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A127080 Infinite square array read by antidiagonals: Q(m, 0) = 1, Q(m, 1) = 1; Q(m, 2k) = (m - 2k + 1)*Q(m+1, 2k-1) - (2k-1)*Q(m+2,2k-2), m*Q(m, 2k+1) = (m - 2k)*Q(m+1, 2k) - 2k(m+1)*Q(m+2, 2k-1). 10

%I #19 Jan 22 2022 23:46:32

%S 1,1,1,1,1,-2,1,1,-1,-5,1,1,0,-4,12,1,1,1,-3,3,43,1,1,2,-2,-4,28,-120,

%T 1,1,3,-1,-9,15,-15,-531,1,1,4,0,-12,4,48,-288,1680,1,1,5,1,-13,-5,75,

%U -105,105,8601,1,1,6,2,-12,-12,72,24,-624,3984,-30240,1,1,7,3,-9,-17,45,105,-735,945,-945,-172965

%N Infinite square array read by antidiagonals: Q(m, 0) = 1, Q(m, 1) = 1; Q(m, 2k) = (m - 2k + 1)*Q(m+1, 2k-1) - (2k-1)*Q(m+2,2k-2), m*Q(m, 2k+1) = (m - 2k)*Q(m+1, 2k) - 2k(m+1)*Q(m+2, 2k-1).

%C Comment from _N. J. A. Sloane_, Jan 29 2020: (Start)

%C It looks like there was a missing 2 in the definition, which I have now corrected. The old definition was:

%C (Wrong!) Infinite square array read by antidiagonals: Q(m, 0) = 1, Q(m, 1) = 1; Q(m, 2k) = (m - 2k + 1)*Q(m+1, 2k-1) - (2k-1)*Q(m+2, k-2), m*Q(m, 2k+1) = (m - 2k)*Q(m+1, 2k) - 2k(m+1)*Q(m+2, 2k-1). (Wrong!) (End)

%D V. van der Noort and N. J. A. Sloane, Paper in preparation, 2007.

%H G. C. Greubel, <a href="/A127080/b127080.txt">Antidiagonals n = 0..100, flattened</a>

%F E.g.f.: Sum_{k >= 0} Q(m,2k) x^k/k! = (1+4x)^((m-1)/2)/(1+2x)^(m/2), Sum_{k >= 0} Q(m,2k+1) x^k/k! = (1+4x)^((m-2)/2)/(1+2x)^((m+1)/2).

%e Array begins:

%e 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ... (A000012)

%e 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ... (A000012)

%e -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, ... (A023444)

%e -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, ... (A023447)

%e 12, 3, -4, -9, -12, -13, -12, -9, -4, 3, ... (A127146)

%e 43, 28, 15, 4, -5, -12, -17, -20, -21, -20, ... (A127147)

%e -120, -15, 48, 75, 72, 45, 0, -57, -120, -183, ... (A127148)

%e -531, -288, -105, 24, 105, 144, 147, 120, 69, 0, ...

%e 1680, 105, -624, -735, -432, 105, 720, 1281, 1680, 1833, ...

%p f:= proc(k) option remember;

%p if `mod`(k,2)=0 then k!/(k/2)!

%p else 2^(k-1)*((k-1)/2)!*add(binomial(2*j, j)/8^j, j=0..((k-1)/2))

%p fi; end;

%p Q:= proc(n, k) option remember;

%p if n=0 then (-1)^binomial(k, 2)*f(k)

%p elif k<2 then 1

%p elif `mod`(k,2)=0 then (n-k+1)*Q(n+1,k-1) - (k-1)*Q(n+2,k-2)

%p else ( (n-k+1)*Q(n+1,k-1) - (k-1)*(n+1)*Q(n+2,k-2) )/n

%p fi; end;

%p seq(seq(Q(n-k, k), k=0..n), n=0..12); # _G. C. Greubel_, Jan 30 2020

%t Q[0, k_]:= Q[0,k]= (-1)^Binomial[k, 2]*If[EvenQ[k], k!/(k/2)!, 2^(k-1)*((k-1)/2)!* Sum[Binomial[2*j, j]/8^j, {j, 0, (k-1)/2}] ];

%t Q[n_, k_]:= Q[n,k]= If[k<2, 1, If[EvenQ[k], (n-k+1)*Q[n+1, k-1] - (k-1)*Q[n+2, k-2], ((n -k+1)*Q[n+1, k-1] - (k-1)*(n+1)*Q[n+2, k-2])/n]];

%t Table[Q[n-k,k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jan 30 2020 *)

%o (Sage)

%o @CachedFunction

%o def f(k):

%o if (mod(k, 2)==0): return factorial(k)/factorial(k/2)

%o else: return 2^(k-1)*factorial((k-1)/2)*sum(binomial(2*j, j)/8^j for j in (0..(k-1)/2))

%o def Q(n,k):

%o if (n==0): return (-1)^binomial(k, 2)*f(k)

%o elif (k<2): return 1

%o elif (mod(k,2)==0): return (n-k+1)*Q(n+1,k-1) - (k-1)*Q(n+2,k-2)

%o else: return ( (n-k+1)*Q(n+1,k-1) - (k-1)*(n+1)*Q(n+2,k-2) )/n

%o [[Q(n-k,k) for k in (0..n)] for n in (0..12)] # _G. C. Greubel_, Jan 30 2020

%Y See A105937 for another version.

%Y Columns give A127137, A127138, A127144, A127145;

%Y Rows give A127146, A127147, A127148.

%K sign,tabl

%O 0,6

%A _N. J. A. Sloane_, Mar 24 2007

%E More terms added by _G. C. Greubel_, Jan 30 2020

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Last modified April 24 13:58 EDT 2024. Contains 371960 sequences. (Running on oeis4.)