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Array read by antidiagonals: A(n,k) = (k+1)^n*(n+k)!/n!.
14

%I #34 Apr 10 2023 14:06:08

%S 1,1,1,1,4,2,1,12,18,6,1,32,108,96,24,1,80,540,960,600,120,1,192,2430,

%T 7680,9000,4320,720,1,448,10206,53760,105000,90720,35280,5040,1,1024,

%U 40824,344064,1050000,1451520,987840,322560,40320

%N Array read by antidiagonals: A(n,k) = (k+1)^n*(n+k)!/n!.

%C A009998/A119502 gives triangle of unreduced coefficients of polynomials defined by A152650/A152656. a(n) gives numerators with denominators n! for each row.

%C Row 0 is A000142. Row 1 is formed from positive members of A001563. Row 2 is A055533. Column 0 is A000012. Column 1 is formed from positive members of A001787. Column 2 is A006043. Column 3 is A006044. - _Omar E. Pol_, Jan 06 2009

%H G. C. Greubel, <a href="/A152818/b152818.txt">Antidiagonals n = 0..50, flattened</a>

%H F. A. Haight, <a href="http://www.jstor.org/stable/2333538">Overflow at a traffic light</a>, Biometrika, 46 (1959), 420-424. See page 422.

%H F. A. Haight, <a href="/A001787/a001787_3.pdf">Overflow at a traffic light</a>, Biometrika, 46 (1959), 420-424. (Annotated scanned copy)

%H F. A. Haight, <a href="/A001787/a001787_2.pdf">Letter to N. J. A. Sloane, n.d.</a>

%F E.g.f. for array as a triangle: exp(x)/(1-t*x*exp(x)) = 1+(1+t)*x+(1+4*t+2*t^2)*x^2/2! + (1+12*t+18*t^2+6*t^3)*x^3/3! + .... E.g.f. is int {z = 0..inf} exp(-z)*F(x,t*z), (x and t chosen sufficiently small for the integral to converge), where F(x,t) = exp(x*(1+t*exp(x))) is the e.g.f. for A154372. - _Peter Bala_, Oct 09 2011

%F From _Peter Bala_, Oct 09 2011: (Start)

%F From the e.g.f., the row polynomials R(n,t) satisfy the recursion R(n,t) = 1 + t*sum {k = 0..n-1} n!/(k!*(n-k-1)!)*R(n-k-1,t). The polynomials 1/n!*R(n,x) are the polynomials P(n,x) of A152650.

%F Sum_{k=0..n} T(n, k) = A072597(n) (antidiagonal sums). (End)

%F From _G. C. Greubel_, Apr 10 2023: (Start)

%F T(n, k) = (k+1)^(n-k) * k! * binomial(n, k) (antidiagonal triangle).

%F Sum_{k=0..n} (-1)^k*T(n, k) = A089148(n). (End)

%e From _Omar E. Pol_, Jan 06 2009: (Start)

%e Array begins:

%e 1, 1, 2, 6, 24, 120, ...

%e 1, 4, 18, 96, 600, 4320, ...

%e 1, 12, 108, 960, 9000, 90720, ...

%e 1, 32, 540, 7680, 105000, 1451520, ...

%e 1, 80, 2430, 53760, 1050000, 19595520, ...

%e 1, 192, 10206, 344064, 9450000, 235146240, ...

%e 1, 448, 40824, 2064384, 78750000, 2586608640, ...

%e 1, 1024, 157464, 11796480, 618750000, 26605117440, ...

%e 1, 2304, 590490, 64880640, 4640625000, 259399895040, ... (End)

%e Antidiagonal triangle:

%e 1;

%e 1, 1;

%e 1, 4, 2;

%e 1, 12, 18, 6;

%e 1, 32, 108, 96, 24;

%e 1, 80, 540, 960, 600, 120;

%e 1, 192, 2430, 7680, 9000, 4320, 720;

%e 1, 448, 10206, 53760, 105000, 90720, 35280, 5040;

%t len= 45; m= 1 + Ceiling[Sqrt[len]]; Sort[Flatten[#, 1] &[MapIndexed[ {(2 +#2[[1]]^2 +(#2[[2]] -1)*#2[[2]] +#2[[1]]*(2*#2[[2]] -3))/ 2, #1}&, Table[(k+1)^n*(n+k)!/n!, {n,0,m}, {k,0,m}], {2}]]][[All, 2]][[1 ;; len]] (* From _Jean-François Alcover_, May 27 2011 *)

%t T[n_, k_]:= (k+1)^(n-k)*k!*Binomial[n, k];

%t Table[T[n,k], {n,0,15}, {k,0,n}]//Flatten (* _G. C. Greubel_, Apr 10 2023 *)

%o (Sage)

%o def A152818_row(n):

%o R.<x> = ZZ[]

%o P = add((n-k+1)^k*x^(n-k+1)*factorial(n)/factorial(k) for k in (0..n))

%o return P.coefficients()

%o for n in (0..12): print(A152818_row(n)) # _Peter Luschny_, May 03 2013

%o (PARI) A(n,k) = (k+1)^n*(n+k)!/n! \\ _Charles R Greathouse IV_, Sep 10 2016

%o (Magma)

%o A152818:= func< n,k | (k+1)^(n-k)*Factorial(k)*Binomial(n,k) >;

%o [A152818(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Apr 10 2023

%Y Cf. A000012, A000142, A001563, A001787, A006043, A006044, A009998.

%Y Cf. A055533, A072597 (antidiagonal sums), A089148, A119502, A152650.

%Y Cf. A152656, A154372.

%K nonn,tabl

%O 0,5

%A _Paul Curtz_, Dec 13 2008

%E Better definition, extended and edited by _Omar E. Pol_ and _N. J. A. Sloane_, Jan 05 2009