A013777 -id:A013777 - cp's OEIS frontend

A262977 a(n) = binomial(4*n-1,n).

1, 3, 21, 165, 1365, 11628, 100947, 888030, 7888725, 70607460, 635745396, 5752004349, 52251400851, 476260169700, 4353548972850, 39895566894540, 366395202809685, 3371363686069236, 31074067324187580, 286845713747883300, 2651487106659130740, 24539426037817994160

Offset: 0

Vladimir Kruchinin, Oct 06 2015

From Gus Wiseman, Sep 28 2022: (Start)
Also the number of integer compositions of 4n with alternating sum 2n, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. These compositions are ranked by A348614. The a(12) = 21 compositions are:
  (6,2)  (1,2,5)  (1,1,5,1)  (1,1,1,1,4)
         (2,2,4)  (2,1,4,1)  (1,1,2,1,3)
         (3,2,3)  (3,1,3,1)  (1,1,3,1,2)
         (4,2,2)  (4,1,2,1)  (1,1,4,1,1)
         (5,2,1)  (5,1,1,1)  (2,1,1,1,3)
                             (2,1,2,1,2)
                             (2,1,3,1,1)
                             (3,1,1,1,2)
                             (3,1,2,1,1)
                             (4,1,1,1,1)
The following pertain to this interpretation:
- The case of partitions is A000712, reverse A006330.
- Allowing any alternating sum gives A013777 (compositions of 4n).
- A011782 counts compositions of n.
- A034871 counts compositions of 2n with alternating sum 2k.
- A097805 counts compositions by alternating (or reverse-alternating) sum.
- A103919 counts partitions by sum and alternating sum (reverse: A344612).
- A345197 counts compositions by length and alternating sum.
(End)

V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.

Cf. A006632, A002293, A004331, A005810, A052203, A224274, A257633.
Cf. A000346, A000984, A001700, A002458, A025047, A058622, A081294, A294175.
Cf. A088218, A163455, A165817.

[Binomial(4*n-1,n): n in [0..20]]; // Vincenzo Librandi, Oct 06 2015

Table[Binomial[4 n - 1, n], {n, 0, 40}] (* Vincenzo Librandi, Oct 06 2015 *)

B(x):=sum(binomial(4*n-1,n-1)*3/(4*n-1)*x^n,n,1,30);
taylor(x*diff(B(x),x,1)/B(x),x,0,20);

a(n) = binomial(4*n-1,n); \\ Michel Marcus, Oct 06 2015

G.f.: A(x)=x*B'(x)/B(x), where B(x) if g.f. of A006632.
a(n) = Sum_{k=0..n}(binomial(n-1,n-k)*binomial(3*n,k)).
a(n) = 3*A224274(n), for n > 0. - Michel Marcus, Oct 12 2015
From Peter Bala, Nov 04 2015: (Start)
The o.g.f. equals f(x)/g(x), where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A005810 (k = 0), A052203 (k = 1), A257633 (k = 2), A224274 (k = 3) and A004331 (k = 4). (End)
a(n) = [x^n] 1/(1 - x)^(3*n). - Ilya Gutkovskiy, Oct 03 2017
a(n) = A071919(3n-1,n+1) = A097805(4n,n+1). - Gus Wiseman, Sep 28 2022
From Peter Bala, Feb 14 2024: (Start)
a(n) = (-1)^n * binomial(-3*n, n).
a(n) = hypergeom([1 - 3*n, -n], [1], 1).
The g.f. A(x) satisfies A(x/(1 + x)^4) = 1/(1 - 3*x). (End)
a(n) = Sum_{k = 0..n} binomial(2*n+k-1, k)*binomial(2*n-k-1, n-k). - Peter Bala, Sep 16 2024
G.f.: 1/(4-3*g) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 17 2025

A348614 Numbers k such that the k-th composition in standard order has sum equal to twice its alternating sum.

Original entry on oeis.org

0, 9, 11, 14, 130, 133, 135, 138, 141, 143, 148, 153, 155, 158, 168, 177, 179, 182, 188, 208, 225, 227, 230, 236, 248, 2052, 2057, 2059, 2062, 2066, 2069, 2071, 2074, 2077, 2079, 2084, 2089, 2091, 2094, 2098, 2101, 2103, 2106, 2109, 2111, 2120, 2129, 2131

Offset: 1

Gus Wiseman, Oct 29 2021

nonn

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			The terms together with their binary indices begin:
    0: ()
    9: (3,1)
   11: (2,1,1)
   14: (1,1,2)
  130: (6,2)
  133: (5,2,1)
  135: (5,1,1,1)
  138: (4,2,2)
  141: (4,1,2,1)
  143: (4,1,1,1,1)
  148: (3,2,3)
  153: (3,1,3,1)
  155: (3,1,2,1,1)
  158: (3,1,1,1,2)

Gus Wiseman, Statistics, classes, and transformations of standard compositions

The unordered case (partitions) is counted by A000712, reverse A006330.
These compositions are counted by A262977.
Except for 0, a subset of A345917 (which is itself a subset of A345913).
A000346 = even-length compositions with alt sum != 0, complement A001700.
A011782 counts compositions.
A025047 counts wiggly compositions, ranked by A345167.
A034871 counts compositions of 2n with alternating sum 2k.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A116406 counts compositions with alternating sum >=0, ranked by A345913.
A138364 counts compositions with alternating sum 0, ranked by A344619.
A345197 counts compositions by length and alternating sum.
Cf. A000984, A002458, A004331, A005810, A224274.
Cf. A008549, A013777, A027306, A058622, A088218, A114121, A120452, A126869, A163493, A294175, A344604.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Select[Range[0,1000],Total[stc[#]]==2*ats[stc[#]]&]

A161736 Denominators of the column sums of the BG2 matrix.

Original entry on oeis.org

1, 9, 75, 1225, 19845, 160083, 1288287, 41409225, 1329696225, 10667118605, 85530896451, 1371086188563, 21972535073125, 176021737014375, 1409850293610375, 90324408810638025, 5786075364399106425, 46326420401234675625, 370882277949065911875, 5938020471163465810125

Offset: 2

Johannes W. Meijer, Jun 18 2009

The BG2 matrix coefficients, see also A008956, are defined by BG2[2m,1] = 2*beta(2m+1) and the recurrence relation BG2[2m,n] = BG2[2m,n-1] - BG2[2m-2,n-1]/(2*n-3)^2 for m = -2, -1, 0, 1, 2, .. and n = 2, 3, .. , with beta(m) = sum((-1)^k/(1+2*k)^m, k=0..infinity). We observe that beta(2m+1) = 0 for m = -1, -2, -3, .. .
A different way to define the matrix coefficients is BG2[2*m,n] = (1/m)*sum(LAMBDA(2*m-2*k,n-1)*BG2[2*k,n], k=0..m-1) with LAMBDA(2*m,n-1) = (1-2^(-2*m))*zeta(2*m)-sum((2*k-1)^(-2*m), k=1..n-1) and BG2[0,n] = Pi/2 for m = 0, 1, 2, .. , and n = 1, 2, 3 .. , with zeta(m) the Riemann zeta function.
The columns sums of the BG2 matrix are defined by sb(n) = sum(BG2[2*m,n], m=0..infinity) for n = 2, 3, .. . For large values of n the value of sb(n) approaches Pi/2.
It is remarkable that if we assume that BG2[2m,1] = 2 for m = 0, 1, .. the columns sums of the modified matrix converge to the original sb(n) values. The first Maple program makes use of this phenomenon and links the sb(n) with the central factorial numbers A008956.
The column sums sb(n) can be linked to other sequences, see the second Maple program.
We observe that the column sums sb(n) of the BG2(n) matrix are related to the column sums sl(n) of the LG2(n) matrix, see A008956, by sb(n) = (-1)^(n+1)*(2*n-1)*sl(n).
a(n+2), for n >= 0, seems to coincide with the numerators belonging to A278145. - Wolfdieter Lang, Nov 16 2016
Suppose that, given values f(x-2*n+1), f(x-2*n+3), ..., f(x-1), f(x+1), ..., f(x+2*n-3), f(x+2*n-1), we approximate f(x) using the first 2*n terms of its Taylor series. Then 1/sb(n+1) is the coefficient of f(x-1) and f(x+1). - Matthew House, Dec 03 2024

			sb(2) = 2; sb(3) = 16/9; sb(4) = 128/75; sb(5) = 2048/1225; etc..

G. C. Greubel, Table of n, a(n) for n = 2..830
R. Arratia, S. Garibaldi, and J. Kilian, Asymptotic distribution for the birthday problem with multiple coincidences, via an embedding of the collision process, arXiv:1310.7055 [math.PR], 2013.

Cf. Numerators A161737.
Cf. A001818, A001902, A008956, A013777, A134372 and A134375.
Cf. A001147, A133221, A161738 and A002474, A278145.

[Denominator((2^(4*n-5)*(Factorial(n-1))^4)/((n-1)*(Factorial(2*n-2))^2)): n in [2..20]]; // G. C. Greubel, Sep 26 2018

nmax := 18; for n from 0 to nmax do A001818(n) := (doublefactorial(2*n-1))^2 od: for n from 0 to nmax do A008956(n, 0):=1 od: for n from 0 to nmax do A008956(n, n) := A001818(n) od: for n from 1 to nmax do for m from 1 to n-1 do A008956(n, m) := (2*n-1)^2*A008956(n-1, m-1) + A008956(n-1, m) od: od: for n from 1 to nmax do for m from 0 to n do s(n, m):=0; s(n, m) := s(n, m)+ sum((-1)^k1*A008956(n, n-k1), k1=0..n-m): od: sb1(n+1) := sum(s(n, k1), k1=1..n) * 2/A001818(n); od: seq(sb1(n), n=2..nmax); # End program 1
nmax1 := nmax; for n from 0 to nmax1 do A001147(n):= doublefactorial(2*n-1) od: for n from 0 to nmax1/2 do A133221(2*n+1) := A001147(n); A133221(2*n) := A001147(n) od: for n from 0 to nmax1 do A002474(n) := 2^(2*n+1)*n!*(n+1)! od: for n from 1 to nmax1 do A161738(n) := ((product((2*n-3-2*k1), k1=0..floor(n/2-1)))) od: for n from 2 to nmax1 do sb2(n) := A002474(n-2) / (A161738(n)*A133221(n-1))^2 od: seq(sb2(n), n=2..nmax1); # End program 2
# Above Maple programs edited by Johannes W. Meijer, Sep 25 2012
r := n -> (1/Pi)*(2*n - 2)*((n - 3/2)!/(n - 1)!)^2: a := n -> numer(simplify(r(n))):
seq(a(n), n = 1..21);  # Peter Luschny, Feb 12 2025

sb[2]=2; sb[n_] := sb[n] = sb[n-1]*4*(n-1)*(n-2)/(2n-3)^2; Table[sb[n] // Denominator, {n, 2, 20}] (* Jean-François Alcover, Aug 14 2017 *)

{a(n) = if( n<2, 0, n--; numerator( binomial( 2*n, n)^2 * n / 2^(n+1) ))}; /* Michael Somos, May 09 2011 */

a(n) = denom(sb(n)) with sb(n) = (2^(4*n-5)*(n-1)!^4)/((n-1)*(2*n-2)!^2) and A161737(n) = numer(sb(n)).
a(n+1) = numerator of C(2*n,n)^2 * n / 2^(n+1). - Michael Somos, May 09 2011
a(n) = A001902(2*n-3). - Mats Granvik, Nov 25 2018
a(n) = numerator((1/Pi)*(2*n - 2)*((n - 3/2)!/(n - 1)!)^2). - Peter Luschny, Feb 13 2025

A195156 a(n) = (16^n-1)/3.

Original entry on oeis.org

0, 5, 85, 1365, 21845, 349525, 5592405, 89478485, 1431655765, 22906492245, 366503875925, 5864062014805, 93824992236885, 1501199875790165, 24019198012642645, 384307168202282325, 6148914691236517205, 98382635059784275285, 1574122160956548404565

Offset: 0

Omar E. Pol, Sep 10 2011

Numbers of A002450 that are multiples of 5. Also sequence found by reading the line from 0, in the direction 0, 5,..., in the square spiral whose edges are the Jacobsthal numbers A001045 and whose vertices are the numbers A000975. This is a semi-diagonal in the spiral.
In binary, these numbers are 101...01 (see A031982). - Alonso del Arte, May 20 2017
0 together with Jacobsthal numbers ending with the decimal digit 5. - Jianing Song, Aug 30 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Bisection of A002450.
Cf. A000975, A001025, A013777, A031982, A131865.
First quadrisection of Jacobsthal numbers A001045; the other quadrisections are A139792 (second), A144864 (third), and A141060 (fourth).

[(16^n-1)/3:n in [0..20]]; // Vincenzo Librandi, Sep 20 2011

A195156:=n->(16^n-1)/3; seq(A195156(k), k=0..50); # Wesley Ivan Hurt, Oct 24 2013

Table[(16^n - 1)/3, {n, 0, 63}] (* Wesley Ivan Hurt, Oct 24 2013 *)

for(n=0,50, print1((16^n - 1)/3, ", ")) \\ G. C. Greubel, Oct 11 2017

From Bruno Berselli, Sep 19 2011: (Start)
G.f.: 5*x/((1-x)*(1-16*x)).
a(n) = A002450(2n) = (16^n-1)/3.
a(n) = 5*A131865(n-1) = a(n-1) + 5*A001025(n-1) = 16*a(n-1) + 5 for n > 0. (End)
From Jianing Song, Aug 30 2022: (Start)
a(n) = A001045(4*n).
a(n+1) - a(n) = 10*A013777(n-1) = 80*A001025(n-1) for n >= 1. (End)
E.g.f.: exp(x)*(exp(15*x) - 1)/3. - Stefano Spezia, Dec 17 2022

New sequence name suggested by Charles R Greathouse IV using Berselli's formula. - Sep 19 2011

A241955 a(n) = 2^(4*n+3) - 1.

Original entry on oeis.org

7, 127, 2047, 32767, 524287, 8388607, 134217727, 2147483647, 34359738367, 549755813887, 8796093022207, 140737488355327, 2251799813685247, 36028797018963967, 576460752303423487, 9223372036854775807, 147573952589676412927, 2361183241434822606847, 37778931862957161709567

Offset: 0

Wassan Letourneur, Aug 09 2014

Jens Kruse Andersen, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (17,-16).

Cf. A000225, A004171, A013777, A241888.

[2^(4*n+3)-1 : n in [0..20]]; // Wesley Ivan Hurt, Aug 15 2014

A241955:=n->2^(4*n+3)-1: seq(A241955(n), n=0..20); # Wesley Ivan Hurt, Aug 15 2014

Table[2^(4n + 3) - 1, {n, 0, 29}]
2^(4 Range[0, 20] + 3) - 1 (* Wesley Ivan Hurt, Aug 15 2014 *)

vector(40, n, 2^(4*n-1)-1) \\ Derek Orr, Aug 11 2014

Vec((8*x+7)/((x-1)*(16*x-1)) + O(x^100)) \\ Colin Barker, Aug 11 2014

a(n) = 2^(4*n+3) - 1 = A000225(4*n+3) = A013777(n) - 1 = 4*A241888(n) + 3.
From Colin Barker, Aug 11 2014: (Start)
a(n) = 17*a(n-1) - 16*a(n-2).
G.f.: (8*x+7)/((x-1)*(16*x-1)). (End)
E.g.f.: exp(x)*(8*exp(15*x) - 1). - Elmo R. Oliveira, Feb 20 2025

A013793 a(n) = 10^(4*n + 3).

Original entry on oeis.org

1000, 10000000, 100000000000, 1000000000000000, 10000000000000000000, 100000000000000000000000, 1000000000000000000000000000, 10000000000000000000000000000000, 100000000000000000000000000000000000, 1000000000000000000000000000000000000000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (10000).

Cf. A004767, A011557, A013715, A013777, A013783, A098608.

[10^(4*n+3): n in [0..10]]; // Vincenzo Librandi, Jun 28 2011

10^(4*Range[0, 10] + 3) (* or *)
NestList[10000*# &, 1000, 10] (* Paolo Xausa, Jul 21 2025 *)

makelist(10^(4*n+3),n,0,20); /* Martin Ettl, Oct 21 2012 */

From Elmo R. Oliveira, Aug 30 2024: (Start)
G.f.: 1000/(1 - 10000*x).
E.g.f.: 1000*exp(10000*x).
a(n) = 10000*a(n-1) for n > 0.
a(n) = A011557(A004767(n)) = A013715(n)*A098608(n+1) = A013777(n)*A013783(n). (End)

A013813 a(n) = 20^(4*n + 3).

Original entry on oeis.org

8000, 1280000000, 204800000000000, 32768000000000000000, 5242880000000000000000000, 838860800000000000000000000000, 134217728000000000000000000000000000, 21474836480000000000000000000000000000000, 3435973836800000000000000000000000000000000000

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences.
Index entries for linear recurrences with constant coefficients, signature (160000).

Subsequence of A009964.

Cf. A004767, A013777, A013793.

[20^(4*n+3): n in [0..15]]; // Vincenzo Librandi, Jul 06 2011

20^(4*Range[0, 10] + 3) (* or *)
NestList[160000*# &, 8000, 10] (* Paolo Xausa, Jul 21 2025 *)

From Elmo R. Oliveira, Jul 13 2025: (Start)
G.f.: 8000/(1-160000*x).
E.g.f.: 8000*exp(160000*x).
a(n) = 160000*a(n-1).
a(n) = A013777(n)*A013793(n) = A009964(A004767(n)). (End)

cp's OEIS Frontend

A262977 a(n) = binomial(4*n-1,n).

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A348614 Numbers k such that the k-th composition in standard order has sum equal to twice its alternating sum.

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A161736 Denominators of the column sums of the BG2 matrix.

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A195156 a(n) = (16^n-1)/3.

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A241955 a(n) = 2^(4*n+3) - 1.

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A013793 a(n) = 10^(4*n + 3).

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A013813 a(n) = 20^(4*n + 3).

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