A007613 -id:A007613 - cp's OEIS frontend

A070775 a(n) = Sum_{k=0..n} binomial(4n,4k).

1, 2, 72, 992, 16512, 261632, 4196352, 67100672, 1073774592, 17179738112, 274878431232, 4398044413952, 70368752566272, 1125899873288192, 18014398643699712, 288230375614840832, 4611686020574871552, 73786976286248271872, 1180591620751771041792, 18889465931341141901312

Offset: 0

Sebastian Gutierrez and Sarah Kolitz (skolitz(AT)mit.edu), May 15 2002

Also the cogrowth sequence of the 16-element group C4 X C4 = . - Sean A. Irvine, Nov 09 2024

Seiichi Manyama, Table of n, a(n) for n = 0..830
Index entries for linear recurrences with constant coefficients, signature (12,64).

Sum_{k=0..n} binomial(b*n,b*k): A000079 (b=1), A081294 (b=2), A007613 (b=3), this sequence (b=4), A070782 (b=5), A070967 (b=6), A094211 (b=7), A070832 (b=8), A094213 (b=9), A070833 (b=10).

a := n -> if n = 0 then 1 else 4^(n - 1)*(2*(-1)^n + 4^n) fi:
seq(a(n), n = 0..19); # Peter Luschny, Jul 02 2022

Table[Sum[Binomial[4n,4k],{k,0,n}],{n,0,30}] (* or *) Join[{1}, LinearRecurrence[{12,64},{2,72},30]] (* Harvey P. Dale, Apr 24 2011 *)

PARI
```
a(n)=sum(k=0,n,binomial(4*n,4*k))
```

N=66; x='x+O('x^N); Vec((1-10*x-16*x^2)/((1-16*x)*(1+4*x))) \\ Seiichi Manyama, Mar 15 2019

a(n) = (1/2)*(-4)^n + (1/4)*16^n for n > 0.
Let b(n) = a(n) - 2^(4n)/4 then b(n+1) = 4*b(n) - Benoit Cloitre, May 27 2004
G.f.: (1 - 10*x - 16*x^2)/((1-16*x)*(1+4*x)). - Seiichi Manyama, Mar 15 2019
G.f.: ((cos(x) + cosh(x))/2)^2 = Sum_{n >= 0} a(n)*x(4*n)/(4*n)!. - Peter Bala, Jun 20 2022

A102518 a(n) = Sum_{k=0..n} binomial(n, k) * Sum_{j=0..k} binomial(3k, 3j).

Original entry on oeis.org

1, 3, 27, 243, 2187, 19683, 177147, 1594323, 14348907, 129140163, 1162261467, 10460353203, 94143178827, 847288609443, 7625597484987, 68630377364883, 617673396283947, 5559060566555523, 50031545098999707, 450283905890997363, 4052555153018976267, 36472996377170786403

Offset: 0

Paul Barry, Jan 13 2005

Binomial transform of A007613.
a(n+1) is the smallest number with a reciprocal with repeating decimal of period a(n). - Matthew Goers, Nov 09 2017
a(n) is the number of walks of 2n steps on the utility graph that start and end at the same vertex (excursions). A001019(n) is the number of 2n+1-step walks on the utility graph that end at one of the 3 adjacent vertices. A013708(n) is the number of 2n+2-step walks that end at one of the 2 remote vertices (at distance 2). The number of n-step walks on the utility (3-regular) graph, summed over all 3 types of final vertices, is 3^n. - R. J. Mathar, Nov 03 2020

Colin Barker, Table of n, a(n) for n = 0..1000
R. J. Mathar, Counting Walks on Finite Graphs, Section 4.
Index entries for linear recurrences with constant coefficients, signature (9).

Cf. A001019, A007613, A013708.

Join[{1},NestList[9#&,3,20]] (* Harvey P. Dale, Feb 03 2021 *)

Vec((1-6*x)/(1-9*x) + O(x^30)) \\ Colin Barker, Mar 17 2016

a(n) = 3^(2*n-1) + 2*0^k/3; a(n+1) = A013708(n).
G.f.: (1-6*x) / (1-9*x). - Colin Barker, Mar 17 2016
E.g.f.: (exp(9*x) + 2)/3. - Stefano Spezia, Jul 09 2024

A070782 a(n) = Sum_{k=0..n} binomial(5n,5k).

Original entry on oeis.org

1, 2, 254, 6008, 215766, 6643782, 215492564, 6863694378, 219993856006, 7035859329512, 225191238869774, 7205634556190798, 230585685502492596, 7378682274243863442, 236118494435702913134, 7555784484021765207768, 241785184867484394069286, 7737125013254912900576822

Offset: 0

Sebastian Gutierrez and Sarah Kolitz (skolitz(AT)mit.edu), May 15 2002

Seiichi Manyama, Table of n, a(n) for n = 0..664
Index entries for linear recurrences with constant coefficients, signature (21,353,-32).

Sum_{k=0..n} binomial(b*n,b*k): A000079 (b=1), A081294 (b=2), A007613 (b=3), A070775 (b=4), this sequence (b=5), A070967 (b=6), A094211 (b=7), A070832 (b=8), A094213 (b=9), A070833 (b=10).

LinearRecurrence[{21,353,-32},{1,2,254},20] (* Harvey P. Dale, Jun 18 2023 *)

PARI
```
a(n)=sum(k=0,n,binomial(5*n,5*k))
```

Vec((1 - 19*x - 141*x^2) / ((1 - 32*x)*(1 + 11*x - x^2)) + O(x^20)) \\ Colin Barker, May 27 2019

a(n) = (1/5)*32^n + (2/5)*(-11/2 + (5/2)*sqrt(5))^n + (2/5)*(-11/2 - (5/2)*sqrt(5))^n.
Let b(n) = a(n) - 2^(5n)/5; then b(n) + 11*b(n-1) - b(n-2) = 0. - Benoit Cloitre, May 27 2004
From Colin Barker, May 27 2019: (Start)
G.f.: (1 - 19*x - 141*x^2) / ((1 - 32*x)*(1 + 11*x - x^2)).
a(n) = 21*a(n-1) + 353*a(n-2) - 32*a(n-3) for n>2.
(End)

A082311 A Jacobsthal sequence trisection.

Original entry on oeis.org

1, 5, 43, 341, 2731, 21845, 174763, 1398101, 11184811, 89478485, 715827883, 5726623061, 45812984491, 366503875925, 2932031007403, 23456248059221, 187649984473771, 1501199875790165, 12009599006321323, 96076792050570581, 768614336404564651, 6148914691236517205

Offset: 0

Paul Barry, Apr 09 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,8).

Cf. A001045, A007613, A015565, A024494, A070366, A081374, A082365, A132805.

[2*8^n/3+(-1)^n/3 : n in [0..30]]; // Vincenzo Librandi, Aug 13 2011

f[n_] := (2*8^n + (-1)^n)/3; Array[f, 25, 0] (* Robert G. Wilson v, Aug 13 2011 *)

x='x+O('x^30); Vec((1-2*x)/((1+x)*(1-8*x))) \\ G. C. Greubel, Sep 16 2018

a(n) = (2*8^n + (-1)^n)/3 = A001045(3*n+1).
From R. J. Mathar, Feb 23 2009: (Start)
a(n) = 7*a(n-1) + 8*a(n-2).
G.f.: (1-2*x)/((1+x)*(1-8*x)). (End)
a(n) = A024494(3*n+1). a(n) = 8*a(n-1) + 3*(-1)^n. Sum of digits = A070366. - Paul Curtz, Nov 20 2007
a(n)= A007613(n) + A132805(n) = A081374(1+3*n). - Paul Curtz, Jun 06 2011
E.g.f.: (cosh(x) + 2*cosh(8*x) - sinh(x) + 2*sinh(8*x))/3. - Stefano Spezia, Jul 15 2024

A070832 a(n) = Sum_{k=0..n} binomial(8n,8k).

Original entry on oeis.org

1, 2, 12872, 1470944, 622116992, 125858012672, 36758056208384, 8793364151263232, 2334899414608412672, 586347560750962049024, 151652224498623981289472, 38612725801339748322639872, 9913426188311626771400228864

Offset: 0

Sebastian Gutierrez and Sarah Kolitz (skolitz(AT)mit.edu), May 15 2002

Seiichi Manyama, Table of n, a(n) for n = 0..415
Index entries for linear recurrences with constant coefficients, signature (136,32880,-552704,-65536).

Sum_{k=0..n} binomial(b*n,b*k): A000079 (b=1), A081294 (b=2), A007613 (b=3), A070775 (b=4), A070782 (b=5), A070967 (b=6), A094211 (b=7), this sequence (b=8), A094213 (b=9), A070833 (b=10).

Table[Sum[Binomial[8n,8k],{k,0,n}],{n,0,15}] (* Harvey P. Dale, Nov 25 2020 *)

a(n)=sum(k=0,n,binomial(8*n,8*k)); \\ Benoit Cloitre, May 27 2004

Vec((1 - 134*x - 20280*x^2 + 207296*x^3 + 8192*x^4) / ((1 - 16*x)*(1 - 256*x)*(1 + 136*x + 16*x^2)) + O(x^15)) \\ Colin Barker, May 27 2019

Let b(n) = a(n)-2^(8*n)/8 then b(n)+120*b(n-1)-2160*b(n-2)-256*b(n-3)=0. - Benoit Cloitre, May 27 2004
a(n) = 1/4*16^n + 1/8*256^n + 1/4*(-68 + 48*sqrt(2))^n + 1/4*(-68-48*sqrt(2))^n.
From Colin Barker, May 27 2019: (Start)
G.f.: (1 - 134*x - 20280*x^2 + 207296*x^3 + 8192*x^4) / ((1 - 16*x)*(1 - 256*x)*(1 + 136*x + 16*x^2)).
a(n) = 21*a(n-1) + 353*a(n-2) - 32*a(n-3) for n>4.
(End)

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 15 2007

A070967 a(n) = Sum_{k=0..n} binomial(6n,6k).

Original entry on oeis.org

1, 2, 926, 37130, 2973350, 174174002, 11582386286, 729520967450, 47006639297270, 2999857885752002, 192222214478506046, 12295976362284182570, 787111112023373201990, 50370558298891875954002, 3223838658635388303336206, 206322355109994528871954490

Offset: 0

Sebastian Gutierrez and Sarah Kolitz (skolitz(AT)mit.edu), May 16 2002

Matthijs Coster, Supercongruences, Thesis, Jun 08, 1988.

Seiichi Manyama, Table of n, a(n) for n = 0..554
Index entries for linear recurrences with constant coefficients, signature (38,1691,-1728).

Sum_{k=0..n} binomial(b*n,b*k): A000079 (b=1), A081294 (b=2), A007613 (b=3), A070775 (b=4), A070782 (b=5), this sequence (b=6), A094211 (b=7), A070832 (b=8), A094213 (b=9), A070833 (b=10).

Table[Sum[Binomial[6n,6k],{k,0,n}],{n,0,20}] (* or *) LinearRecurrence[ {38,1691,-1728},{1,2,926,37130},30] (* Harvey P. Dale, Jun 19 2021 *)

PARI
```
a(n)=sum(k=0,n,binomial(6*n,6*k))
```

a(n)=if(n<0,0,(2*(-27)^n+2+64^n+0^n)/6)

a(n)=if(n<0,0,polsym(x*(x-64)*(x+27)^2*(x-1)^2,n)[n+1]/6)

G.f.: (1-36x-841x^2+288x^3)/((1-x)*(1+27x)*(1-64x)).
a(n) = ((-27)^n + 1)/3 + (64^n + 0^n)/6.
Let b(n) = a(n)-2^(6n)/6 then b(n)+26*b(n-1)-27*b(n-2) = 0. - Benoit Cloitre, May 27 2004

A070833 a(n) = Sum_{k=0..n} binomial(10n,10k).

Original entry on oeis.org

1, 2, 184758, 60090032, 139541849878, 94278969044262, 126648421364527548, 111019250117021378442, 125257104438594491956518, 121088185204450642433930072, 128442558588779813655233443038, 128767440665677943753184267342902

Offset: 0

Sebastian Gutierrez and Sarah Kolitz (skolitz(AT)mit.edu), May 15 2002

Seiichi Manyama, Table of n, a(n) for n = 0..332
Index entries for linear recurrences with constant coefficients, signature (522,587797,-75135226,-392963125,3200000).

Sum_{k=0..n} binomial(b*n,b*k): A000079 (b=1), A081294 (b=2), A007613 (b=3), A070775 (b=4), A070782 (b=5), A070967 (b=6), A094211 (b=7), A070832 (b=8), A094213 (b=9), this sequence (b=10). Cf. A000032.

a(n) = sum(k=0, n, binomial(10*n, 10*k)); \\ Michel Marcus, Mar 15 2019

Vec((1 - 520*x - 404083*x^2 + 37605988*x^3 + 117888625*x^4 - 320000*x^5) / ((1 - 1024*x)*(1 - 123*x + x^2)*(1 + 625*x + 3125*x^2)) + O(x^15)) \\ Colin Barker, Mar 15 2019

a(n) = 1/10*1024^n+1/5*(-625/2+275/2*sqrt(5))^n+1/5*(-625/2-275/2*sqrt(5))^n+1/5*(123/2+55/2*sqrt(5))^n+1/5*(123/2-55/2*sqrt(5))^n.
G.f.: (1 - 520*x - 404083*x^2 + 37605988*x^3 + 117888625*x^4 - 320000*x^5) / ((1 - 1024*x)*(1 - 123*x + x^2)*(1 + 625*x + 3125*x^2)). - Colin Barker, Mar 15 2019
a(2*n) = (2^(20*n-1) + Lucas(20*n) + 5^(5*n)*Lucas(10*n))/5, for n>0 and for Lucas(n) = A000032(n). - Greg Dresden, Feb 04 2023

A094211 a(n) = Sum_{k=0..n} binomial(7n,7k).

Original entry on oeis.org

1, 2, 3434, 232562, 42484682, 4653367842, 644032289258, 79450506979090, 10353832741654602, 1313930226050847938, 168883831255263816554, 21573903987107973878962, 2764126124873404346104778, 353643666623193292098680930, 45276535087893983968685884906

Offset: 0

Benoit Cloitre, May 27 2004

Seiichi Manyama, Table of n, a(n) for n = 0..474
Index entries for linear recurrences with constant coefficients, signature (71,7585,-36991,-128).

Sum_{k=0..n} binomial(b*n,b*k): A000079 (b=1), A081294 (b=2), A007613 (b=3), A070775 (b=4), A070782 (b=5), A070967 (b=6), this sequence (b=7), A070832 (b=8), A094213 (b=9), A070833 (b=10).

A094211:=n->add(binomial(7*n,7*k), k=0..n): seq(A094211(n), n=0..20); # Wesley Ivan Hurt, Feb 16 2017

Table[Sum[Binomial[7n,7k],{k,0,n}],{n,0,20}] (* or *) LinearRecurrence[ {71,7585,-36991,-128},{1,2,3434,232562},20] (* Harvey P. Dale, May 06 2012 *)

PARI
```
a(n)=sum(k=0,n,binomial(7*n,7*k))
```

Let b(n) = a(n) - 2^(7n)/7; then b(n) + 57*b(n-1) - 289*b(n-2) - b(n-3) = 0.
a(n) = 71*a(n-1) + 7585*a(n-2) - 36991*a(n-3) - 128*a(n-4); a(0)=1, a(1)=2, a(2)=3434, a(3)=232562. - Harvey P. Dale, May 06 2012
G.f.: (1 - 69*x - 4293*x^2 + 10569*x^3) / ((1 - 128*x)*(1 + 57*x - 289*x^2 - x^3)). - Colin Barker, May 27 2019
a(n) = (1 + 2*(s*(3 - 4*s^2))^(7*n) + 2*(-1)^n*((1 - 2*s^2)^(7*n) + s^(7*n))) * 2^(7*n)/7, where s = sin(Pi/14). - Vaclav Kotesovec, Apr 17 2023

A094213 a(n) = Sum_{k=0..n} binomial(9n,9k).

Original entry on oeis.org

1, 2, 48622, 9373652, 9263421862, 3433541316152, 2140802758677844, 984101481334553024, 536617781178725122150, 265166261617029717011822, 138567978655457801631498052, 70126939586658252408697345838, 36144812798331420987905742371116

Offset: 0

Benoit Cloitre, May 27 2004

Seiichi Manyama, Table of n, a(n) for n = 0..369
Index entries for linear recurrences with constant coefficients, signature (265,139823,-6826204,-6965249,512).

Sum_{k=0..n} binomial(b*n,b*k): A000079 (b=1), A081294 (b=2), A007613 (b=3), A070775 (b=4), A070782 (b=5), A070967 (b=6), A094211 (b=7), A070832 (b=8), this sequence (b=9), A070833 (b=10).

Table[Sum[Binomial[9n,9k],{k,0,n}],{n,0,15}] (* Harvey P. Dale, Jul 14 2019 *)

PARI
```
a(n)=sum(k=0,n,binomial(9*n,9*k))
```

Vec((1 - 263*x - 91731*x^2 + 3035380*x^3 + 1547833*x^4) / ((1 + x)*(1 - 512*x)*(1 + 246*x - 13605*x^2 + x^3)) + O(x^15)) \\ Colin Barker, May 27 2019

Let b(n) = a(n)-2^(9*n)/9 then b(n)+246*b(n-1)-13605*b(n-2)+b(n-3)+(-1)^n*3078=0.
Conjectures from Colin Barker, May 27 2019: (Start)
G.f.: (1 - 263*x - 91731*x^2 + 3035380*x^3 + 1547833*x^4) / ((1 + x)*(1 - 512*x)*(1 + 246*x - 13605*x^2 + x^3)).
a(n) = 265*a(n-1) + 139823*a(n-2) - 6826204*a(n-3) - 6965249*a(n-4) + 512*a(n-5) for n>4. (End)
a(n) ~ (1/9)*exp(n*9*log(2)) (conjecture). - Bill McEachen, Aug 11 2025

A216316 G.f.: 1/( (1-8x)(1+x)^2 )^(1/3).

Original entry on oeis.org

1, 2, 13, 80, 538, 3740, 26650, 193160, 1417945, 10511450, 78533629, 590485208, 4463274232, 33886781840, 258260802232, 1974759985952, 15143163422794, 116417053435316, 896996316176290, 6925241271855296, 53562550587963052, 414948608904171464, 3219356873886333676

Offset: 0

Paul D. Hanna, Sep 03 2012

			G.f.: A(x) = 1 + 2*x + 13*x^2 + 80*x^3 + 538*x^4 + 3740*x^5 + 26650*x^6 +...
where 1/A(x)^3 = 1 - 6*x - 15*x^2 - 8*x^3.
The logarithm of the g.f. begins:
log(A(x)) = 2*x + 22*x^2/2 + 170*x^3/3 + 1366*x^4/4 + 10922*x^5/5 + 87382*x^6/6 +...+ A007613(n)*x^n/n +...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A004987, A007613, A216317, A216357, A216358.

CoefficientList[Series[1/((1-8*x)*(1+x)^2)^(1/3), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)

{a(n)=polcoeff(1/( (1-8*x)*(1+x)^2 +x*O(x^n) )^(1/3),n)}

{a(n)=local(A=1+x); A=exp(sum(m=1, n+1, sum(j=0, m, binomial(3*m, 3*j))*x^m/m +x*O(x^n))); polcoeff(A, n)}
for(n=0, 31, print1(a(n), ", "))

G.f.: exp( Sum_{n>=1} A007613(n)*x^n/n ), where A007613(n) = Sum_{k=0..n} binomial(3*n,3*k).
Recurrence: n*a(n) = (7*n-5)*a(n-1) + 8*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ Gamma(2/3)*2^(3*n+1)/(3^(5/6)*Pi*n^(2/3)). - Vaclav Kotesovec, Oct 20 2012
Inverse binomial transform of A004987. - Peter Bala, Jul 02 2023

cp's OEIS Frontend

A070775 a(n) = Sum_{k=0..n} binomial(4*n,4*k).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

A102518 a(n) = Sum_{k=0..n} binomial(n, k) * Sum_{j=0..k} binomial(3k, 3j).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A070782 a(n) = Sum_{k=0..n} binomial(5*n,5*k).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

A082311 A Jacobsthal sequence trisection.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A070832 a(n) = Sum_{k=0..n} binomial(8*n,8*k).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Formula

Extensions

A070967 a(n) = Sum_{k=0..n} binomial(6*n,6*k).

Views

Author

Keywords

References

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

Formula

A070833 a(n) = Sum_{k=0..n} binomial(10*n,10*k).

Views

Author

Keywords

Links

Crossrefs

Programs

PARI

A070775 a(n) = Sum_{k=0..n} binomial(4n,4k).

A070782 a(n) = Sum_{k=0..n} binomial(5n,5k).

A070832 a(n) = Sum_{k=0..n} binomial(8n,8k).

A070967 a(n) = Sum_{k=0..n} binomial(6n,6k).

A070833 a(n) = Sum_{k=0..n} binomial(10n,10k).

A094211 a(n) = Sum_{k=0..n} binomial(7n,7k).

A094213 a(n) = Sum_{k=0..n} binomial(9n,9k).

A216316 G.f.: 1/( (1-8x)(1+x)^2 )^(1/3).