A000581 -id:A000581 - cp's OEIS frontend

A096754 Triangle read by rows giving coefficients of the trigonometric expansion of Cos(n*x).

1, 1, 0, -1, 1, 0, -3, 1, 0, -6, 0, 1, 1, 0, -10, 0, 5, 1, 0, -15, 0, 15, 0, -1, 1, 0, -21, 0, 35, 0, -7, 1, 0, -28, 0, 70, 0, -28, 0, 1, 1, 0, -36, 0, 126, 0, -84, 0, 9, 1, 0, -45, 0, 210, 0, -210, 0, 45, 0, -1, 1, 0, -55, 0, 330, 0, -462, 0, 165, 0, -11, 1, 0, -66, 0, 495, 0, -924, 0, 495, 0, -66, 0, 1, 1, 0, -78, 0, 715

Offset: 1

Robert G. Wilson v, Jul 07 2004

T(n,k)=cos(n,k)*cos(pi*k/2) begins {1}, {1,0}, {1,0,-1}, {1,0,-3,0},... - Paul Barry, May 21 2006

			The trigonometric expansion of Cos(4x) = Cos[x]^4 - 6*Cos[x]^2*Sin[x]^2 + Sin[x]^4, therefore the fourth row is 1, 0, -6, 0, 1.
The trigonometric expansion of Cos(5x) = Cos[x]^5 - 10*Cos[x]^3*Sin[x]^2 + 5*Cos[x]*Sin[x]^4, therefore the fifth row of the triangle is 1, 0, -10, 0, 5
The table begins:
1
1 0 -1
1 0 -3
1 0 -6 0 1
1 0 -10 0 5
1 0 -15 0 15 0 -1
1 0 -21 0 35 0 -7
1 0 -28 0 70 0 -28 0 1

Clark Kimberling, Polynomials associated with reciprocation, JIS 12 (2009) 09.3.4, section 5.

Another version of the triangle in A034839. Cf. A095704.
First column is A000012 = C(n, 0), third column is A000217 = C(n, 2), fifth column is A000332 = C(n, 4), seventh column is A000579 = C(n, 6), ninth column is A000581 = C(n, 8).
A001287 = C(n, 10), A010965 = C(n, 12), A010967 = C(n, 14), A010969 = C(n, 16), A010971 = C(n, 18),
A010973 = C(n, 20), A010975 = C(n, 22), A010977 = C(n, 24), A010979 = C(n, 26), A010981 = C(n, 28),
A010983 = C(n, 30), A010985 = C(n, 32), A010987 = C(n, 34), A010989 = C(n, 36), A010991 = C(n, 38),
A010993 = C(n, 40), A010995 = C(n, 42), A010997 = C(n, 44), A010999 = C(n, 46), A011001 = C(n, 48),
A017714 = C(n, 50), A017716 = C(n, 52), A017718 = C(n, 54), A017720 = C(n, 56), etc.

Flatten[Table[ Plus @@ CoefficientList[ TrigExpand[ Cos[n*x]], { Cos[x], Sin[x]}], {n, 13}]]

A116690 a(n) = C(n,8) + C(n,7) + C(n,6) + C(n,5) + C(n,4) + C(n,3) + C(n,2) + C(n,1).

Original entry on oeis.org

0, 1, 3, 7, 15, 31, 63, 127, 255, 510, 1012, 1980, 3796, 7098, 12910, 22818, 39202, 65535, 106761, 169765, 263949, 401929, 600369, 880969, 1271625, 1807780, 2533986, 3505698, 4791322, 6474540, 8656936, 11460948, 15033172, 19548045

Offset: 0

Jonathan Vos Post, Mar 15 2006

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

[n*(n+1)*(25584 - 9604*n + 5264*n^2 - 1295*n^3 + 231*n^4 - 21*n^5 + n^6 ) /40320: n in [0..30]]; // G. C. Greubel, Nov 25 2017

seq(sum(binomial(n,k),k=1..8),n=0..33); # Zerinvary Lajos, Dec 14 2007

Table[n*(n + 1)*(25584 - 9604*n + 5264*n^2 - 1295*n^3 + 231*n^4 - 21*n^5 + n^6)/40320, {n, 0, 50}] (* G. C. Greubel, Nov 25 2017 *)
Table[Total[Binomial[n,Range[8]]],{n,0,40}] (* Harvey P. Dale, Aug 14 2023 *)

for(n=0,30, print1(n*(n+1)*(25584 - 9604*n + 5264*n^2 - 1295*n^3 + 231*n^4 - 21*n^5 + n^6 ) /40320, ", ")) \\ G. C. Greubel, Nov 25 2017

[binomial(n,2)+binomial(n,4)+binomial(n,6)+binomial(n,8) for n in range(1, 35)] # Zerinvary Lajos, May 17 2009

[binomial(n,2)+binomial(n,4)+binomial(n,6)+binomial(n,8)+binomial(n,1)+binomial(n,3)+binomial(n,5)+binomial(n,7)for n in range(0, 34)] # Zerinvary Lajos, May 17 2009

a(n) = A000581(n) + A000580(n) + A000579(n) + A000389(n) + A000332(n) + A000292(n) + A000217(n) + n. a(n) = A000581(n) + A116082(n).
G.f. ( -x*(2*x^2 - 2*x + 1)*(2*x^4 - 4*x^3 + 6*x^2 - 4*x + 1) ) / (x-1)^9. - R. J. Mathar, Oct 21 2011
a(n) = n*(n+1)*(25584 - 9604*n + 5264*n^2 - 1295*n^3 + 231*n^4 - 21*n^5 + n^6)/40320. - G. C. Greubel, Nov 25 2017

A122960 Triangle T(n,k), 0 <= k <= n, read by rows given by [0, 1, -1, -1, 1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 0, 3, 1, 0, 1, 0, 6, 1, 0, 0, 5, 0, 10, 1, 0, 1, 0, 15, 0, 15, 1, 0, 0, 7, 0, 35, 0, 21, 1, 0, 1, 0, 28, 0, 70, 0, 28, 1, 0, 0, 9, 0, 84, 0, 126, 0, 36, 1, 0, 1, 0, 45, 0, 210, 0, 210, 0, 45, 1

Offset: 0

Philippe Deléham, Oct 26 2006

T(n,k) = binomial (n,n-k+1) if (n-k) is an odd number (see A000217, A000332, A000579, A000581, ...). T(n,k)= 0 if (n-k)=2x with x > 0 (see A000004). T(n,n)=1 (see A000012).

			Triangle begins:
  1;
  0, 1;
  0, 1, 1;
  0, 0, 3,  1;
  0, 1, 0,  6,  1;
  0, 0, 5,  0, 10,   1;
  0, 1, 0, 15,  0,  15,   1;
  0, 0, 7,  0, 35,   0,  21,   1;
  0, 1, 0, 28,  0,  70,   0,  28,  1;
  0, 0, 9,  0, 84,   0, 126,   0, 36,  1;
  0, 1, 0, 45,  0, 210,   0, 210,  0, 45, 1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A000004, A000012, A000217, A000332, A000579, A000581, A007318.

T:= proc(n, k) option remember;
      if k<0 or k>n then 0
    elif k=n then 1
    elif n=2 and k=1 then 1
    elif k=0 then 0
    else 3*T(n-1, k-1) + T(n-2, k) - 3*T(n-2, k-2) - T(n-3, k-1) + T(n-3, k-3)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Feb 17 2020

With[{m = 10}, CoefficientList[CoefficientList[Series[(1-2*x*y-x^2+x^2*y^2+
x^2*y)/(1-3*x*y-x^2+3*x^2*y^2+x^3*y-x^3*y^3), {x, 0 , m}, {y, 0, m}], x], y]] // Flatten (* Georg Fischer, Feb 17 2020 *)

T(n, k) = if(k<0 || k>n, 0, if(k==n, 1, if(n==2 && k==1, 1, if(k==0, 0, 3*T(n-1, k-1) + T(n-2, k) - 3*T(n-2, k-2) - T(n-3, k-1) + T(n-3, k-3) )))); \\ G. C. Greubel, Feb 17 2020

@CachedFunction
def T(n, k):
    if (k<0 or k>n): return 0
    elif (k==n): return 1
    elif (n==2 and k==1): return 1
    elif (k==0): return 0
    else: return 3*T(n-1, k-1) + T(n-2, k) - 3*T(n-2, k-2) - T(n-3, k-1) + T(n-3, k-3)
print([[T(n, k) for k in (0..n)] for n in (0..10)]) # G. C. Greubel, Feb 17 2020

Sum_{k=0..n} T(n,k) = A011782(n).
Sum_{k=0..n} 2^k*T(n,k) = A083323(n).
Sum_{k=0..n} 2^(n-k)*T(n,k) = A122983(n).
G.f.: (1 - 2*x*y - x^2 + x^2*y^2 + x^2*y)/(1 - 3*x*y - x^2 + 3*x^2*y^2 + x^3*y - x^3*y^3). - Philippe Deléham, Nov 09 2013
T(n,k) = 3*T(n-1,k-1) + T(n-2,k) - 3*T(n-2,k-2) - T(n-3,k-1) + T(n-3,k-3), T(0,0) = T(1,1) = T(2,1) = T(2,2) = 1, T(1,0) = T(2,0) = 0, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 09 2013

a(12) corrected by Georg Fischer, Feb 17 2020

A165618 a(n) = binomial(n+8,8) - 1.

Original entry on oeis.org

0, 8, 44, 164, 494, 1286, 3002, 6434, 12869, 24309, 43757, 75581, 125969, 203489, 319769, 490313, 735470, 1081574, 1562274, 2220074, 3108104, 4292144, 5852924, 7888724, 10518299, 13884155, 18156203, 23535819, 30260339, 38608019, 48903491

Offset: 0

Enrique Pérez Herrero, Sep 22 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
E. Pérez Herrero, Binomial Matrix (I) partitions, Psychedelic Geometry Blogspot, 09/22/09
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000096, A000581, A062748, A063258, A062988, A124089, A124090, A035927.

Table[ -1 + Binomial[n + 8, 8], {n, 0, 30}]
LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{0,8,44,164,494,1286,3002,6434,12869},40] (* Harvey P. Dale, Nov 18 2013 *)

vector(100,n,binomial(n+7,8)-1) \\ Charles R Greathouse IV, May 27 2011

a(n) = binomial(n+8,8) - 1 = A000581(n+8) - 1.
a(n) = Sum_{r=1..n} binomial(8,r)*binomial(n,r).
a(n) = n(n+9)(n^6 + 27n^5 + 303n^4 + 1809n^3 + 6168n^2 + 11772n + 12176)/40320.

Edited by Charles R Greathouse IV, May 27 2011

A201458 Expansion of 1/((1-2x)(1-3x+3x^2)(1-4x+6x^2-4x^3)).

Original entry on oeis.org

1, 9, 46, 175, 551, 1520, 3811, 8921, 19922, 43211, 92363, 196608, 419295, 897565, 1926458, 4135255, 8854359, 18875392, 40024059, 84417521, 177221602, 370688979, 773342163, 1610612736, 3350668423, 6964989333, 14466833194, 30021724351, 62233946303

Offset: 0

Bruno Berselli, Dec 01 2011

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9,-35,76,-98,72,-24).

Cf. for type of g.f.: A099855, with 1/((1-2*x+2*x^2)*(1-4*x+4*x^2)); A000581, with 1/((1-x)^2*(1-x)^3*(1-x)^4).

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/((1-2*x)*(1-3*x+3*x^2)*(1-4*x+6*x^2-4*x^3))));

CoefficientList[Series[1/((1-2*x)*(1-3*x+3*x^2)*(1-4*x+6*x^2-4*x^3)), {x, 0, 30}], x]
LinearRecurrence[{9,-35,76,-98,72,-24},{1,9,46,175,551,1520},30] (* Harvey P. Dale, Feb 01 2012 *)

makelist(coeff(taylor(1/((1-2*x)*(1-3*x+3*x^2)*(1-4*x+6*x^2-4*x^3)), x, 0, n), x, n), n, 0, 29);

Vec(1/((1-2*x)*(1-3*x+3*x^2)*(1-4*x+6*x^2-4*x^3))+O(x^30))

G.f.: 1/((1-2*x)^2*(1-2*x+2*x^2)*(1-3*x+3*x^2)) = 1/((1-2*x+2*x^2)*(1-3*x+3*x^2)*(1-4*x+4*x^2)).
a(n) = 9*a(n-1)-35*a(n-2)+76*a(n-3)-98*a(n-4)+72*a(n-5)-24*a(n-6) for a(-5)=a(-4)=a(-3)=a(-2)=a(-1)=0, a(0)=1.
a(n) = 8*2^n*(n+1)+2*((1-i)^(n-1)+(1+i)^(n-1))+((3+i*sqrt(3))/2)^(n+4)+((3-i*sqrt(3))/2)^(n+4), where i=sqrt(-1).

A206294 Riordan array (1, x/(1-x)^3).

Original entry on oeis.org

1, 0, 1, 0, 3, 1, 0, 6, 6, 1, 0, 10, 21, 9, 1, 0, 15, 56, 45, 12, 1, 0, 21, 126, 165, 78, 15, 1, 0, 28, 252, 495, 364, 120, 18, 1, 0, 36, 462, 1287, 1365, 680, 171, 21, 1, 0, 45, 792, 3003, 4368, 3060, 1140, 231, 24, 1

Offset: 0

Philippe Deléham, Feb 05 2012

The convolution triangle of the triangular numbers A000217. - Peter Luschny, Oct 07 2022

			Triangle begins:
1
0, 1
0, 3, 1
0, 6, 6, 1
0, 10, 21, 9, 1
0, 15, 56, 45, 12, 1
0, 21, 126, 165, 78, 15, 1
0, 28, 252, 495, 364, 120, 18, 1
0, 36, 462, 1287, 1365, 680, 171, 21, 1
0, 45, 792, 3003, 4368, 3060, 1140, 231, 24, 1
0, 55, 1287, 6435, 12376, 11628, 5985, 1771, 300, 27, 1
0, 66, 2002, 12870, 31824, 38760, 26324, 10626, 2600, 378, 30, 1

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. Columns: A000007, A000217 (triangular numbers), A000389, A000581, A001288, A010967..(+3)..A011000, A017714..(+3)..A017762.
Row sums are A052529.
Cf. A127893.

# Uses function PMatrix from A357368.
PMatrix(10, n -> n * (n + 1) / 2); # Peter Luschny, Oct 07 2022

Table[If[n == 0 && k == 0 , 1, Binomial[n - 1 + 2 k, n - k]], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Nov 25 2017 *)

{T(n,k)=polcoeff(1/(1-x+x*O(x^(n-k)))^(3*k),n-k)}

{T(n,k)=polcoeff(polcoeff((1-x)^3/((1-x)^3-y*x +x*O(x^n)),n,x),k,y)}
for(n=0,12,for(k=0,n,print1(T(n,k),", "));print(""))

Triangle T(n,k), read by rows, given by (0, 3, -1, 2/3, -1/6, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.
T(n,0) = 0^n, T(n,k) = C(n-1+2k, n-k) for k > 0.
T(n,n) = 1, T(k+1,k) = 3*k = A008585(k), T(k+2,k) = A081266(k).
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A052529(n), A052910(n) for x = 0, 1, 2 respectively.
G.f.: (1-x)^3/((1-x)^3-y*x).

A270950 Number of distinct cardinalities of orbits of lattice points under the automorphism group of the n-dimensional integer lattice.

Original entry on oeis.org

0, 1, 2, 5, 9, 12, 20, 29, 40, 53, 76, 99, 132, 172, 216, 270, 341, 424, 532, 660, 810, 983, 1210, 1446, 1750, 2111, 2508, 2975, 3569, 4197, 4948, 5807, 6817, 7963, 9351, 10863, 12604, 14598, 16892, 19439, 22472, 25780, 29588, 33892, 38800, 44206, 50463, 57297, 65086, 73919, 83842, 94510

Offset: 0

Philippe A.J.G. Chevalier, Mar 26 2016

A finite number of orbits partition hypercubic shells of infinity norm s in the n-dimensional integer lattice. The number of orbits is given by C(n+s-1,s). The number of distinct cardinalities of the orbits of lattice points under the automorphism group of the n-dimensional integer lattice is found under the condition that n <= s.

A new connection was discovered using the partition of the dimension 'n'. These partitions create a base set of cardinalities. Each of these cardinalities can be subjected to the process of prime factorization. The prime factorization yields the exponents of the primes that form lattice points in a new integer lattice of dimension 'n'. These lattice points become elements of a set A. The unique summands of a specific partition of 'n' give the multipliers of the base vector (1,0^n) that need to be subtracted from the specific partition representative element of set A. The cardinality of the set A increases until all the specific partitions of 'n' have been processed. This augmented set A* has the correct cardinality. This method is much faster than the brute force technique. - Philippe A.J.G. Chevalier, Jun 24 2022

			For n=0 the a(0)=0.
For n=3 we have the following distinct cardinalities of the orbits 6, 8, 12, 24, 48 and thus a(3)=5.
For n=4 we have the distinct cardinalities of the orbits 8, 16, 24, 32, 48, 64, 96, 192, 384 and thus a(4)=9.
For n=5 we have the distinct cardinalities of the orbits 10, 32, 40, 160, 240, 320, 480, 640, 960, 1920, 3840 and thus a(5)=12.

Philippe A.J.G. Chevalier, Table of n, a(n) for n = 0..54

Cf. A000217, A000292, A000332, A000389, A000579, A000580, A000581, A000582, A001287.

a(17) corrected and a(18)-a(51) from Philippe A.J.G. Chevalier, Jun 24 2022

A275514 Triangle read by rows: the coefficient [t^k] of the Ehrhart polynomial of the 2-hypersimplex in dimension n.

Original entry on oeis.org

1, 1, -1, 1, 0, 0, 1, 2, 1, 0, 1, 5, 5, 0, 0, 1, 9, 15, 1, 0, 0, 1, 14, 35, 7, 0, 0, 0, 1, 20, 70, 28, 1, 0, 0, 0, 1, 27, 126, 84, 9, 0, 0, 0, 0, 1, 35, 210, 210, 45, 1, 0, 0, 0, 0, 1, 44, 330, 462, 165, 11, 0, 0, 0, 0, 0, 1, 54, 495, 924, 495, 66, 1, 0, 0, 0, 0, 0, 1, 65, 715

Offset: 1

R. J. Mathar, Jul 31 2016

			The triangle starts in row n=1 with coefficients 0<=k<n as:
1;
1, -1;
1, 0, 0;
1, 2, 1, 0;
1, 5, 5, 0, 0;
1, 9, 15, 1, 0, 0;
1, 14, 35, 7, 0, 0, 0;
1, 20, 70, 28, 1, 0, 0, 0;
1, 27, 126, 84, 9, 0, 0, 0, 0;
1, 35, 210, 210, 45, 1, 0, 0, 0, 0;
1, 44, 330, 462, 165, 11, 0, 0, 0, 0, 0;
1, 54, 495, 924, 495, 66, 1, 0, 0, 0, 0, 0;
1, 65, 715, 1716, 1287, 286, 13, 0, 0, 0, 0, 0, 0;

Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. II. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 10.
Nan Li, Ehrhart h*-vectors of hypersimplices, Discr. Comp. Geom. 48 (2012) 847-878, Eq. (1.1)

Cf. A010054 (coefficients for the 1-hypersimplex), A258993, A000295 (row sums), A000096 (column k=1), A000332 (k=2), A000579 (k=3), A000581 (k=4), A001287 (k=5).

subki := proc(n,r,l)
    local i,t;
    add(t^i,i=0..l-1) ;
    %^n ;
    expand(%) ;
    coeff(%,t,r) ;
end proc:
hstard := proc(d,k,n)
    add((-1)^i*binomial(n,i)*subki(n, (k-i)*d-i,k-i) ,i=0..k-1) ;
end proc:
A275514 := proc(n,k)
    hstard(k,2,n)
end proc:

A303987 Triangle read by rows: T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0, k = 0..n.

Original entry on oeis.org

1, 1, 4, 1, 36, 36, 1, 144, 900, 400, 1, 400, 8100, 19600, 4900, 1, 900, 44100, 313600, 396900, 63504, 1, 1764, 176400, 2822400, 9922500, 7683984, 853776, 1, 3136, 571536, 17640000, 133402500, 276623424, 144288144, 11778624, 1, 5184, 1587600, 85377600, 1200622500, 5194373184, 7070119056, 2650190400, 165636900

Offset: 0

Wolfdieter Lang, May 14 2018

The row sums of this triangle are b(n) = A005259(n), for n >= 0. This sequence b was used in R. Apéry's 1979 proof of the irrationality of Zeta(3). See A005259 for references and links.

Row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k = hypergeometric([-n, -n, n+1, n+1], [1, 1, 1], x), hence b(n) = hypergeometric([-n, -n, n+1, n+1], [1, 1, 1], 1) (see the formula in A005259 given by K. A. Penson. This is the solution to Exercise 2.14 of the Koepf reference given there, p. 29).

			The triangle T begins:
n\k  0    1       2        3          4          5          6          7 ...
0:   1
1:   1    4
2:   1   36      36
3:   1  144     900      400
4:   1  400    8100    19600       4900
5:   1  900   44100   313600     396900      63504
6:   1 1764  176400  2822400    9922500    7683984     853776
7:   1 3136  571536 17640000  133402500  276623424  144288144   11778624
----------------------------------------------------------------------------
row n = 8:   1 5184 1587600 85377600 1200622500 5194373184 7070119056 2650190400 165636900,
row n = 9: 1 8100 3920400 341510400 8116208100 63631071504 176752976400 169612185600 47869064100 2363904400,
row n = 10: 1 12100 8820900 1177862400 44188244100 572679643536 2828047622400 5446435737600 3877394192100 853369488400 34134779536.
...

Cf. A000984, A002894, A005259, A008459, A063007.

The column sequences (without zeros) are A000012, A035287(n+1) = 4*A000217(n)^2, 36*A288876, 400*A000579(n+6)^2, 4900*A000581(n+8)^2, 63504*A001287(n+10)^2, ...

Flat(List([0..10],n->List([0..n],k->(Binomial(n,k)*Binomial(n+k,k))^2))); # Muniru A Asiru, May 15 2018

T[n_, k_] := (Gamma[k + n + 1]/(Gamma[k + 1]^2*Gamma[-k + n + 1]))^2;
Flatten[Table[T[n, k], {n, 0, 8}, {k, 0, n}]] (* Peter Luschny, May 14 2018 *)

T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0 and k = 0..n.
T(n, k) = (binomial(n+k, 2*k)*cbi(k))^2, with cbi(k) = A000984(k) = binomial(2*k, k), and cbi(k)^2 = A002894(k).
G.f. for column sequences (without leading zeros):
  cbi(k)^2*P2(2*k, x)/(1 - x)^(4*k+1), with the row polynomials of A008459 (Pascal entries squared) P2(2*k, x) = Sum_{j=0..2*k} A008459(2*k, j)*x^j. For a proof see the general comment in A288876 on the diagonals and columns of A008459.

A306571 Number of ways to write n as w(w+1) + C(x+3,4) + C(y+5,6) + C(z+7,8) with w,x,y,z nonnegative integers, where C(m,k) denotes the binomial coefficient m!/(k!(m-k)!).

Original entry on oeis.org

1, 3, 4, 4, 3, 2, 3, 6, 7, 5, 4, 4, 6, 7, 7, 6, 6, 6, 6, 5, 4, 8, 9, 6, 3, 4, 4, 6, 7, 7, 6, 8, 5, 5, 4, 8, 10, 9, 5, 5, 5, 5, 8, 10, 10, 9, 8, 7, 9, 9, 7, 8, 8, 7, 7, 6, 7, 12, 12, 8, 2, 3, 6, 11, 9, 8, 9, 7, 2, 4, 5, 8, 13, 14, 8, 6, 6, 8, 9, 9, 11, 8, 7, 7, 10, 9, 10, 11, 8, 7, 9, 11, 13, 11, 8, 5, 6, 7, 10, 10, 13

Offset: 0

Zhi-Wei Sun, Feb 24 2019

nonn

Conjecture: a(n) > 0 for all n >= 0. In other words, each n = 0,1,2,... can be written as 2*C(w,2) + C(x,4) + C(y,6) + C(z,8) with w,x,y,z positive integers.
We have verified a(n) > 0 for all n = 0..2*10^7.
Note that 10413917 is the least positive integer not representable as w^2 + C(x,4) + C(y,6) + C(z,8) with w,x,y,z nonnegative integers.
See also A306477 for a similar conjecture.

			a(0) = 1 with 0 = 0*1 + C(3,4) + C(5,6) + C(7,8).
a(60) = 2 with 60 = 0*1 + C(6,4) + C(5,6) + C(10,8) = 5*6 + C(4,4) + C(8,6) + C(8,8).
a(220544) = 1 with 220544 = 151*152 + C(48,4) + C(14,6) + C(9,8).
a(809165) = 1 with 809165 = 295*296 + C(63,4) + C(10,6) + C(20,8).
a(16451641) = 1 with 16451641 = 2256*2257 + C(130,4) + C(12,6) + C(10,8).

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Cf. A000332, A000579, A000581, A002378, A306459, A306460, A306462, A306471, A306477.

f[m_,n_]:=f[m,n]=Binomial[m+n-1,m];
TQ[n_]:=TQ[n]=IntegerQ[Sqrt[4n+1]];
tab={};Do[r=0;Do[If[f[8,z]>n,Goto[cc]];Do[If[f[6,y]>n-f[8,z],Goto[bb]];Do[If[f[4,x]>n-f[8,z]-f[6,y],Goto[aa]];If[TQ[n-f[8,z]-f[6,y]-f[4,x]],r=r+1],{x,0,n-f[8,z]-f[6,y]}];Label[aa],{y,0,n-f[8,z]}];Label[bb],{z,0,n}];Label[cc];tab=Append[tab,r],{n,0,100}];Print[tab]

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