Barry E. Williams - cp's OEIS frontend

User: Barry E. Williams

Barry E. Williams has authored 175 sequences. Here are the ten most recent ones:

A056854 a(n) = Lucas(4*n).

2, 7, 47, 322, 2207, 15127, 103682, 710647, 4870847, 33385282, 228826127, 1568397607, 10749957122, 73681302247, 505019158607, 3461452808002, 23725150497407, 162614600673847, 1114577054219522, 7639424778862807, 52361396397820127, 358890350005878082, 2459871053643326447

Offset: 0

Barry E. Williams, Aug 29 2000

a(n) and b(n) := A004187(n) are the nonnegative proper and improper solutions of the Pell equation a(n)^2 - 5*(3*b(n))^2 = +4. See the cross-reference to A004187 below. - Wolfdieter Lang, Jun 26 2013

Lucas numbers of the form n^2-2. - Michel Lagneau, Aug 11 2014

			Pell equation: n = 0, 2^2 - 45*0^2 = +4 (improper);  n = 1, 7^2 - 5*(3*1)^2 = +4; n=2, 47^2 - 5*(3*7)^2 = +4. - _Wolfdieter Lang_, Jun 26 2013

R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

G. C. Greubel, Table of n, a(n) for n = 0..1190
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Dale Gerdemann, Combinatorial proofs of Zeckendorf family identities, Fib. Quart. 46/47 (2009) 249. [_N. J. A. Sloane_, Dec 05 2009]
A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
Tanya Khovanova, Recursive Sequences
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A004187, A005248.
Cf. quadrisection of A000032: this sequence (first), A056914 (second), A246453 (third, without 11), A288913 (fourth).
Cf. Lucas(k*n): A000032 (k = 1), A005248 (k = 2), A014448 (k = 3), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

[Lucas(4*n): n in [0..100]]; // Vincenzo Librandi, Apr 14 2011

a[0] = 2; a[1] = 7; a[n_] := 7a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 19}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{7,-1},{2,7},25] (* or *) LucasL[4*Range[0,25]] (* Harvey P. Dale, Aug 08 2011 *)

a(n)=if(n<0,0,polsym(1-7*x+x^2,n)[n+1])

a(n)=if(n<0,0,2*subst(poltchebi(n),x,7/2))

[lucas_number2(n,7,1) for n in range(27)] #Zerinvary Lajos, Jun 25 2008

a(n) = 7*a(n-1) - a(n-2) with a(0)=2, a(1)=7.
a(n) = A000032(4*n), where A000032 = Lucas numbers.
a(n) = 7*S(n-1, 7) - 2*S(n-2, 7) = S(n, 7) - S(n-2, 7) = 2*T(n, 7/2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U(n, x), resp. T(n, x), are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 7) = A004187(n), n>=0. See A049310 and A053120.
a(n) = ((7+sqrt(45))/2)^n + ((7-sqrt(45))/2)^n.
G.f.: (2-7x)/(1-7x+x^2).
a(n) = A005248(2*n); bisection of A005248.
a(n) = Fibonacci(8*n)/Fibonacci(4*n), n>0. - Gary Detlefs, Dec 26 2010
a(n) = 2 + 5*Fibonacci(2*n)^2 = 2 + 5*A049684(n), n >= 0. This is in Koshy's book (reference under A065563) 15. on p. 88. Compare with the above Chebyshev T formula. - Wolfdieter Lang, Aug 27 2012
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(7 - 3*sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.14242 42709 40138 85949 ... = 2 + 1/(7 + 1/(47 + 1/(322 + ...))).
Also F(-alpha) = 0.85670 72882 04563 14901 ... has the continued fraction representation 1 - 1/(7 - 1/(47 - 1/(322 - ...))) and the simple continued fraction expansion 1/(1 + 1/((7-2) + 1/(1 + 1/((47-2) + 1/(1 + 1/((322-2) + 1/(1 + ...))))))). Cf. A005248.
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((7^2-4) + 1/(1 + 1/((47^2-4) + 1/(1 + 1/((322^2-4) + 1/(1 + ...))))))).
Added Oct 13 2019: 1/2 + (1/2)*F(alpha)/F(-alpha) = 1.16675297774947414828... has the simple continued fraction expansion 1 + 1/((7 - 2) + 1/(1 + 1/((322 - 2) + 1/(1 + 1/(15127 - 2) + 1/(1 + ...))))). (End)
a(n) = Fibonacci(4*n+2) - Fibonacci(4*n-2), where Fibonacci(-2) = -1. - Bruno Berselli, May 25 2015
a(n) = sqrt(45*(A004187(n))^2+4).
From Peter Bala, Oct 13 2019: (Start)
a(n) = F(4*n+4)/F(4) - F(4*n-4)/F(4) = A004187(n+1) - A004187(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^4 = [2, 3; 3, 5].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) (mod p^k) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
5*Sum_{n >= 1} 1/(a(n) - 9/a(n)) = 1: (9 = Lucas(4)+2 and 5 = Lucas(4)-2)
9*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 5/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4)*( theta_3((7-3*sqrt(5))/2)^2 - 1 ), where theta_3(q) = 1 + 2*Sum_{n >= 1} q^n^2. Cf. A153415.
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4)*( 1 - theta_3((3*sqrt(5)-7)/2)^2 ).
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 7*x^2 + 48*x^3 + 329*x^4 + ... is the o.g.f. for A004187. (End)
E.g.f.: 2*exp(7*x/2)*cosh(3*sqrt(5)*x/2). - Stefano Spezia, Oct 18 2019
a(2k+1)/7 is the numerator of the continued fraction [3*sqrt(5), 3*sqrt(5), ..., 3*sqrt(5)] with 2k copies of 3*sqrt(5), for k>0. - Greg Dresden and Tracy Z. Wu, Sep 10 2020
a(n) = Sum_{k>=1} Lucas(2*n*k)/(Lucas(2*n)^k). - Diego Rattaggi, Jan 20 2025

More terms from James Sellers, Aug 31 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A056918 a(n) = 9*a(n-1)-a(n-2); a(0)=2, a(1)=9.

Original entry on oeis.org

2, 9, 79, 702, 6239, 55449, 492802, 4379769, 38925119, 345946302, 3074591599, 27325378089, 242853811202, 2158358922729, 19182376493359, 170483029517502, 1515164889164159, 13466000972959929, 119678843867475202

Offset: 0

Barry E. Williams, Aug 21 2000

All nonnegative integer solutions of Pell equation a(n)^2 - 77*b(n)^2 = +4 together with b(n)=A018913(n), n>=0. - Wolfdieter Lang, Aug 31 2004

Except for the first term, positive values of x (or y) satisfying x^2 - 9xy + y^2 + 77 = 0. - Colin Barker, Feb 13 2014

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
A. F. Horadam, Special Properties of the Sequence W(n){a,b; p,q}, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (9,-1).

Cf. A018913. a(n)=sqrt(77*A018913(n)^2 + 4). A005248.

a056918 n = a056918_list !! n
a056918_list = 2 : 9 :
   zipWith (-) (map (* 9) $ tail a056918_list) a056918_list
-- Reinhard Zumkeller, Jan 06 2013

a[0] = 2; a[1] = 9; a[n_] := 9a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)

[lucas_number2(n,9,1) for n in range(23)] # Zerinvary Lajos, Jun 25 2008

a(n) = 9*S(n-1, 9) - 2*S(n-2, 9) = S(n, 9) - S(n-2, 9) = 2*T(n, 9/2), with S(n, x) := U(n, x/2) (see A049310), S(-1, x) := 0, S(-2, x) := -1. S(n-1, 9)=A018913(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind.
a(n) = {9*[((9+sqrt(77))/2)^n - ((9-sqrt(77))/2)^n] - 2*[((9+sqrt(77))/2)^(n-1) - ((9-sqrt(77))/2)^(n-1)]}/sqrt(77).
G.f.: (2-9*x)/(1-9*x+x^2).
a(n) = ap^n + am^n, with ap := (9+sqrt(77))/2 and am := (9-sqrt(77))/2.
G.f.: (2-9*x)/(1-9*x+x^2). - Philippe Deléham, Nov 03 2008
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = product {n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(9 - sqrt(77)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.11095 50589 89701 91909 ... = 2 + 1/(9 + 1/(79 + 1/(702 + ...))).
Also F(-alpha) = 0.88873 23915 40314 47623 ... has the continued fraction representation 1 - 1/(9 - 1/(79 - 1/(702 - ...))) and the simple continued fraction expansion 1/(1 + 1/((9-2) + 1/(1 + 1/((79-2) + 1/(1 + 1/((702-2) + 1/(1 + ...))))))). F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((9^2-4) + 1/(1 + 1/((79^2-4) + 1/(1 + 1/((702^2-4) + 1/(1 + ...))))))). Cf. A005248.
(End)

More terms from James Sellers, Sep 07 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A056115 a(n) = n*(n+11)/2.

Original entry on oeis.org

0, 6, 13, 21, 30, 40, 51, 63, 76, 90, 105, 121, 138, 156, 175, 195, 216, 238, 261, 285, 310, 336, 363, 391, 420, 450, 481, 513, 546, 580, 615, 651, 688, 726, 765, 805, 846, 888, 931, 975, 1020, 1066, 1113, 1161, 1210, 1260, 1311, 1363, 1416, 1470, 1525

Offset: 0

Barry E. Williams, Jul 04 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A001477, A055999, A056000, A056119.

Third column of Pascal (1, 6) triangle A096956.

List([0..50], n-> n*(n+11)/2 ); # G. C. Greubel, Jan 18 2020

[n*(n+11)/2: n in [0..50]]; // G. C. Greubel, Jan 18 2020

((2*Range[0,50]+11)^2 -11^2)/8 (* G. C. Greubel, Jan 18 2020 *)

a(n)=n*(n+11)/2; \\ Joerg Arndt, Oct 25 2014

[n*(n+11)/2 for n in (0..50)] # G. C. Greubel, Jan 18 2020

G.f.: x*(6-5*x)/(1-x)^3.
a(n) = A000096(n) + 4*A001477(n) = A056000(n) + A001477(n) = A056119(n) - A001477(n). - Zerinvary Lajos, Oct 01 2006
a(n) = A126890(n,5) for n>4. - Reinhard Zumkeller, Dec 30 2006
Equals A119412/2. - Zerinvary Lajos, Feb 12 2007
If we define f(n,i,a) = Sum_{k=0..n-i} ( binomial(n,k)*stirling1(n-k,i) *Product_{j=0..k-1} (-a-j) ), then a(n) = -f(n,n-1,6), for n>=1. - Milan Janjic, Dec 20 2008
a(n) = a(n-1) + n + 5 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
Sum_{n>=1} 1/a(n) = 83711/152460. - R. J. Mathar, Jul 14 2012
a(n) = 6*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
E.g.f.: x*(12 + x)*exp(x)/2. - G. C. Greubel, Jan 18 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/11 - 20417/152460. - Amiram Eldar, Jan 10 2021

A056118 a(n) = (11n+5)(n+4)(n+3)(n+2)*(n+1)/120.

Original entry on oeis.org

1, 16, 81, 266, 686, 1512, 2982, 5412, 9207, 14872, 23023, 34398, 49868, 70448, 97308, 131784, 175389, 229824, 296989, 378994, 478170, 597080, 738530, 905580, 1101555, 1330056, 1594971, 1900486, 2251096, 2651616, 3107192, 3623312

Offset: 0

Barry E. Williams, Jul 04 2000

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A055268.

List([0..40], n-> (11*n+5)*Binomial(n+4, 4)/5 ); # G. C. Greubel, Jan 17 2020

[(11*n+5)*Binomial(n+4, 4)/5: n in [0..40]]; // G. C. Greubel, Jan 17 2020

seq( (11*n+5)*binomial(n+4, 4)/5, n=0..40); # G. C. Greubel, Jan 17 2020

Table[((11n+5)Times@@(n+Range[4]))/120,{n,0,40}] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1}, {1,16,81,266,686,1512}, 40] (* Harvey P. Dale, Oct 18 2013 *)
Table[11*Binomial[n+5,5] -8*Binomial[n+4,4], {n,0,40}] (* G. C. Greubel, Jan 17 2020 *)

vector(41, n, (11*n-6)*binomial(n+3,4)/5 ) \\ G. C. Greubel, Jan 17 2020

[(11*n+5)*binomial(n+4, 4)/5 for n in (0..40)] # G. C. Greubel, Jan 17 2020

a(n) = (11*n+5)*binomial(n+4,4)/5.
G.f.: (1+10*x)/(1-x)^6.
a(0)=1, a(1)=16, a(2)=81, a(3)=266, a(4)=686, a(5)=1512; for n>5, 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). - Harvey P. Dale, Oct 18 2013
From G. C. Greubel, Jan 17 2020: (Start)
a(n) = 11*binomial(n+5,5) - 8*binomial(n+4,4).
E.g.f.: (360 +2760*x +3720*x^2 +1560*x^3 +235*x^4 +11*x^5)*exp(x)/120. (End)

A056120 a(n) = (3^3)*4^(n-3) with a(0)=1, a(1)=1 and a(2)=7.

Original entry on oeis.org

1, 1, 7, 27, 108, 432, 1728, 6912, 27648, 110592, 442368, 1769472, 7077888, 28311552, 113246208, 452984832, 1811939328, 7247757312, 28991029248, 115964116992, 463856467968, 1855425871872

Offset: 0

Barry E. Williams, Jul 05 2000

For n>=3, a(n) is equal to the number of functions f:{1,2,...,n}->{1,2,3,4} such that for fixed, different x_1, x_2, x_3 in {1,2,...,n} and fixed y_1, y_2, y_3 in {1,2,3,4} we have f(x_i)<>y_i, (i=1,2,...,n). - Milan Janjic, May 13 2007

G. C. Greubel, Table of n, a(n) for n = 0..1000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (4).

Cf. A055841.

First differences of A002063.

Concatenation([1,1,7], List([3..25], n-> 27*4^(n-3) )); # G. C. Greubel, Jan 18 2020

[1,1,7] cat [27*4^(n-3): n in [3..25]]; // G. C. Greubel, Jan 18 2020

1,1,7, seq( 27*4^(n-3), n=3..25); # G. C. Greubel, Jan 18 2020

Table[If[n<2, 1, If[n==2, 7, 27*4^(n-3)]], {n,0,25}] (* G. C. Greubel, Jan 18 2020 *)

vector(26, n, if(n<2, 1, if(n==2, 7, 27*4^(n-3))) ) \\ G. C. Greubel, Jan 18 2020

[1,1,7]+[27*4^(n-3) for n in (3..25)] # G. C. Greubel, Jan 18 2020

a(n) = 4*a(n-1) + (-1)^n*binomial(3, 3-n).
G.f.: (1-x)^3/(1-4*x).
E.g.f.: (37 - 44*x + 8*x^2 + 27*exp(4*x))/64. - G. C. Greubel, Jan 18 2020

a(21) corrected by R. J. Mathar, Dec 03 2014

A056122 a(n) = (8n+9)C(n+8,8)/9.

Original entry on oeis.org

1, 17, 125, 605, 2255, 7007, 19019, 46475, 104390, 218790, 432718, 814606, 1469650, 2554930, 4299130, 7027834, 11195503, 17425375, 26558675, 39714675, 58363305, 84412185, 120310125, 169169325, 234908700, 322420956, 437766252

Offset: 0

Barry E. Williams, Jul 06 2000

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

Cf. A056001.
Cf. A093565 ((8, 1) Pascal, column m=9). Partial sums of A056001.
Cf. similar sequences listed in A254142.

List([0..40], n-> (8*n+9)*Binomial(n+8,8)/9); # G. C. Greubel, Aug 29 2019

[(8*n+9)*Binomial(n+8,8)/9: n in [0..40]]; // G. C. Greubel, Aug 29 2019

seq((8*n+9)*binomial(n+8,8)/9, n=0..40); # G. C. Greubel, Aug 29 2019

Table[(8n+9) Binomial[n+8,8]/9,{n,0,40}]  (* Harvey P. Dale, Mar 09 2011 *)

vector(40, n, (8*n+1)*binomial(n+7,8)/9) \\ G. C. Greubel, Aug 29 2019

[(8*n+9)*binomial(n+8,8)/9 for n in (0..40)] # G. C. Greubel, Aug 29 2019

G.f.: (1+7*x)/(1-x)^10.
a(n) = (362880 + 1308816*n + 1939788*n^2 + 1550548*n^3 + 740313*n^4 + 220416*n^5 + 41202*n^6 + 4692*n^7 + 297*n^8 + 8*n^9)/362880. - Harvey P. Dale, Mar 09 2011
E.g.f.: (362880 +5806080*x +16692480*x^2 +16934400*x^3 +7832160*x^4 + 1862784*x^5 +239904 x^6 +16704*x^7 +585*x^8 +8*x^9)*exp(x)/362880. - G. C. Greubel, Aug 29 2019

A056123 a(n) = 3*a(n-1) - a(n-2) with a(0)=1, a(1)=11.

Original entry on oeis.org

1, 11, 32, 85, 223, 584, 1529, 4003, 10480, 27437, 71831, 188056, 492337, 1288955, 3374528, 8834629, 23129359, 60553448, 158530985, 415039507, 1086587536, 2844723101, 7447581767, 19498022200, 51046484833, 133641432299

Offset: 0

Barry E. Williams, Jul 06 2000

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A000032, A000045, A055850.

List([0..30], n-> Fibonacci(2*n+2) +8*Fibonacci(2*n) ); # G. C. Greubel, Jan 17 2020

[Fibonacci(2*n+2) +8*Fibonacci(2*n): n in [0..30]]; // G. C. Greubel, Jan 17 2020

with(combinat); seq( fiboacci(2*n+2) +8*fibonacci(2*n), n=0..30); # G. C. Greubel, Jan 17 2020

Table[Fibonacci[2*n+2] +8*Fibonacci[2*n], {n,0,30}] (* G. C. Greubel, Jan 17 2020 *)

vector(31, n, fibonacci(2*n) +8*fibonacci(2*n-2) ) \\ G. C. Greubel, Jan 17 2020

[fibonacci(2*n+2) +8*fibonacci(2*n) for n in (0..30)] # G. C. Greubel, Jan 17 2020

a(n) = {11*[((3+sqrt(5))/2)^n - ((3-sqrt(5))/2)^n] - [((3+sqrt(5))/2)^(n-1) - ((3-sqrt(5))/2)^(n-1)]}/sqrt(5).
G.f.: (1+8*x)/(1-3*x+x^2).
a(n) = 6*Lucas(2n+1) - Fibonacci(2n+5).
From G. C. Greubel, Jan 17 2020: (Start)
a(n) = Fibonacci(2*n+2) + 8*Fibonacci(2*n).
E.g.f.: exp(3*t/2)*( cosh(sqrt(5)*t/2) + (19/sqrt(5))*sinh(sqrt(5)*t/2) ). (End)

A056124 a(n) = 3*a(n-1) - a(n-2) + 8 with a(0)=1, a(1)=11.

Original entry on oeis.org

1, 11, 40, 117, 319, 848, 2233, 5859, 15352, 40205, 105271, 275616, 721585, 1889147, 4945864, 12948453, 33899503, 88750064, 232350697, 608302035, 1592555416, 4169364221, 10915537255, 28577247552, 74816205409

Offset: 0

Barry E. Williams, Jul 07 2000

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000032, A000045, A055850 (first differences).

List([0..30], n-> Fibonacci(2*n+2) + 8*Fibonacci(2*n+1) - 8 ); # G. C. Greubel, Jan 19 2020

[Fibonacci(2*n+2) + 8*Fibonacci(2*n+1) - 8: n in [0..30]]; // G. C. Greubel, Jan 19 2020

with(combinat); seq( fibonacci(2*n+2) + 8*fibonacci(2*n+1) - 8, n=0..30); # G. C. Greubel, Jan 19 2020

LinearRecurrence[{4,-4,1},{1,11,40},30] (* Harvey P. Dale, Mar 25 2015 *)

vector(31, n, fibonacci(2*n) +8*fibonacci(2*n-1) -8 ) \\ G. C. Greubel, Jan 19 2020

[fibonacci(2*n+2) + 8*fibonacci(2*n+1) - 8 for n in (0..30)] # G. C. Greubel, Jan 19 2020

a(n) = ( 19*(((3+sqrt(5))/2)^n - ((3-sqrt(5))/2)^n) - 9*(((3+sqrt(5))/2)^(n-1) - ((3-sqrt(5))/2)^(n-1)) )/sqrt(5) - 8.
G.f.: (1+7*x)/((1-x)*(1-3*x+x^2)).
a(n) = Fibonacci(2*n+5) + 2*Lucas(2*n) - 8.
From G. C. Greubel, Jan 19 2020: (Start)
a(n) = Fibonacci(2*n+2) + 8*Fibonacci(2*n+1) - 8.
E.g.f.: exp(3*x/2)*( 9*cosh(sqrt(5)*x/2) - (11/sqrt(5))*sinh(sqrt(5)*x/2) ) - 8*exp(x). (End)

A056125 a(n) = (5n + 4)binomial(n+7,7)/4.

Original entry on oeis.org

1, 18, 126, 570, 1980, 5742, 14586, 33462, 70785, 140140, 262548, 469404, 806208, 1337220, 2151180, 3368244, 5148297, 7700814, 11296450, 16280550, 23088780, 32265090, 44482230, 60565050, 81516825, 108548856, 143113608, 186941656

Offset: 0

Barry E. Williams, Jul 07 2000

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A052254.
Cf. A093645 ((10, 1) Pascal, column m=8).
Partial sums of A052254.

List([0..30], n-> (5*n+4)*Binomial(n+7,7)/4 ); # G. C. Greubel, Jan 19 2020

[(5*n+4)*Binomial(n+7,7)/4: n in [0..30]]; // G. C. Greubel, Jan 19 2020

seq( (5*n+4)*binomial(n+7,7)/4, n=0..30); # G. C. Greubel, Jan 19 2020

Table[((5n+4)Binomial[n+7,7])/4,{n,0,30}] (* or *) LinearRecurrence[{9,-36,84, -126,126,-84,36,-9,1},{1,18,126,570,1980,5742,14586,33462,70785},30] (* Harvey P. Dale, Jan 18 2013 *)

vector(31, n, (5*n-1)*binomial(n+6,7)/4 ) \\ G. C. Greubel, Jan 19 2020

[(5*n+4)*binomial(n+7,7)/4 for n in (0..30)] # G. C. Greubel, Jan 19 2020

G.f.: (1+9*x)/(1-x)^9.
a(0)=1, a(1)=18, a(2)=126, a(3)=570, a(4)=1980, a(5)=5742, a(6)=14586, a(7)=33462, a(8)=70785, a(n) = 9*a(n-1) -36*a(n-2) +84*a(n-3) -126*a(n-4) + 126*a(n-5) -84*a(n-6) +36*a(n-7) -9*a(n-8) +a(n-9). - Harvey P. Dale, Jan 18 2013
From G. C. Greubel, Jan 19 2020: (Start)
a(n) = 10*binomial(n+8,8) - 9*binomial(n+7,7).
E.g.f.: (20160 + 342720*x + 917280*x^2 + 823200*x^3 + 323400*x^4 + 62328*x^5 + 6076*x^6 + 284*x^7 + 5*x^8)*exp(x)/20160. (End)

A056126 a(n) = n*(n + 17)/2.

Original entry on oeis.org

0, 9, 19, 30, 42, 55, 69, 84, 100, 117, 135, 154, 174, 195, 217, 240, 264, 289, 315, 342, 370, 399, 429, 460, 492, 525, 559, 594, 630, 667, 705, 744, 784, 825, 867, 910, 954, 999, 1045, 1092, 1140, 1189, 1239, 1290, 1342, 1395, 1449, 1504, 1560, 1617, 1675

Offset: 0

Barry E. Williams, Jul 07 2000

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A001477, A051942, A056000, A056121.

Cf. A001008, A002805, A098849, A120071, A132760, A132761, A132765.

List([0..50], n-> n*(n+17)/2 ); # G. C. Greubel, Jan 19 2020

[n*(n+17)/2: n in [0..50]]; // G. C. Greubel, Jan 19 2020

seq( n*(n+17)/2, n=0..50); # G. C. Greubel, Jan 19 2020

Table[n(n+17)/2,{n,0,50}] (* Harvey P. Dale, Apr 25 2011 *)

a(n)=n*(n+17)/2 \\ Charles R Greathouse IV, Sep 24 2015

[n*(n+17)/2 for n in (0..50)] # G. C. Greubel, Jan 19 2020

G.f.: x*(9-8*x)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A126890(n,8) for n>7. - Reinhard Zumkeller, Dec 30 2006
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*stirling1(n-k,i)* Product_{j=0..k-1} (-a-j), then a(n) = -f(n,n-1,9), for n>=1. - Milan Janjic, Dec 20 2008
a(n) = a(n-1) + n + 8 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
a(n) = 9*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
E.g.f.: x*(18 + x)*exp(x)/2. - G. C. Greubel, Jan 19 2020
From Amiram Eldar, Jan 10 2021: (Start)
Sum_{n>=1} 1/a(n) = 2*A001008(17)/(17*A002805(17)) = 42142223/104144040.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/17 - 1768477/20828808. (End)

cp's OEIS Frontend

User: Barry E. Williams

A056854 a(n) = Lucas(4*n).

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A056918 a(n) = 9*a(n-1)-a(n-2); a(0)=2, a(1)=9.

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A056115 a(n) = n*(n+11)/2.

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A056118 a(n) = (11*n+5)*(n+4)*(n+3)*(n+2)*(n+1)/120.

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A056120 a(n) = (3^3)*4^(n-3) with a(0)=1, a(1)=1 and a(2)=7.

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A056122 a(n) = (8*n+9)*C(n+8,8)/9.

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A056118 a(n) = (11n+5)(n+4)(n+3)(n+2)*(n+1)/120.

A056122 a(n) = (8n+9)C(n+8,8)/9.

A056125 a(n) = (5n + 4)binomial(n+7,7)/4.