A034266 -id:A034266 - cp's OEIS frontend

A093564 (7,1) Pascal triangle.

1, 7, 1, 7, 8, 1, 7, 15, 9, 1, 7, 22, 24, 10, 1, 7, 29, 46, 34, 11, 1, 7, 36, 75, 80, 45, 12, 1, 7, 43, 111, 155, 125, 57, 13, 1, 7, 50, 154, 266, 280, 182, 70, 14, 1, 7, 57, 204, 420, 546, 462, 252, 84, 15, 1, 7, 64, 261, 624, 966, 1008, 714, 336, 99, 16, 1, 7, 71, 325, 885

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(7;n,m) gives in the columns m>=1 the figurate numbers based on A016993, including the 9-gonal numbers A001106, (see the W. Lang link).
This is the seventh member, d=7, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-3, for d=1..6.
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+6*z)/(1-(1+x)*z).
The SW-NE diagonals give A022097(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 6. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins
  [1];
  [7,  1];
  [7,  8,  1];
  [7, 15,  9,  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
W. Lang, First 10 rows and array of figurate numbers .

Row sums: A000079(n+2), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 6 for n=2 and 0 otherwise.
The column sequences give for m=1..9: A016993, A001106 (9-gonal), A007584, A051740, A051877, A050403, A027818, A034266, A055994.
Cf. A093565 (d=8).

a093564 n k = a093564_tabl !! n !! k
a093564_row n = a093564_tabl !! n
a093564_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [7, 1]
-- Reinhard Zumkeller, Sep 01 2014

N:= 20: # to get the first N rows
T:=Matrix(N,N):
T[1,1]:= 1:
for m from 2 to N do
T[m,1]:= 7:
T[m,2..m]:= T[m-1,1..m-1] + T[m-1,2..m];
od:
for m from 1 to N do
convert(T[m,1..m],list)
od; # Robert Israel, Dec 28 2014

a(n, m)=F(7;n-m, m) for 0<= m <= n, otherwise 0, with F(7;0, 0)=1, F(7;n, 0)=7 if n>=1 and F(7;n, m):=(7*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)=7 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+6*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 6*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)*(7 + 15*x + 9*x^2/2! + x^3/3!) = 7 + 22*x + 46*x^2/2! + 80*x^3/3! + 125*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).

A055994 Expansion of (1+6x)/(1-x)^10.

Original entry on oeis.org

1, 16, 115, 550, 2035, 6292, 17017, 41470, 92950, 194480, 384098, 722228, 1301690, 2261000, 3801710, 6210644, 9887999, 15382400, 23434125, 35027850, 51456405, 74397180, 106002975, 149009250, 206859900, 283853856, 385314996, 517788040

Offset: 0

Barry E. Williams, Jun 04 2000

Partial sums of A034266. - Vladimir Joseph Stephan Orlovsky, Jun 25 2009

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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1)

Cf. A034266.

Cf. A093564 ((7, 1) Pascal, column m=9). Partial sums of A034266.

[((7*n+9)*Binomial(n+8,8))/9: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014

CoefficientList[Series[(1 + 6 x)/(1 - x)^10, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 30 2014 *)
LinearRecurrence[{10,-45,120,-210,252,-210,120,-45,10,-1},{1,16,115,550,2035,6292,17017,41470,92950,194480},30] (* Harvey P. Dale, Sep 07 2022 *)

a(n) = (7n+9)*C(n+8, 8)/9.

G.f.: (1+6x)/(1-x)^10.

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A093564 (7,1) Pascal triangle.

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

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A055994 Expansion of (1+6x)/(1-x)^10.

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