A127147 -id:A127147 - cp's OEIS frontend

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), mQ(m, 2k+1) = (m - 2k)Q(m+1, 2k) - 2k(m+1)*Q(m+2, 2k-1).

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

Offset: 0

N. J. A. Sloane, Mar 24 2007

Comment from N. J. A. Sloane, Jan 29 2020: (Start)
It looks like there was a missing 2 in the definition, which I have now corrected.  The old definition was:
(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)

			Array begins:
     1,    1,    1,    1,    1,   1,   1,    1,    1,    1, ... (A000012)
     1,    1,    1,    1,    1,   1,   1,    1,    1,    1, ... (A000012)
    -2,   -1,    0,    1,    2,   3,   4,    5,    6,    7, ... (A023444)
    -5,   -4,   -3,   -2,   -1,   0,   1,    2,    3,    4, ... (A023447)
    12,    3,   -4,   -9,  -12, -13, -12,   -9,   -4,    3, ... (A127146)
    43,   28,   15,    4,   -5, -12, -17,  -20,  -21,  -20, ... (A127147)
  -120,  -15,   48,   75,   72,  45,   0,  -57, -120, -183, ... (A127148)
  -531, -288, -105,   24,  105, 144, 147,  120,   69,    0, ...
  1680,  105, -624, -735, -432, 105, 720, 1281, 1680, 1833, ...

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

G. C. Greubel, Antidiagonals n = 0..100, flattened

See A105937 for another version.
Columns give A127137, A127138, A127144, A127145;
Rows give A127146, A127147, A127148.

f:= proc(k) option remember;
      if `mod`(k,2)=0 then k!/(k/2)!
    else 2^(k-1)*((k-1)/2)!*add(binomial(2*j, j)/8^j, j=0..((k-1)/2))
      fi; end;
Q:= proc(n, k) option remember;
      if n=0 then (-1)^binomial(k, 2)*f(k)
    elif k<2 then 1
    elif `mod`(k,2)=0 then (n-k+1)*Q(n+1,k-1) - (k-1)*Q(n+2,k-2)
    else ( (n-k+1)*Q(n+1,k-1) - (k-1)*(n+1)*Q(n+2,k-2) )/n
      fi; end;
seq(seq(Q(n-k, k), k=0..n), n=0..12); # G. C. Greubel, Jan 30 2020

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}] ];
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]];
Table[Q[n-k,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 30 2020 *)

@CachedFunction
def f(k):
    if (mod(k, 2)==0): return factorial(k)/factorial(k/2)
    else: return 2^(k-1)*factorial((k-1)/2)*sum(binomial(2*j, j)/8^j for j in (0..(k-1)/2))
def Q(n,k):
    if (n==0): return (-1)^binomial(k, 2)*f(k)
    elif (k<2): return 1
    elif (mod(k,2)==0): return (n-k+1)*Q(n+1,k-1) - (k-1)*Q(n+2,k-2)
    else: return ( (n-k+1)*Q(n+1,k-1) - (k-1)*(n+1)*Q(n+2,k-2) )/n
[[Q(n-k,k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 30 2020

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).

More terms added by G. C. Greubel, Jan 30 2020

A154599 a(n) = 2n^2 + 20n + 8.

Original entry on oeis.org

30, 56, 86, 120, 158, 200, 246, 296, 350, 408, 470, 536, 606, 680, 758, 840, 926, 1016, 1110, 1208, 1310, 1416, 1526, 1640, 1758, 1880, 2006, 2136, 2270, 2408, 2550, 2696, 2846, 3000, 3158, 3320, 3486, 3656, 3830, 4008, 4190, 4376, 4566, 4760, 4958, 5160

Offset: 1

Vincenzo Librandi, Jan 12 2009

Tenth diagonal of A144562.

2*a(n) + 84 is a square.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A067076, A127147, A144562, A153238.

I:=[30, 56, 86]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 26 2012

LinearRecurrence[{3, -3, 1}, {30, 56, 86}, 50] (* Vincenzo Librandi, Feb 26 2012 *)
Table[2n^2+20n+8,{n,50}] (* Harvey P. Dale, Jun 15 2019 *)

for(n=1, 40, print1(2*n^2+20*n+8", ")); \\ Vincenzo Librandi, Feb 26 2012

[2*n^2+20*n+8 for n in range(1,41)] # G. C. Greubel, May 30 2024

From R. J. Mathar, Jan 05 2011: (Start)
a(n) = 2*A127147(n+13).
G.f.: 2*x*(5-4*x)*(3-x)/(1-x)^3. (End)
From Amiram Eldar, Feb 25 2023: (Start)
Sum_{n>=1} 1/a(n) = 79/952 - cot(sqrt(21)*Pi)*Pi/(4*sqrt(21)).
Sum_{n>=1} (-1)^(n+1)/a(n) = 2851/14280 - cosec(sqrt(21)*Pi)*Pi/(4*sqrt(21)). (End)
E.g.f.: 2*(-4 + (4 + 11*x + x^2)*exp(x)). - G. C. Greubel, May 30 2024

A164582 a(n) = 5*a(n - 1) - a(n - 2), with n>2, a(1)=2, a(2)=3.

Original entry on oeis.org

2, 3, 13, 62, 297, 1423, 6818, 32667, 156517, 749918, 3593073, 17215447, 82484162, 395205363, 1893542653, 9072507902, 43468996857, 208272476383, 997893385058, 4781194448907, 22908078859477, 109759199848478, 525887920382913, 2519680402066087

Offset: 1

Vincenzo Librandi, Aug 17 2009

From Klaus Purath, Aug 18 2024: (Start)
For any two consecutive terms (x,y), x^2 - 5xy + y^2 = -17 = A127147(6) always applies. In general, the following applies to all recurrences (t) with constant coefficients (5,-1) and t(0) = 2 and two consecutive terms (x,y): x^2 - 5xy + y^2 = A127147(t(1)+3) for any integer t(1). This includes and interprets the Feb 08 2014 comment on A003501 by Colin Barker.
By analogy to this, for three consecutive terms (x,y,z) of any recurrence (t) of the  form (5,-1) with t(0) = 2: y^2 - xz = A127147(t(1)+3).
a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/6, where (t) is any third order recurrence with constant coefficients (6,-6,1) and initial values t(0) = x, t(1) = x + 2, t(2) = x + 5 for any integer x.
a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/4, where (t) is any third order recurrence with constant coefficients (4,4,-1) and initial values t(0) = x, t(1) = 2 - x, t(2) = x + 1 for any integer x. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-1).

[n le 2 select n+1 else 5*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 12 2013

CoefficientList[Series[(2 - 7 x) / (1 - 5 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 12 2013 *)
LinearRecurrence[{5,-1},{2,3},30] (* Harvey P. Dale, Apr 06 2016 *)

Vec(x*(2 - 7*x) / (1 - 5*x + x^2) + O(x^30)) \\ Colin Barker, Nov 08 2017

a(n) = 5*a(n-1) - a(n-2) = 2*A004254(n) - 7*A004254(n-1).
G.f.: x*(2-7*x) / (1-5*x+x^2).
a(n) = (2^(-1-n)*((5+sqrt(21))^n*(-31+7*sqrt(21)) + (5-sqrt(21))^n*(31+7*sqrt(21)))) / sqrt(21). - Colin Barker, Nov 08 2017
a(n) = (a(n-1)^2 + 17)/a(n-2). - Klaus Purath, Aug 30 2020

Extended by R. J. Mathar, Aug 19 2009

cp's OEIS Frontend

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).

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A154599 a(n) = 2*n^2 + 20*n + 8.

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A164582 a(n) = 5*a(n - 1) - a(n - 2), with n>2, a(1)=2, a(2)=3.

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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), mQ(m, 2k+1) = (m - 2k)Q(m+1, 2k) - 2k(m+1)*Q(m+2, 2k-1).

A154599 a(n) = 2n^2 + 20n + 8.