the polynomial Qbar(2,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials.
a(n) = (1/4^n) * Sum_{k = 0..n} (2*k + 1)^4*binomial(2*n + 1, n - k).
a(n-1) = (2/4^n) * binomial(2*n,n) * ( 1 + 3^4*(n - 1)/(n + 1) + 5^4*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^4*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)
From Amiram Eldar, Feb 27 2022: (Start)
Sum_{n>=0} 1/a(n) = (sqrt(3)*Pi + 3*log(3))/8.
Sum_{n>=0} (-1)^n/a(n) = Pi/8 + sqrt(3)*log(2+sqrt(3))/4. (End)
E.g.f.: exp(x)*(1 + 20*x + 12*x^2). - Stefano Spezia, Apr 16 2022
a(n) = A016754(n) + 4*A014105(n). - Leo Tavares, May 20 2022

More terms from Patrick De Geest

Better description from N. J. A. Sloane

A160485 Triangle of the RBS1 polynomial coefficients.

Original entry on oeis.org

1, 1, -2, 1, -8, 12, 1, -2, 60, -120, 1, -128, -168, 0, 1680, 1, 2638, 7320, -5040, -25200, -30240, 1, -98408, -300828, 52800, 1053360, 1330560, 665280, 1, 5307118, 17914260, 2522520, -56456400, -90810720, -60540480, -17297280

Offset: 1

Johannes W. Meijer, May 24 2009, Jul 06 2009, Sep 19 2012

In A160480 we defined the BS1 matrix by BS1[2*m-1,n=1] = 2*beta(2*m) and the recurrence relation BS1 [2*m-1,n] = (2*n-3)/(2*n-2)*(BS1[2*m-1,n-1]- BS1[2*m-3,n-1]/(2*n-3)^2), for positive and negative values of m and n= 1, 2, .. . As usual beta(m) = sum((-1)^k/(1+2*k)^m, k=0..infinity). It is well-known that BS1[1-2*m,n=1] = euler(2*m-2) for m = 1, 2, .., with euler(2*m-2) the Euler numbers A000364. These values together with the recurrence relation lead to BS1[ -1,n] = 1 for n = 1, 2, .. .
We discovered that the n-th term of the row coefficients BS1[1-2*m,n] for m = 1, 2, .., can be generated with the rather simple polynomials RBS1(1-2*m,n). Our discovery was enabled by the recurrence relation for the RBS1(1-2*m,n) polynomials which we derived from the recurrence relation for the BS[2*m-1,n] coefficients and the fact that RBS1(-1,n) = 1.
The RBS1 polynomials and the polynomials defined by sequence A083061 are related by a shift of +-1/2 and scaling by a power of 2 (see arXiv link). - Richard P. Brent, Jul 15 2014

			The first few rows of the triangle are:
[1]
[1, -2]
[1, -8, 12]
[1, -2, 60, -120]
[1, -128, -168, 0, 1680]
The first few RBS1(1-2*m,n) polynomials are:
RBS1(-1,n) = 1
RBS1(-3,n) = 1 - 2*n
RBS1(-5,n) = 1 - 8*n + 12*n^2
RBS1(-7,n) = 1 - 2*n + 60*n^2 - 120*n^3
From _Peter Bala_, Jan 22 2019: (Start)
Qbar(r,n) = binomial(2*n+2,n+1)/(2^(2*n+1)) * Sum_{k = 0..n} binomial(n,k)/binomial(n+k+1,k)*(2*k + 1)^(2*r):
Case r = 2: Qbar(2,n) = binomial(2*n+2,n+1)/2^(2*n+1) * ( 1 + 3^4*n/(n+2) + 5^4*n*(n-1)/((n+2)*(n+3)) + 7^4*n*(n-1)*(n-2)/((n+2)*(n+3)*(n+4)) + ... ) = 12*n^2 + 8*n + 1, valid for n a nonnegative integer (when the series terminates). The identity is also valid for complex n with real part greater than 1 (provided the factor binomial(2*n,n) is replaced with the appropriate expression involving the gamma function).
Case r = 3: Qbar(3,n) = binomial(2*n+2,n+1)/(2^(2*n+1)) * ( 1 + 3^6*n/(n+2) + 5^6*n*(n-1)/((n+2)*(n+3)) + 7^6*n*(n-1)*(n-2)/((n+2)*(n+3)*(n+4)) + ... ) = 120*n^3 + 60*n^2 + 2*n + 1, valid for n a nonnegative integer. The identity is also valid for complex n with real part greater than 2.
Note, the case r = 0 is equivalent to the identity 1 = binomial(2*n,n)/2^(2*n-1) * ( 1 + (n-1)/(n+1) + (n-1)*(n-2)/((n+1)*(n+2)) + (n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... ), which is valid for complex n with real part greater than 0. This identity was found by Ramanujan. See Example 6, Chapter 10 in Berndt. (End)

B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, Chapter 10, p. 21.

P. Bala, A245244 and A160485 and some hypergeometric series evaluations of Ramanujan
R. P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.

A160480 is the Beta triangle.
A009389(2*n) equals the second left hand column divided by 2.
A001813 equals the first right hand column.
The absolute values of the row sums equal the Euler numbers A000364.
Cf. A083061, A036970, A245244.

nmax := 8; mmax := nmax: A(1, 1) := 1: RBS1(n, 2) := (2*n-1)^2*1-(2*n)*(2*n-1)*1: for m from 3 to mmax do for k from 0 to m-1 do A(m-1, k+1) := coeff(RBS1(n, m-1), n, k) od; RBS1(n+1, m-1) := 0: for k from 0 to m-1 do RBS1(n+1, m-1) := RBS1(n+1, m-1) + A(m-1, k+1)*(n+1)^k od: RBS1(n, m) := (2*n-1)^2*RBS1(n, m-1)-(2*n)*(2*n-1) * RBS1(n+1, m-1) od: for k from 0 to nmax-1 do A(nmax, k+1) := coeff(RBS1(n, nmax), n, k) od: seq(seq(A(n, m), m=1..n), n=1..nmax);

RBS1(1-2*m,n) = (2*n-1)^2*RBS1(3-2*m,n)-(2*n)*(2*n-1)*RBS1(3-2*m,n+1) for m = 2, 3, .., with RBS1(-1,n) =1 for n = 1, 2, .. .
From Peter Bala, Jan 22 2019: (Start)
The row polynomials RBS1 of the triangle are related to the polynomials Qbar(r,n), r = 0,1,2,..., introduced by Brent by Qbar(r,n) = RBS1(-2*r-1,-n).
Recurrence: Qbar(r+1,n) = (2*n + 1)^2*Qbar(r,n) - 2*n(2*n + 1)*Qbar(r,n-1) with Qbar(0,n) = 1 (Brent, equation 19, p.7).
Qbar(r,n) = binomial(2*n + 2,n + 1)/(2^(2*n + 1)) * Sum_{k = 0..n} binomial(n,k)/binomial(n + k + 1,k)*(2*k + 1)^(2*r); this follows easily from the above recurrence. Two examples are given below.
Qbar(r,n) = 1/4^n * Sum_{k = 0..n} binomial(2*n + 1,n - k)*(2*k + 1)^(2*r) For related polynomial sequences see A036970, A083061 and A245244. (End)

A272126 a(n) = 120n^3 + 60n^2 + 2*n + 1.

Original entry on oeis.org

1, 183, 1205, 3787, 8649, 16511, 28093, 44115, 65297, 92359, 126021, 167003, 216025, 273807, 341069, 418531, 506913, 606935, 719317, 844779, 984041, 1137823, 1306845, 1491827, 1693489, 1912551, 2149733, 2405755, 2681337, 2977199, 3294061, 3632643, 3993665

Offset: 0

Vincenzo Librandi, Apr 25 2016

This is the polynomial Qbar(3,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials. - Peter Bala, Jan 22 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A014641, A158673, A160485, A245244, A272127, A272129.

[120*n^3 + 60*n^2 + 2*n + 1: n in [0..50]];

Table[120 n^3 + 60 n^2 + 2 n + 1, {n, 0, 40}]
LinearRecurrence[{4,-6,4,-1},{1,183,1205,3787},40] (* Harvey P. Dale, Nov 08 2020 *)

a(n) = 120*n^3 + 60*n^2 + 2*n + 1; \\ Altug Alkan, Apr 30 2016

O.g.f.: (1 + 179*x + 479*x^2 + 61*x^3)/(1-x)^4.
E.g.f.: (1 + 182*x + 420*x^2 + 120*x^3)*exp(x).
a(n) = (2*n+1)*(60*n^2+1).
a(n) = (2*n+1) * A158673(n).
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3.
See page 7 in Brent's paper:
a(n) = (2*n+1)^2*A014641(n) - 2*n*(2*n+1)*A014641(n-1).
A272127(n) = (2*n+1)^2*a(n) - 2*n*(2*n+1)*a(n-1).
From Peter Bala, Jan 22 2019: (Start)
a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^6 * binomial(2*n + 1, n - k).
a(n-1) = 2/4^n * binomial(2*n,n) * ( 1 + 3^6*(n - 1)/(n + 1) + 5^6*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^6*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)

A272127 a(n) = 1680n^4 - 168n^2 + 128*n + 1.

Original entry on oeis.org

1, 1641, 26465, 134953, 427905, 1046441, 2172001, 4026345, 6871553, 11010025, 16784481, 24577961, 34813825, 47955753, 64507745, 85014121, 110059521, 140268905, 176307553, 218881065, 268735361, 326656681, 393471585, 470046953, 557289985, 656148201, 767609441

Offset: 0

Vincenzo Librandi, Apr 25 2016

This is the polynomial Qbar(4,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials. - Peter Bala, Jan 21 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A014641, A272126, A160485, A245244, A272129.

[1680*n^4-168*n^2+128*n+1: n in [0..50]];

Table[1680 n^4 - 168 n^2 + 128 n + 1, {n, 0, 30}]
LinearRecurrence[{5,-10,10,-5,1},{1,1641,26465,134953,427905},30] (* Harvey P. Dale, Nov 27 2017 *)

a(n) = 1680*n^4 - 168*n^2 + 128*n + 1; \\ Altug Alkan, Apr 30 2016

O.g.f.: (1+1636*x+18270*x^2+19028*x^3+1385*x^4)/(1-x)^5.
E.g.f.: (1+1640*x+11592*x^2+10080*x^3+1680*x^4)*exp(x).
a(n) = (2*n+1)*(840*n^3-420*n^2+126*n+1).
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5), for n>4
See page 7 in Brent's paper:
a(n) = (2*n+1)^2*A272126(n) - 2*n*(2*n+1)*A272126(n-1).
A272128(n) = (2*n+1)^2*a(n) - 2*n*(2*n+1)*a(n-1).
From Peter Bala, Jan 22 2019: (Start)
a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^8 * binomial(2*n + 1, n - k).
a(n-1) = 2/4^n * binomial(2*n,n) * ( 1 + 3^8*(n - 1)/(n + 1) + 5^8*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^8*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)

A272129 a(n) = 32n^2 - 56n + 25.

Original entry on oeis.org

25, 1, 41, 145, 313, 545, 841, 1201, 1625, 2113, 2665, 3281, 3961, 4705, 5513, 6385, 7321, 8321, 9385, 10513, 11705, 12961, 14281, 15665, 17113, 18625, 20201, 21841, 23545, 25313, 27145, 29041, 31001, 33025, 35113, 37265, 39481, 41761, 44105, 46513, 48985

Offset: 0

Vincenzo Librandi, Apr 26 2016

Subsequence of A001844.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014 (page 16).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001844, A272126, A272127, A272128, A272131, A272132, A272133, A245244.

Magma
```
[32*n^2 - 56*n + 25: n in [0..50]];
```

[32*n^2-56*n+25$n=0..40]; # Muniru A Asiru, Jan 28 2019

Table[32 n^2 - 56 n + 25, {n, 0, 40}]
LinearRecurrence[{3,-3,1},{25,1,41},50] (* Harvey P. Dale, Jul 03 2018 *)

lista(nn) = for(n=0, nn, print1(32*n^2-56*n+25, ", ")); \\ Altug Alkan, Apr 26 2016

O.g.f.: (25 - 74*x + 113*x^2)/(1-x)^3.
E.g.f.: (25 - 24*x + 32*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
n*a(n) = 1 + 3^5*(n-1)/(n+1) + 5^5*((n-1)*(n-2))/((n+1)*(n+2)) + ... for n >= 1. See A245244. - Peter Bala, Jan 19 2019

A272132 a(n) = 6144n^4 - 29184n^3 + 52416n^2 - 41840n + 12465.

Original entry on oeis.org

12465, 1, 3281, 68385, 388849, 1305665, 3307281, 7029601, 13255985, 22917249, 37091665, 57004961, 84030321, 119688385, 165647249, 223722465, 295877041, 384221441, 491013585, 618658849, 769710065, 946867521, 1152978961, 1391039585, 1664192049, 1975726465

Offset: 0

Vincenzo Librandi, Apr 26 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014 (page 16).
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A272131, A272133, A245244.

[6144*n^4 - 29184*n^3 + 52416*n^2 - 41840*n + 12465: n in [0..40]];

[6144*n^4-29184*n^3+52416*n^2-41840*n+12465$n=0..30]; # Muniru A Asiru, Jan 28 2019

Table[6144 n^4 - 29184 n^3 + 52416 n^2 - 41840 n + 12465, {n, 0, 40}]
LinearRecurrence[{5,-10,10,-5,1},{12465,1,3281,68385,388849},30] (* Harvey P. Dale, Aug 06 2022 *)

lista(nn) = for(n=0, nn, print1(6144*n^4-29184*n^3+52416*n^2-41840*n+12465, ", ")); \\ Altug Alkan, Apr 26 2016

O.g.f.: (12465 - 62324*x + 127926*x^2 - 72660*x^3 + 142049*x^4)/(1-x)^5.
E.g.f.: (12465 - 12464*x + 7872*x^2 + 7680*x^3 + 6144*x^4)*exp(x).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
See page 7 in Brent's paper:
a(n) = (2*n-1)^2*A272131(n) - 4*(n-1)^2*A272131(n-1).
A272133(n) = (2*n-1)^2*a(n) - 4*(n-1)^2*a(n-1).
n*a(n) = 1 + 3^9*(n-1)/(n+1) + 5^9*((n-1)*(n-2))/((n+1)*(n+2)) + ... for n >= 1. See A245244. - Peter Bala, Jan 19 2019

A272131 a(n) = 384n^3 - 1184n^2 + 1228*n - 427.

Original entry on oeis.org

-427, 1, 365, 2969, 10117, 24113, 47261, 81865, 130229, 194657, 277453, 380921, 507365, 659089, 838397, 1047593, 1288981, 1564865, 1877549, 2229337, 2622533, 3059441, 3542365, 4073609, 4655477, 5290273, 5980301, 6727865, 7535269, 8404817, 9338813, 10339561

Offset: 0

Vincenzo Librandi, Apr 26 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014 (page 16).
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A272129, A272132, A245244.

[384*n^3 - 1184*n^2 + 1228*n - 427: n in [0..50]];

[384*n^3-1184*n^2+1228*n-427$n=0..35]; # Muniru A Asiru, Jan 28 2019

Table[384 n^3 - 1184 n^2 + 1228 n - 427, {n, 0, 40}]
LinearRecurrence[{4,-6,4,-1},{-427,1,365,2969},40] (* Harvey P. Dale, Aug 24 2024 *)

lista(nn) = for(n=0, nn, print1(384*n^3-1184*n^2+1228*n-427, ", ")); \\ Altug Alkan, Apr 26 2016

O.g.f.: (-427 + 1709*x - 2201*x^2 + 3223*x^3)/(1-x)^4.
E.g.f.: (-427 + 428*x - 32*x^2 + 384*x^3)*exp(x).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3.
See page 7 in Brent's paper:
a(n) = (2*n-1)^2*A272129(n) - 4*(n-1)^2*A272129(n-1).
A272132(n) =  (2*n-1)^2*a(n) - 4*(n-1)^2*a(n-1).
n*a(n) = 1 + 3^7*(n-1)/(n+1) + 5^7*((n-1)*(n-2))/((n+1)*(n+2)) + ... for n >= 1. See A245244. - Peter Bala, Jan 19 2019

A272133 a(n) = 122880n^5 - 829440n^4 + 2258688n^3 - 3076288n^2 + 2079892*n - 555731.

Original entry on oeis.org

-555731, 1, 29525, 1657129, 16591741, 80872529, 269614501, 711754105, 1604794829, 3229552801, 5964902389, 10302521801, 16861638685, 26403775729, 39847496261, 58283149849, 82987617901, 115439059265, 157331655829, 210590358121, 277385630909, 360148198801

Offset: 0

Vincenzo Librandi, Apr 26 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16)
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A272132, A245244.

[122880*n^5 - 829440*n^4 + 2258688*n^3 -3076288*n^2 + 2079892*n - 555731: n in [0..30]];

[122880*n^5-829440*n^4+2258688*n^3-3076288*n^2+2079892*n-555731$n=0..30]; # Muniru A Asiru, Jan 28 2019

Table[122880 n^5 - 829440 n^4 + 2258688 n^3 - 3076288 n^2 + 2079892 n - 555731, {n, 0, 40}]
LinearRecurrence[{6,-15,20,-15,6,-1},{-555731,1,29525,1657129,16591741,80872529},30] (* Harvey P. Dale, Feb 10 2021 *)

lista(nn) = for(n=0, nn, print1(122880*n^5 - 829440*n^4 + 2258688*n^3 - 3076288*n^2 + 2079892*n - 555731, ", ")); \\ Altug Alkan, Apr 26 2016

O.g.f.: (-555731 + 3334387*x - 8306446*x^2 + 12594614*x^3 - 1244143*x^4 + 8922919*x^5)/(1-x)^6.
E.g.f.: (-555731 + 555732*x - 263104*x^2 + 354048*x^3 + 399360*x^4 + 122880*x^5)*exp(x).
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).
a(n) = (2*n-1)^2*A272132(n) - 4*(n-1)^2*A272132(n-1), see page 7 in Brent's paper.
n*a(n) = 1 + 3^11*(n-1)/(n+1) + 5^11*((n-1)*(n-2))/((n+1)*(n+2)) + ... for n >= 1. See A245244. - Peter Bala, Jan 19 2019

A245683 Array T(n,k) read by antidiagonals, where T(0,k) = -A226158(k) and T(n+1,k) = 2*T(n,k+1) - T(n,k).

Original entry on oeis.org

0, 2, 1, 0, 1, 1, -6, -3, -1, 0, 0, -3, -3, -2, -1, 50, 25, 11, 4, 1, 0, 0, 25, 25, 18, 11, 6, 3, -854, -427, -201, -88, -35, -12, -3, 0, 0, -427, -427, -314, -201, -118, -65, -34, -17, 24930, 12465, 6019, 2796, 1241, 520, 201, 68, 17, 0

Offset: 0

Paul Curtz, Jul 29 2014

Take T(n,k) = -A226158(k) and its transform via T(n+1,k) = 2*T(n,k+1) - T(n,k):
0,       1,    1,    0,   -1,    0,   3,   0, -17, ...
2,       1,   -1,   -2,    1,    6,  -3, -34, ...     = A230324
0,      -3,   -3,    4,   11,  -12, -65, ...
-6,     -3,   11,   18,  -35, -118, ...
0,      25,   25,  -88, -201, ...
50,     25, -201, -314, ...
0,    -427, -427, ...
-854, -427, ...
0, ...
Every row is alternatively an autosequence of the first kind, see A226158, and of the second kind, see A190339.
The second column is twice 1, -3, 25, -427, 12465, ... = (-1)^n*A009843(n) which is in the third column. See A132049(n), numerators of Euler's formula for Pi from the Bernoulli numbers, A243963 and A245244. Hence a link between the Genocchi numbers and Pi.
a(n) is the triangle of the increasing antidiagonals.

			Triangle a(n):
   0,
   2,  1,
   0,  1,  1,
  -6, -3, -1,  0,
   0, -3, -3, -2, -1,
  50, 25, 11,  4,  1,  0,
  etc.

Cf. A226158, A230324, A009843, A132049, A243963, A245244.

t[0, 0] = 0; t[0, 1] = 1; t[0, k_] := -k*EulerE[k-1, 0]; t[n_, k_] := t[n, k] = -t[n-1, k] + 2*t[n-1, k+1]; Table[t[n-k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 04 2014 *)

cp's OEIS Frontend

A014641 Odd octagonal numbers: (2n+1)*(6n+1).

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A160485 Triangle of the RBS1 polynomial coefficients.

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A272126 a(n) = 120*n^3 + 60*n^2 + 2*n + 1.

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A272127 a(n) = 1680*n^4 - 168*n^2 + 128*n + 1.

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A272129 a(n) = 32*n^2 - 56*n + 25.

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A272132 a(n) = 6144*n^4 - 29184*n^3 + 52416*n^2 - 41840*n + 12465.

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A272131 a(n) = 384*n^3 - 1184*n^2 + 1228*n - 427.

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A272126 a(n) = 120n^3 + 60n^2 + 2*n + 1.

A272127 a(n) = 1680n^4 - 168n^2 + 128*n + 1.

A272129 a(n) = 32n^2 - 56n + 25.

A272132 a(n) = 6144n^4 - 29184n^3 + 52416n^2 - 41840n + 12465.

A272131 a(n) = 384n^3 - 1184n^2 + 1228*n - 427.

A272133 a(n) = 122880n^5 - 829440n^4 + 2258688n^3 - 3076288n^2 + 2079892*n - 555731.