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Search: a245244 -id:a245244
Displaying 1-9 of 9 results found. page 1
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A014641 Odd octagonal numbers: (2n+1)*(6n+1). +10
14
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
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
LINKS
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).
FORMULA
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
MAPLE
A014641:=n->(2*n+1)*(6*n+1); seq(A014641(n), n=0..50); # Wesley Ivan Hurt, Jun 08 2014
MATHEMATICA
Table[(2n + 1)(6n + 1), {n, 0, 49}] (* Harvey P. Dale, Mar 24 2011 *)
PROG
(Magma) [ (2*n+1)*(6*n+1) : n in [0..50] ]; // Wesley Ivan Hurt, Jun 08 2014
(PARI) a(n)=(2*n+1)*(6*n+1) \\ Charles R Greathouse IV, Jun 17 2017
(GAP) List([0..50], n->(2*n+1)*(6*n+1)); # Muniru A Asiru, Feb 05 2019
CROSSREFS
Cf. A000567, A001082, A014642, A014793, A014794, A243201, A289873.
Cf. A160485, A245244.
Cf. A016754, A014105.
KEYWORD
nonn,easy
AUTHOR
Mohammad K. Azarian, Dec 11 1999
EXTENSIONS
More terms from Patrick De Geest
Better description from N. J. A. Sloane
STATUS
approved
A160485 Triangle of the RBS1 polynomial coefficients. +10
12
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
InA160480 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
REFERENCES
B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, Chapter 10, p. 21.
LINKS
Table of n, a(n) for n=1..36.
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.
FORMULA
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)
EXAMPLE
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)
MAPLE
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);
CROSSREFS
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.
KEYWORD
easy,sign,tabl
AUTHOR
Johannes W. Meijer, May 24 2009, Jul 06 2009, Sep 19 2012
STATUS
approved
A272126 a(n) = 120*n^3 + 60*n^2 + 2*n + 1. +10
4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
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
LINKS
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).
FORMULA
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)
MATHEMATICA
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 *)
PROG
(Magma) [120*n^3 + 60*n^2 + 2*n + 1: n in [0..50]];
(PARI) a(n) = 120*n^3 + 60*n^2 + 2*n + 1; \\ Altug Alkan, Apr 30 2016
CROSSREFS
Cf. A014641, A158673, A160485, A245244, A272127, A272129.
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Apr 25 2016
STATUS
approved
A272127 a(n) = 1680*n^4 - 168*n^2 + 128*n + 1. +10
4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
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
LINKS
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).
FORMULA
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)
MATHEMATICA
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 *)
PROG
(Magma) [1680*n^4-168*n^2+128*n+1: n in [0..50]];
(PARI) a(n) = 1680*n^4 - 168*n^2 + 128*n + 1; \\ Altug Alkan, Apr 30 2016
CROSSREFS
Cf. A014641, A272126, A160485, A245244, A272129.
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Apr 25 2016
STATUS
approved
A272129 a(n) = 32*n^2 - 56*n + 25. +10
4
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
Subsequence of A001844.
LINKS
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).
FORMULA
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
MAPLE
[32*n^2-56*n+25$n=0..40]; # Muniru A Asiru, Jan 28 2019
MATHEMATICA
Table[32 n^2 - 56 n + 25, {n, 0, 40}]
LinearRecurrence[{3, -3, 1}, {25, 1, 41}, 50] (* Harvey P. Dale, Jul 03 2018 *)
PROG
(Magma) [32*n^2 - 56*n + 25: n in [0..50]];
(PARI) lista(nn) = for(n=0, nn, print1(32*n^2-56*n+25, ", ")); \\ Altug Alkan, Apr 26 2016
CROSSREFS
Cf. A001844, A272126, A272127, A272128, A272131, A272132, A272133, A245244.
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Apr 26 2016
STATUS
approved
A272132 a(n) = 6144*n^4 - 29184*n^3 + 52416*n^2 - 41840*n + 12465. +10
4
12465, 1, 3281, 68385, 388849, 1305665, 3307281, 7029601, 13255985, 22917249, 37091665, 57004961, 84030321, 119688385, 165647249, 223722465, 295877041, 384221441, 491013585, 618658849, 769710065, 946867521, 1152978961, 1391039585, 1664192049, 1975726465 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014 (page 16).
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
O.g.f.: (12465 - 62324*x + 127926*x^2 - 72660*x^3 + 142049*x^4)/(1-x)^5.
E.g.f.: (12465 - 12464*x + 7872*x^2 + 7680*x^3 + 6144*x^4)*exp(x).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
See page 7 in Brent's paper:
a(n) = (2*n-1)^2*A272131(n) - 4*(n-1)^2*A272131(n-1).
A272133(n) = (2*n-1)^2*a(n) - 4*(n-1)^2*a(n-1).
n*a(n) = 1 + 3^9*(n-1)/(n+1) + 5^9*((n-1)*(n-2))/((n+1)*(n+2)) + ... for n >= 1. See A245244. - Peter Bala, Jan 19 2019
MAPLE
[6144*n^4-29184*n^3+52416*n^2-41840*n+12465$n=0..30]; # Muniru A Asiru, Jan 28 2019
MATHEMATICA
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 *)
PROG
(Magma) [6144*n^4 - 29184*n^3 + 52416*n^2 - 41840*n + 12465: n in [0..40]];
(PARI) lista(nn) = for(n=0, nn, print1(6144*n^4-29184*n^3+52416*n^2-41840*n+12465, ", ")); \\ Altug Alkan, Apr 26 2016
CROSSREFS
Cf. A272131, A272133, A245244.
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Apr 26 2016
STATUS
approved
A272131 a(n) = 384*n^3 - 1184*n^2 + 1228*n - 427. +10
3
-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 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
LINKS
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).
FORMULA
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
MAPLE
[384*n^3-1184*n^2+1228*n-427$n=0..35]; # Muniru A Asiru, Jan 28 2019
MATHEMATICA
Table[384 n^3 - 1184 n^2 + 1228 n - 427, {n, 0, 40}]
PROG
(Magma) [384*n^3 - 1184*n^2 + 1228*n - 427: n in [0..50]];
(PARI) lista(nn) = for(n=0, nn, print1(384*n^3-1184*n^2+1228*n-427, ", ")); \\ Altug Alkan, Apr 26 2016
CROSSREFS
Cf. A272129, A272132, A245244.
KEYWORD
sign,easy
AUTHOR
Vincenzo Librandi, Apr 26 2016
STATUS
approved
A272133 a(n) = 122880*n^5 - 829440*n^4 + 2258688*n^3 - 3076288*n^2 + 2079892*n - 555731. +10
3
-555731, 1, 29525, 1657129, 16591741, 80872529, 269614501, 711754105, 1604794829, 3229552801, 5964902389, 10302521801, 16861638685, 26403775729, 39847496261, 58283149849, 82987617901, 115439059265, 157331655829, 210590358121, 277385630909, 360148198801 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,1
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16)
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
FORMULA
O.g.f.: (-555731 + 3334387*x - 8306446*x^2 + 12594614*x^3 - 1244143*x^4 + 8922919*x^5)/(1-x)^6.
E.g.f.: (-555731 + 555732*x - 263104*x^2 + 354048*x^3 + 399360*x^4 + 122880*x^5)*exp(x).
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) = (2*n-1)^2*A272132(n) - 4*(n-1)^2*A272132(n-1), see page 7 in Brent's paper.
n*a(n) = 1 + 3^11*(n-1)/(n+1) + 5^11*((n-1)*(n-2))/((n+1)*(n+2)) + ... for n >= 1. See A245244. - Peter Bala, Jan 19 2019
MAPLE
[122880*n^5-829440*n^4+2258688*n^3-3076288*n^2+2079892*n-555731$n=0..30]; # Muniru A Asiru, Jan 28 2019
MATHEMATICA
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 *)
PROG
(Magma) [122880*n^5 - 829440*n^4 + 2258688*n^3 -3076288*n^2 + 2079892*n - 555731: n in [0..30]];
(PARI) 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
CROSSREFS
Cf. A272132, A245244.
KEYWORD
sign,easy
AUTHOR
Vincenzo Librandi, Apr 26 2016
STATUS
approved
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). +10
0
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
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.
LINKS
Table of n, a(n) for n=0..54.
EXAMPLE
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.
MATHEMATICA
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 *)
CROSSREFS
Cf. A226158, A230324, A009843, A132049, A243963, A245244.
KEYWORD
sign,tabl
AUTHOR
Paul Curtz, Jul 29 2014
STATUS
approved
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