A045543 -id:A045543 - cp's OEIS frontend

A054335 A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).

1, 2, 1, 6, 4, 1, 20, 16, 6, 1, 70, 64, 30, 8, 1, 252, 256, 140, 48, 10, 1, 924, 1024, 630, 256, 70, 12, 1, 3432, 4096, 2772, 1280, 420, 96, 14, 1, 12870, 16384, 12012, 6144, 2310, 640, 126, 16, 1, 48620, 65536, 51480, 28672, 12012, 3840, 924, 160, 18, 1

Offset: 0

Wolfdieter Lang, Mar 13 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. for the row polynomials p(n,x) (increasing powers of x) is 1/(sqrt(1-4*z)-x*z).
The column sequences are for m=0..20: A000984, A000302 (powers of 4), A002457, A002697, A002802, A038845, A020918, A038846, A020920, A040075, A020922, A045543, A020924, A054337, A020926, A054338, A020928, A054339, A020930, A054340, A020932.
Riordan array (1/sqrt(1-4*x),x/sqrt(1-4*x)). - Paul Barry, May 06 2009
The matrix inverse is apparently given by deleting the leftmost column from A206022. - R. J. Mathar, Mar 12 2013

			Triangle begins:
    1;
    2,    1;
    6,    4,   1;
   20,   16,   6,   1;
   70,   64,  30,   8,  1;
  252,  256, 140,  48, 10,  1;
  924, 1024, 630, 256, 70, 12, 1; ...
Fourth row polynomial (n=3): p(3,x) = 20 + 16*x + 6*x^2 + x^3.
From _Paul Barry_, May 06 2009: (Start)
Production matrix begins
    2,   1;
    2,   2,  1;
    0,   2,  2,  1;
   -2,   0,  2,  2,  1;
    0,  -2,  0,  2,  2,  1;
    4,   0, -2,  0,  2,  2, 1;
    0,   4,  0, -2,  0,  2, 2, 1;
  -10,   0,  4,  0, -2,  0, 2, 2, 1;
    0, -10,  0,  4,  0, -2, 0, 2, 2, 1; (End)

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.

Cf. A000984, A035324, A054336.

Row sums: A026671.

T:= function(n, k)
    if k mod 2=0 then return Binomial(2*n-k, n-Int(k/2))*Binomial(n-Int(k/2),Int(k/2))/Binomial(k,Int(k/2));
    else return 4^(n-k)*Binomial(n-Int((k-1)/2)-1, Int((k-1)/2));
    fi;
  end;
Flat(List([0..10], n-> List([0..n], k-> T(n, k) ))); # G. C. Greubel, Jul 20 2019

T:= func< n, k | (k mod 2) eq 0 select Binomial(2*n-k, n-Floor(k/2))* Binomial(n-Floor(k/2),Floor(k/2))/Binomial(k,Floor(k/2)) else 4^(n-k)*Binomial(n-Floor((k-1)/2)-1, Floor((k-1)/2)) >;
[[T(n,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jul 20 2019

A054335 := proc(n,k)
    if k <0 or k > n then
        0 ;
    elif type(k,odd) then
        kprime := floor(k/2) ;
        binomial(n-kprime-1,kprime)*4^(n-k) ;
    else
        kprime := k/2 ;
        binomial(2*n-k,n-kprime)*binomial(n-kprime,kprime)/binomial(k,kprime) ;
    end if;
end proc: # R. J. Mathar, Mar 12 2013
# Uses function PMatrix from A357368. Adds column 1,0,0,0,... to the left.
PMatrix(10, n -> binomial(2*(n-1), n-1)); # Peter Luschny, Oct 19 2022

Flatten[ CoefficientList[#1, x] & /@ CoefficientList[ Series[1/(Sqrt[1 - 4*z] - x*z), {z, 0, 9}], z]] (* or *)
a[n_, k_?OddQ] := 4^(n-k)*Binomial[(2*n-k-1)/2, (k-1)/2]; a[n_, k_?EvenQ] := (Binomial[n-k/2, k/2]*Binomial[2*n-k, n-k/2])/Binomial[k, k/2]; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 08 2011, updated Jan 16 2014 *)

T(n, k) = if(k%2==0, binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2), 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2));
for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Jul 20 2019

def T(n, k):
    if (mod(k,2)==0): return binomial(2*n-k, n-k/2)*binomial(n-k/2,k/2)/binomial(k,k/2)
    else: return 4^(n-k)*binomial(n-(k-1)/2-1, (k-1)/2)
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Jul 20 2019

a(n, 2*k+1) = binomial(n-k-1, k)*4^(n-2*k-1), a(n, 2*k) = binomial(2*(n-k), n-k)*binomial(n-k, k)/binomial(2*k, k), k >= 0, n >= m >= 0; a(n, m) := 0 if n

Column recursion: a(n, m)=2*(2*n-m-1)*a(n-1, m)/(n-m), n>m >= 0, a(m, m) := 1.

G.f. for column m: cbie(x)*(x*cbie(x))^m, with cbie(x) := 1/sqrt(1-4*x).

G.f.: 1/(1-x*y-2*x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction). - Paul Barry, May 06 2009

Sum_{k>=0} T(n,2*k)*(-1)^k*A000108(k) = A000108(n+1). - Philippe Deléham, Jan 30 2012

Sum_{k=0..floor(n/2)} T(n-k,n-2*k) = A098615(n). - Philippe Deléham, Feb 01 2012

T(n,k) = 4*T(n-1,k) + T(n-2,k-2) for k>=1. - Philippe Deléham, Feb 02 2012

Vertical recurrence: T(n,k) = 1*T(n-1,k-1) + 2*T(n-2,k-1) + 6*T(n-3,k-1) + 20*T(n-4,k-1) + ... for k >= 1 (the coefficients 1, 2, 6, 20, ... are the central binomial coefficients A000984). - Peter Bala, Oct 17 2015

A050982 5-idempotent numbers.

Original entry on oeis.org

1, 30, 525, 7000, 78750, 787500, 7218750, 61875000, 502734375, 3910156250, 29326171875, 213281250000, 1510742187500, 10458984375000, 70971679687500, 473144531250000, 3105010986328125, 20091247558593750, 128360748291015625, 810699462890625000

Offset: 5

Eric W. Weisstein

Number of n-permutations of 6 objects: t,u,v,z,x, y with repetition allowed, containing exactly five u's. Example: a(6)=30 because we have uuuuut, uuuutu, uuutuu, uutuuu, utuuuu, tuuuuu, uuuuuv, uuuuvu, uuuvuu, uuvuuu, uvuuuu, vuuuuu, uuuuuz, uuuuzu, uuuzuu, uuzuuu, uzuuuu, zuuuuu, uuuuux, uuuuxu, uuuxuu, uuxuuu, uxuuuu, xuuuuu, uuuuuy, uuuuyu, uuuyuu, uuyuuu, uyuuuu, yuuuuu. - Zerinvary Lajos, Jun 16 2008

Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 91, #43.

Vincenzo Librandi, Table of n, a(n) for n = 5..1000
Eric Weisstein's World of Mathematics, Idempotent Number.
Index entries for linear recurrences with constant coefficients, signature (30,-375,2500,-9375,18750,-15625).

Cf. A001788, A036216, A040075, A050988, A050989, A000389, A054849, A036219, A045543, A036084, A140404, A000389, A054849, A036219, A045543, A036084, A140404.

[Binomial(n, 5)*5^(n-5): n in [5..25]]; // Vincenzo Librandi, Aug 12 2017

seq(binomial(n, 5)*5^(n-5), n=5..32); # Zerinvary Lajos, Jun 16 2008

CoefficientList[Series[1 / (1 - 5 x)^6, {x, 0, 33}], x] (* Vincenzo Librandi, Aug 12 2017 *)

a(n)=binomial(n, 5)*5^(n-5) \\ Charles R Greathouse IV, Sep 03 2011

a(n) = C(n, 5)*5^(n-5).
G.f.: x^5/(1-5*x)^6. - Zerinvary Lajos, Aug 06 2008
From Amiram Eldar, Apr 17 2022: (Start)
Sum_{n>=5} 1/a(n) = 6400*log(5/4) - 17125/12.
Sum_{n>=5} (-1)^(n+1)/a(n) = 32400*log(6/5) - 23625/4. (End)

A140404 a(n) = binomial(n+5, 5)*7^n.

Original entry on oeis.org

1, 42, 1029, 19208, 302526, 4235364, 54353838, 652246056, 7419298887, 80787921214, 848273172747, 8636963213424, 85649885199788, 830145041167176, 7886377891088172, 73606193650156272, 676256904160810749, 6126091955339109138, 54794489156088698401, 484498640959100070072

Offset: 0

Zerinvary Lajos, Jun 16 2008

With a different offset, number of n-permutations of 8 objects:r,s,t,u,v,z,x,y with repetition allowed, containing exactly five (5) u's. Example: a(1)=42 because we have
uuuuur, uuuuru, uuuruu, uuruuu, uruuuu, ruuuuu
uuuuus, uuuusu, uuusuu, uusuuu, usuuuu, suuuuu,
uuuuut, uuuutu, uuutuu, uutuuu, utuuuu, tuuuuu,
uuuuuv, uuuuvu, uuuvuu, uuvuuu, uvuuuu, vuuuuu,
uuuuuz, uuuuzu, uuuzuu, uuzuuu, uzuuuu, zuuuuu,
uuuuux, uuuuxu, uuuxuu, uuxuuu, uxuuuu, xuuuuu,
uuuuuy, uuuuyu, uuuyuu, uuyuuu, uyuuuu, yuuuuu.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (42,-735,6860,-36015,100842,-117649).

Cf. A000389, A036084, A036219, A045543, A050982, A054849.

[7^n* Binomial(n+5, 5): n in [0..20]]; // Vincenzo Librandi, Oct 12 2011

Maple
```
seq(binomial(n+5,5)*7^n,n=0..17);
```

Table[Binomial[n+5,5]7^n,{n,0,20}] (* or *) LinearRecurrence[ {42,-735,6860,-36015,100842,-117649},{1,42,1029,19208,302526,4235364},21] (* Harvey P. Dale, Sep 08 2011 *)

a(n)=binomial(n+5,5)*7^n \\ Charles R Greathouse IV, Oct 07 2015

G.f.: 1/(1-7*x)^6. - Zerinvary Lajos, Aug 06 2008
a(n) = 42*a(n-1) - 735*a(n-2) + 6860*a(n-3) - 36015*a(n-4) + 100842*a(n-5) - 117649*a(n-6). - Harvey P. Dale, Sep 08 2011
From Amiram Eldar, Aug 28 2022: (Start)
Sum_{n>=0} 1/a(n) = 45360*log(7/6) - 27965/4.
Sum_{n>=0} (-1)^n/a(n) = 143360*log(8/7) - 229705/12. (End)

A172978 a(n) = binomial(n+10, 10)*4^n.

Original entry on oeis.org

1, 44, 1056, 18304, 256256, 3075072, 32800768, 318636032, 2867724288, 24216338432, 193730707456, 1479398129664, 10848919617536, 76776969601024, 526470648692736, 3509804324618240, 22813728110018560, 144934272698941440, 901813252348968960, 5505807224867389440

Offset: 0

Zerinvary Lajos, Feb 06 2010

Vincenzo Librandi, Table of n, a(n) for n = 0..157
Index entries for linear recurrences with constant coefficients, signature (44,-880,10560,-84480,473088,-1892352,5406720,-10813440,14417920,-11534336,4194304).

Cf. A002697, A038845, A038846, A040075, A045543, A054337, A054338, A054339, A054340

[Binomial(n+10, 10)*4^n: n in [0..30]]; // Vincenzo Librandi, Jun 06 2011

Table[Binomial[n + 10, 10]*4^n, {n, 0, 20}]

From Amiram Eldar, Mar 27 2022: (Start)
G.f.: 1/(1 - 4*x)^11.
Sum_{n>=0} 1/a(n) = 14269429/63 - 787320*log(4/3).
Sum_{n>=0} (-1)^n/a(n) = 78125000*log(5/4) - 1098284605/63. (End)

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A045622 Convolution of A000108 (Catalan numbers) with A045543.

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A054335 A convolution triangle of numbers based on A000984 (central binomial coefficients of even order).

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A050982 5-idempotent numbers.

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A140404 a(n) = binomial(n+5, 5)*7^n.

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A172978 a(n) = binomial(n+10, 10)*4^n.

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