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A093645 (10,1) Pascal triangle. 30

%I

%S 1,10,1,10,11,1,10,21,12,1,10,31,33,13,1,10,41,64,46,14,1,10,51,105,

%T 110,60,15,1,10,61,156,215,170,75,16,1,10,71,217,371,385,245,91,17,1,

%U 10,81,288,588,756,630,336,108,18,1,10,91,369,876,1344,1386,966,444,126,19,1

%N (10,1) Pascal triangle.

%C The array F(10;n,m) gives in the columns m >= 1 the figurate numbers based on A017281, including the 12-gonal numbers A051624 (see the W. Lang link).

%C This is the tenth member, d=10, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-5 and A093644 for d=1..9.

%C 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+9*z)/(1-(1+x)*z).

%C The SW-NE diagonals give A022100(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k), n >= 1, with n=0 value 9. Observation by _Paul Barry_, Apr 29 2004. Proof via recursion relations and comparison of inputs.

%D Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.

%D Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch. 5, pp. 109-122.

%H Reinhard Zumkeller, <a href="/A093645/b093645.txt">Rows n = 0..125 of triangle, flattened</a>

%H W. Lang, <a href="/A093645/a093645.txt">First 10 rows and array of figurate numbers </a>.

%F a(n, m) = F(10;n-m, m) for 0 <= m <= n, else 0, with F(10;0, 0)=1, F(10;n, 0)=10 if n >= 1 and F(10;n, m):=(10*n+m)*binomial(n+m-1, m-1)/m if m >= 1.

%F Recursion: a(n, m)=0 if m > n, a(0, 0)=1; a(n, 0)=10 if n >= 1; a(n, m) = a(n-1, m) + a(n-1, m-1).

%F G.f. column m (without leading zeros): (1+9*x)/(1-x)^(m+1), m >= 0.

%F T(n, k) = C(n, k) + 9*C(n-1, k). - _Philippe Deléham_, Aug 28 2005

%F exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(10 + 21*x + 12*x^2/2! + x^3/3!) = 10 + 31*x + 64*x^2/2! + 110*x^3/3! + 170*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

%e Triangle begins

%e 1;

%e 10, 1;

%e 10, 11, 1;

%e 10, 21, 12, 1;

%e ...

%t t[0, 0] = 1; t[n_, k_] := Binomial[n, k] + 9*Binomial[n-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 05 2013, after _Philippe Deléham_ *)

%o (Haskell)

%o a093645 n k = a093645_tabl !! n !! k

%o a093645_row n = a093645_tabl !! n

%o a093645_tabl = [1] : iterate

%o (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [10, 1]

%o -- _Reinhard Zumkeller_, Aug 31 2014

%Y Row sums: 1 for n=0 and A005015(n-1), n >= 1, alternating row sums are 1 for n=0, 9 for n=2 and 0 otherwise.

%Y The column sequences give for m=1..9: A017281, A051624 (12-gonal), A007587, A051799, A051880, A050406, A052254, A056125, A093646.

%K nonn,easy,tabl

%O 0,2

%A _Wolfdieter Lang_, Apr 22 2004

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Last modified April 14 22:47 EDT 2021. Contains 342971 sequences. (Running on oeis4.)