A034265 -id:A034265 - cp's OEIS frontend

A034261 Infinite square array f(a,b) = C(a+b,b+1)(ab+a+1)/(b+2), a, b >= 0, read by antidiagonals. Equivalently, triangular array T(n,k) = f(k,n-k), 0 <= k <= n, read by rows.

0, 0, 1, 0, 1, 3, 0, 1, 5, 6, 0, 1, 7, 14, 10, 0, 1, 9, 25, 30, 15, 0, 1, 11, 39, 65, 55, 21, 0, 1, 13, 56, 119, 140, 91, 28, 0, 1, 15, 76, 196, 294, 266, 140, 36, 0, 1, 17, 99, 300, 546, 630, 462, 204, 45, 0, 1, 19, 125, 435, 930, 1302, 1218, 750, 285, 55

Offset: 0

Clark Kimberling

f(h,k) = number of paths consisting of steps from (0,0) to (h,k) using h unit steps right, k+1 unit steps up and 1 unit step down, in some order, with first step not down and no repeated points.

			Triangle begins:
  0;
  0, 1;
  0, 1, 3;
  0, 1, 5,  6;
  0, 1, 7, 14, 10;
  ...
As a square array,
  [ 0  0  0   0   0 ...]
  [ 1  1  1   1   1 ...]
  [ 3  5  7   9  11 ...]
  [ 6 14 25  39  56 ...]
  [10 30 65 119 196 ...]
  [...      ...     ...]

Cf. A001787 (row sums), A000330(n) = f(n,1).

Cf. A034263, A034264, A034265, A034267 - A034275 for diagonals n -> f(n,n+k), for several fixed k.

A034261 := proc(n, k) binomial(n, n-k+1)*(k+(k-1)/(k-n-2)); end proc; # argument indices of the triangle

Flatten[Table[Binomial[n,n-k+1](k+(k-1)/(k-n-2)),{n,0,15},{k,0,n}]] (* Harvey P. Dale, Jan 11 2013 *)

f(h,k)=binomial(h+k,k+1)*(k*h+h+1)/(k+2)

tabl(nn) = for (n=0, nn, for (k=0, n, print1(binomial(n, n-k+1)*(k+(k-1)/(k-n-2)), ", ")); print()); \\ Michel Marcus, Mar 20 2015

Another formula: f(h,k) = binomial(h+k,k+1) + Sum{C(i+j-1, j)*C(h+k-i-j, k-j+1): i=1, 2, ..., h-1, j=1, 2, ..., k+1}

Entry revised by N. J. A. Sloane, Apr 21 2000. The formula for f in the definition was found by Michael Somos.

Edited by M. F. Hasler, Nov 08 2017

A093563 (6,1)-Pascal triangle.

Original entry on oeis.org

1, 6, 1, 6, 7, 1, 6, 13, 8, 1, 6, 19, 21, 9, 1, 6, 25, 40, 30, 10, 1, 6, 31, 65, 70, 40, 11, 1, 6, 37, 96, 135, 110, 51, 12, 1, 6, 43, 133, 231, 245, 161, 63, 13, 1, 6, 49, 176, 364, 476, 406, 224, 76, 14, 1, 6, 55, 225, 540, 840, 882, 630, 300, 90, 15, 1, 6, 61, 280, 765, 1380

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(6;n,m) gives in the columns m >= 1 the figurate numbers based on A016921, including the octagonal numbers A000567, (see the W. Lang link).
This is the sixth member, d=6, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-2, for d=1..5.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+5*z)/(1-(1+x)*z).
The SW-NE diagonals give A022096(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

			Triangle begins
  1;
  6,  1;
  6,  7,  1;
  6, 13,  8,  1;
  6, 19, 21,  9,  1;
  6, 25, 40, 30, 10,  1;
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Wolfdieter Lang, First 10 rows and array of figurate numbers

Row sums: A005009(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 5 for n=2 and 0 else.
Cf. A007318, A093564 (d=7), A228196, A228576.
The column sequences give for m=1..9: A016921, A000567 (octagonal), A002414, A002419, A051843, A027810, A034265, A054487, A055848.

a093563 n k = a093563_tabl !! n !! k
a093563_row n = a093563_tabl !! n
a093563_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [6, 1]
-- Reinhard Zumkeller, Aug 31 2014

lim = 11; s = Series[(1 + 5*x)/(1 - x)^(m + 1), {x, 0, lim}]; t = Table[ CoefficientList[s, x], {m, 0, lim}]; Flatten[ Table[t[[j - k + 1, k]], {j, lim + 1}, {k, j, 1, -1}]] (* Jean-François Alcover, Sep 16 2011, after g.f. *)

from math import comb, isqrt
def A093563(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+5*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m)=F(6;n-m, m) for 0<= m <= n, otherwise 0, with F(6;0, 0)=1, F(6;n, 0)=6 if n>=1 and F(6;n, m):= (6*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=6 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+5*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 5*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(6 + 13*x + 8*x^2/2! + x^3/3!) = 6 + 19*x + 40*x^2/2! + 70*x^3/3! + 110*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A254142 a(n) = (9n+10)binomial(n+9,9)/10.

Original entry on oeis.org

1, 19, 154, 814, 3289, 11011, 32032, 83512, 199342, 442442, 923780, 1830764, 3468374, 6317234, 11113784, 18958808, 31461815, 50930165, 80613390, 125014890, 190285095, 284712285, 419329560, 608658960, 871616460, 1232604516, 1722822024, 2381824984

Offset: 0

Bruno Berselli, Jan 26 2015

Partial sums of A056003.

If n is of the form 8*k+2*(-1)^k-1 or 8*k+2*(-1)^k-2 then a(n) is odd.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A000581, A001287, A056003.

Cf. sequences of the type (k*n+k+1)*binomial(n+k,k)/(k+1): A000217 (k=1), A000330 (k=2), A001296 (k=3), A034263 (k=4), A051946 (k=5), A034265 (k=6), A034266 (k=7), A056122 (k=8), this sequence (k=9).

List([0..30], n-> (9*n+10)*Binomial(n+9,9)/10); # G. C. Greubel, Aug 28 2019

[(9*n+10)*Binomial(n+9,9)/10: n in [0..30]];

seq((9*n+10)*binomial(n+9,9)/10, n=0..30); # G. C. Greubel, Aug 28 2019

Table[(9n+10)Binomial[n+9, 9]/10, {n, 0, 30}]

vector(30, n, n--; (9*n+10)*binomial(n+9, 9)/10)

[(9*n+10)*binomial(n+9,9)/10 for n in (0..30)]

G.f.: (1 + 8*x)/(1-x)^11.
a(n) = Sum_{i=0..n} (i+1)*A000581(i+8).
a(n+1) = 8*A001287(n+10) + A001287(n+11).

A054487 a(n) = (3n+4)binomial(n+7, 7)/4.

Original entry on oeis.org

1, 14, 90, 390, 1320, 3762, 9438, 21450, 45045, 88660, 165308, 294372, 503880, 833340, 1337220, 2089164, 3187041, 4758930, 6970150, 10031450, 14208480, 19832670, 27313650, 37153350, 49961925, 66475656, 87576984, 114316840

Offset: 0

Barry E. Williams, May 06 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A034265.

Cf. A093563 ((6, 1) Pascal, column m=8).

List([0..40], n-> (3*n+4)*Binomial(n+7, 7)/4 ); # G. C. Greubel, Jan 19 2020

[((3*n+4)*Binomial(n+7,7))/4: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014

seq( (3*n+4)*binomial(n+7,7)/4, n=0..40); # G. C. Greubel, Jan 19 2020

CoefficientList[Series[(1+5x)/(1-x)^9, {x,0,40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
Table[6*Binomial[n+8,8] -5*Binomial[n+7,7], {n,0,40}] (* G. C. Greubel, Jan 19 2020 *)
LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,14,90,390,1320,3762,9438,21450,45045},30] (* Harvey P. Dale, Jul 19 2022 *)

a(n) = (3*n+4)*binomial(n+7, 7)/4; \\ Michel Marcus, Sep 07 2017

[(3*n+4)*binomial(n+7, 7)/4 for n in (0..40)] # G. C. Greubel, Jan 19 2020

G.f.: (1+5*x)/(1-x)^9.
From G. C. Greubel, Jan 19 2020: (Start)
a(n) = 6*binomial(n+8, 8) - 5*binomial(n+7, 7).
E.g.f.: (20160 +262080*x +635040*x^2 +540960*x^3 +205800*x^4 +38808*x^5 +3724*x^6 +172*x^7 +3*x^8)*exp(x)/20160. (End)
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). - Wesley Ivan Hurt, Jun 07 2021

Corrected and extended by James Sellers, May 10 2000

A136526 Coefficients polynomials B(x, n) = ((1 + a + b)x - c)B(x, n-1) - abB(x, n-2) with a = 3, b = 2, and c = 0.

Original entry on oeis.org

1, 0, 1, -6, 0, 6, 0, -42, 0, 36, 36, 0, -288, 0, 216, 0, 468, 0, -1944, 0, 1296, -216, 0, 4536, 0, -12960, 0, 7776, 0, -4104, 0, 38880, 0, -85536, 0, 46656, 1296, 0, -51840, 0, 311040, 0, -559872, 0, 279936, 0, 32400, 0, -544320, 0, 2379456, 0, -3639168, 0, 1679616

Offset: 0

Roger L. Bagula, Mar 23 2008

			Triangle begins as:
     1;
     0,     1;
    -6,     0,      6;
     0,   -42,      0,    36;
    36,     0,   -288,     0,    216;
     0,   468,      0, -1944,      0,   1296;
  -216,     0,   4536,     0, -12960,      0,    7776;
     0, -4104,      0, 38880,      0, -85536,       0, 46656;
  1296,     0, -51840,     0, 311040,      0, -559872,     0, 279936;

Harry Hochstadt, The Functions of Mathematical Physics, Dover, New York, 1986, page 93

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000567, A002414, A002419, A016921, A027810, A030192, A034265, A051843, A054487, A055848, A081106, A136531.

f:= func< n,k | k eq 0 select (-1)^Floor(n/2) else (-1)^Floor((n-k)/2)*6^Floor((k-1)/2)*(1/k)*(6*Floor((n-k)/2) +k)*Binomial(Floor((n-k)/2) +k-1, k-1) >;
A136526:= func< n,k | ((n+k+1) mod 2)*6^Floor(n/2)*f(n,k) >;
[A136526(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 22 2022

(* First program *)
a= (b+1)/(b-1); c=0; b=2;
B[x_, n_]:= B[x, n]= If[n<2, x^n, ((1+a+b)*x -c)*B[x, n-1] -a*b*B[x, n-2]];
Table[CoefficientList[B[x,n], x], {n,0,10}]//Flatten
(* Second program *)
B[x_, n_]:= 6^(n/2)*(ChebyshevU[n, Sqrt[3/2]*x] -(5*x/Sqrt[6])*ChebyshevU[n-1, Sqrt[3/2]*x]);
Table[CoefficientList[B[x, n], x]/6^Floor[n/2], {n,0,16}]//Flatten (* G. C. Greubel, Sep 22 2022 *)

def f(n,k):
    if (k==0): return (-1)^(n//2)
    else: return (-1)^((n-k)//2)*6^((k-1)//2)*(1/k)*(6*((n-k)//2) + k)*binomial(((n-k)//2) +k-1, k-1)
def A136526(n,k): return ((n+k+1)%2)*6^(n//2)*f(n,k)
flatten([[A136526(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 22 2022

T(n, k) = coefficients of the polynomials defined by B(x, n) = ((1 + a + b)*x - c)*B(x, n - 1) - a*b*B(x, n - 2) with B(x, 0) = 1, B(x, 1) = x, a = 3, b = 2, and c = 0.
From G. C. Greubel, Sep 22 2022: (Start)
T(n, k) = coefficients of the polynomials defined by B(x, n) = 6^(n/2)*(ChebyshevU(n, sqrt(3/2)*x) - (5*x/sqrt(6))*ChebyshevU(n-1, sqrt(3/2)*x)).
T(n, k) = (1/2)*(1+(-1)^(n+k))*6^floor(n/2)*f(n, k), where f(n, k) = (-1)^floor((n -k)/2)*6^floor((k-1)/2)*(1/k)*(6*floor((n-k)/2) + k)*binomial(floor((n-k)/2) + k -1, k-1), for k >= 1, and (-1)^floor(n/2) for k = 0.
T(n, 0) = (1/2)*(1+(-1)^n)*(-6)^floor(n/2).
T(n, 1) = (1/2)*(1-(-1)^n)*(-6)^floor((n-1)/2)*A016921(floor((n-1)/2)), n >= 1.
T(n, 2) = (1/2)*(1+(-1)^n)*(-1)^(1+Floor((n+1)/2))*6^floor((n+1)/2)*A000567(floor( (n+1)/2)), n >= 2.
T(n, 3) = (1/2)*(1-(-1)^n)*(-6)^floor((n+1)/2)*A002414(floor((n-1)/2)), n >= 3.
T(n, 4) = (3/2)*(1+(-1)^n)*(-6)^floor((n+1)/2)*A002419(floor((n-1)/2)), n >= 4.
T(n, 5) = 18*(1-(-1)^n)*(-6)^floor((n-1)/2)*A051843(floor((n-3)/2)), n >= 5.
T(n, n) = 6^(n-1) + (5/6)*[n=0].
T(n, n-2) = -6*A081106(n-2), n >= 2.
Sum_{k=0..n} T(n, k) = -6*A030192(n-3), n>= 0.
Sum_{k=0..floor(n/2)} T(n-k, k) = [n=0] - 5*[n=2].
G.f.: (1 - 5*x*y)/(1 - 6*x*y + 6*y^2). (End)

Edited by G. C. Greubel, Sep 22 2022

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A050485 Duplicate of A034265.

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A034261 Infinite square array f(a,b) = C(a+b,b+1)*(a*b+a+1)/(b+2), a, b >= 0, read by antidiagonals. Equivalently, triangular array T(n,k) = f(k,n-k), 0 <= k <= n, read by rows.

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A093563 (6,1)-Pascal triangle.

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A254142 a(n) = (9*n+10)*binomial(n+9,9)/10.

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A054487 a(n) = (3*n+4)*binomial(n+7, 7)/4.

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A136526 Coefficients polynomials B(x, n) = ((1 + a + b)*x - c)*B(x, n-1) - a*b*B(x, n-2) with a = 3, b = 2, and c = 0.

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A034261 Infinite square array f(a,b) = C(a+b,b+1)(ab+a+1)/(b+2), a, b >= 0, read by antidiagonals. Equivalently, triangular array T(n,k) = f(k,n-k), 0 <= k <= n, read by rows.

A254142 a(n) = (9n+10)binomial(n+9,9)/10.

A054487 a(n) = (3n+4)binomial(n+7, 7)/4.

A136526 Coefficients polynomials B(x, n) = ((1 + a + b)x - c)B(x, n-1) - abB(x, n-2) with a = 3, b = 2, and c = 0.