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A081581
Pascal-(1,6,1) array.
12
1, 1, 1, 1, 8, 1, 1, 15, 15, 1, 1, 22, 78, 22, 1, 1, 29, 190, 190, 29, 1, 1, 36, 351, 848, 351, 36, 1, 1, 43, 561, 2339, 2339, 561, 43, 1, 1, 50, 820, 5006, 9766, 5006, 820, 50, 1, 1, 57, 1128, 9192, 28806, 28806, 9192, 1128, 57, 1, 1, 64, 1485, 15240, 68034, 116208, 68034, 15240, 1485, 64, 1
OFFSET
0,5
COMMENTS
One of a family of Pascal-like arrays. A007318 is equivalent to the (1,0,1)-array. A008288 is equivalent to the (1,1,1)-array. Rows include A016993, A081591, A081592. Coefficients of the row polynomials in the Newton basis are given by A013614.
LINKS
Vincenzo Librandi, Rows n = 0..100, flattened
FORMULA
Square array T(n, k) defined by T(n, 0) = T(0, k) = 1, T(n, k) = T(n, k-1) + 6*T(n-1, k-1) + T(n-1, k).
Rows are the expansions of (1+6*x)^k/(1-x)^(k+1).
T(n, k) = Hypergeometric2F1([-k, k-n], [1], 7). - Jean-François Alcover, May 24 2013
E.g.f. for the n-th subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} binomial(n,k)*(7*x)^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 + 14*x + 49*x^2/2) = 1 + 15*x + 78*x^2/2! + 190*x^3/3! + 351*x^4/4! + 561*x^5/5! + .... - Peter Bala, Mar 05 2017
From G. C. Greubel, May 26 2021: (Start)
T(n, k, m) = Sum_{j=0..n-k} binomial(k,j)*binomial(n-j,k)*m^j, for m = 6.
Sum_{k=0..n} T(n, k, 6) = A083099(n+1). (End)
EXAMPLE
Rows start as:
1, 1, 1, 1, 1, ... A000012;
1, 8, 15, 22, 29, ... A016993;
1, 15, 78, 190, 351, ... A081591;
1, 22, 190, 848, 2339, ...
1, 29, 351, 2339, 9766, ...
The triangle starts as:
1;
1, 1;
1, 8, 1;
1, 15, 15, 1;
1, 22, 78, 22, 1;
1, 29, 190, 190, 29, 1;
1, 36, 351, 848, 351, 36, 1;
1, 43, 561, 2339, 2339, 561, 43, 1;
MATHEMATICA
Table[Hypergeometric2F1[-k, k-n, 1, 7], {n, 0, 10}, {k, 0, n}]//Flatten (* Jean-François Alcover, May 24 2013 *)
PROG
(PARI) t(n, k) = sum(j=0, n-k, binomial(n-k, j)*binomial(k, j)*7^j) \\ Michel Marcus, May 24 2013
(Magma)
A081581:= func< n, k, q | (&+[Binomial(k, j)*Binomial(n-j, k)*q^j: j in [0..n-k]]) >;
[A081581(n, k, 6): k in [0..n], n in [0..12]]; // G. C. Greubel, May 26 2021
(Sage) flatten([[hypergeometric([-k, k-n], [1], 7).simplify() for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2021
CROSSREFS
Cf. Pascal (1,m,1) array: A123562 (m = -3), A098593 (m = -2), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081579 (m = 4), A081580 (m = 5), A081582 (m = 7), A143683 (m = 8).
Sequence in context: A131067 A157170 A143679 * A174125 A051425 A051469
KEYWORD
easy,nonn,tabl
AUTHOR
Paul Barry, Mar 23 2003
STATUS
approved