OFFSET
1,8
COMMENTS
Related identity: Sum_{n=-oo..+oo} x^n * (1 - x^n)^n = 0.
Related identity: Sum_{k=-oo..+oo} x^k * (1 - x^k)^(n+k) = Sum_{k=-oo..oo, k<>0} (-1)^(k-n) * x^(k*(k-n-1)) / (1 - x^k)^(k-n) for n >= 0.
In row n of this triangle, the index k in T(n,k) goes from k = 0 to k = floor((n+1)/2)^2.
Each row sum appears to equal zero.
Conjectural formulas for the terms of this triangle:
(C.1) T(n,0) = -1 for n >= 1.
(C.2) T(n,1) = 1 for n >= 1.
(C.3) T(n,2) = -(n-2) for n >= 2.
(C.4) T(n,3) = n*(n-3)/2 + 2 for n >= 3.
(C.5) T(n,4) = -(n-3)*(n^2 - 3*n + 8)/6 for n >= 4.
(C.6) T(2*n-1, n^2) = (-1)^(n-1) for n >= 1.
(C.7) T(2*n-1, n^2-1) = (-1)^n*2 for n >= 2.
(C.8) T(2*n, n^2) = (-1)^(n-1)*n for n >= 1.
(C.9) T(2*n, n^2-1) = (-1)^n*2*n for n >= 2.
(C.10) T(p,p) = 2 for odd prime p.
(C.11) T(n, prime(pi(n)+1)) = 0 for n >= 5; the first zero in row n, T(n,k) = 0, appears to occur at k equal to the next prime after n for n >= 5. Thus, T(5,7) = 0, T(6,7) = 0, T(7,11) = 0, T(8,11) = 0, T(11,13) = 0, T(13,17) = 0, etc.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 1..11100 (first 50 rows of this triangle), in which row r begins at n = r*(r^2 + 14 - 3*(r mod 2))/12 for r = 1..50; that is, rows begin at positions n = [1, 3, 5, 10, 15, 25, 35, 52, 69, 95, 121, 158, ...].
FORMULA
The terms of this irregular triangle satisfy the following formulas.
(1) Sum_{k=0..floor((n+1)/2)^2} T(n,k) * x^(-k) equals the power series expansion of Sum_{k=-oo..oo} x^k * (1 - x^k)^(n+k) after being truncated to include only nonpositive powers of x, for n >= 1 (by definition).
(2) Sum_{k=0..floor((n+1)/2)^2} T(n,k) = 0 for n >= 1 (conjecture).
(3) Sum_{k=0..floor((n+1)/2)^2} k * T(n,k) = 1 for n >= 1 (conjecture).
(4) Sum_{k=0..floor((n+1)/2)^2} k^2 * T(n,k) = -(2*n-5) for n >= 2 (conjecture).
(5) Sum_{k=0..floor((n+1)/2)^2} -(-1)^k * T(n,k) = 2*round(2^n/3) = 2*A001045(n) for n >= 1 (conjecture), where A001045 is the Jacobsthal numbers (g.f. x/((1+x)*(1-2*x))).
(6) Sum_{k=0..floor((n+1)/2)^2} -(-1)^k * k * T(n,k), n >= 1, forms a sequence with g.f. x*(1-x+8*x^2)/((1+x)^2*(1-2*x)^2) (conjecture), and begins [1, 1, 13, 25, 81, 181, 453, 1025, ...].
(7) Sum_{k=0..floor((n+1)/2)^2} T(n,k)^2 = 2*A391811(n) for n >= 1.
EXAMPLE
This irregular triangle, in which row n consists of terms T(n,k) where k = 0..floor((n+1)/2)^2, begins:
1: [-1, 1];
2: [-1, 1];
3: [-1, 1, -1, 2, -1];
4: [-1, 1, -2, 4, -2];
5: [-1, 1, -3, 7, -6, 2, 1, 0, -2, 1];
6: [-1, 1, -4, 11, -13, 6, 3, 0, -6, 3];
7: [-1, 1, -5, 16, -24, 16, 1, 2, -13, 6, 3, 0, -3, 0, 0, 2, -1];
8: [-1, 1, -6, 22, -40, 36, -10, 8, -24, 10, 12, 0, -12, 0, 0, 8, -4];
9: [-1, 1, -7, 29, -62, 71, -40, 29, -47, 17, 31, 0, -34, 0, 5, 21, -12, 0, -3, 0, 1, 2, 0, 0, -2, 1];
...
and is formed from the coefficients of the following row functions.
1: (-x + 1)/x
2: (-x + 1)/x
3: (-x^4 + x^3 - x^2 + 2*x - 1)/x^4
4: (-x^4 + x^3 - 2*x^2 + 4*x - 2)/x^4
5: (-x^9 + x^8 - 3*x^7 + 7*x^6 - 6*x^5 + 2*x^4 + x^3 - 2*x + 1)/x^9
6: (-x^9 + x^8 - 4*x^7 + 11*x^6 - 13*x^5 + 6*x^4 + 3*x^3 - 6*x + 3)/x^9
7: (-x^16 + x^15 - 5*x^14 + 16*x^13 - 24*x^12 + 16*x^11 + x^10 + 2*x^9 - 13*x^8 + 6*x^7 + 3*x^6 - 3*x^4 + 2*x - 1)/x^16
8: (-x^16 + x^15 - 6*x^14 + 22*x^13 - 40*x^12 + 36*x^11 - 10*x^10 + 8*x^9 - 24*x^8 + 10*x^7 + 12*x^6 - 12*x^4 + 8*x - 4)/x^16
9: (-x^25 + x^24 - 7*x^23 + 29*x^22 - 62*x^21 + 71*x^20 - 40*x^19 + 29*x^18 - 47*x^17 + 17*x^16 + 31*x^15 - 34*x^13 + 5*x^11 + 21*x^10 - 12*x^9 - 3*x^7 + x^5 + 2*x^4 - 2*x + 1)/x^25
...
Define the doubly infinite series
S(n) = Sum_{k=-oo..+oo} x^k * (1 - x^k)^(n+k), where S(0) = 0.
Then the above row functions are obtained from the portion of S(n) that consists of only the nonpositive powers of x. Examples of these series are
S(1) = 1/x - 1 + x - 2*x^2 + 3*x^3 - 3*x^4 + x^5 + x^6 + x^7 - 7*x^8 + 10*x^9 - 6*x^10 + x^11 + x^13 - 8*x^14 + 23*x^15 - 25*x^16 + x^17 + ...
S(2) = S(1) = 1/x - 1 + x - 2*x^2 + 3*x^3 - 3*x^4 + x^5 + x^6 + x^7 + ...
S(3) = -1/x^4 + 2/x^3 - 1/x^2 + 1/x - 1 + x - 3*x^2 + 7*x^3 - 9*x^4 + 3*x^5 + 5*x^6 + x^7 - 17*x^8 + 16*x^9 + x^11 - 10*x^12 + x^13 - 9*x^14 + 47*x^15 - 46*x^16 + x^17 + ...
S(4) = -2/x^4 + 4/x^3 - 2/x^2 + 1/x - 1 + x - 4*x^2 + 11*x^3 - 15*x^4 + 5*x^5 + 9*x^6 + x^7 - 27*x^8 + 22*x^9 + 6*x^10 + x^11 - 20*x^12 + x^13 - 10*x^14 + 71*x^15 - 67*x^16 + x^17 + ...
S(5) = 1/x^9 - 2/x^8 + 1/x^6 + 2/x^5 - 6/x^4 + 7/x^3 - 3/x^2 + 1/x - 1 + x - 5*x^2 + 16*x^3 - 26*x^4 + 16*x^5 + 7*x^6 + 3*x^7 - 43*x^8 + 29*x^9 + 26*x^10 + x^11 - 52*x^12 + x^13 - 2*x^14 + 116*x^15 - 98*x^16 + x^17 + ...
S(6) = 3/x^9 - 6/x^8 + 3/x^6 + 6/x^5 - 13/x^4 + 11/x^3 - 4/x^2 + 1/x - 1 + x - 6*x^2 + 22*x^3 - 42*x^4 + 36*x^5 - x^6 + 7*x^7 - 65*x^8 + 37*x^9 + 60*x^10 + x^11 - 106*x^12 + x^13 + 15*x^14 + 182*x^15 - 139*x^16 + x^17 + ...
S(7) = -1/x^16 + 2/x^15 - 3/x^12 + 3/x^10 + 6/x^9 - 13/x^8 + 2/x^7 + 1/x^6 + 16/x^5 - 24/x^4 + 16/x^3 - 5/x^2 + 1/x - 1 + x - 7*x^2 + 29*x^3 - 64*x^4 + 71*x^5 - 29*x^6 + 29*x^7 - 103*x^8 + 48*x^9 + 115*x^10 + x^11 - 203*x^12 + x^13 + 71*x^14 + 277*x^15 - 216*x^16 + x^17 + ...
etc.
PROG
(PARI) \\ generate the row polynomial from the series
{R(n) = my(N=((n+1)\2)^2); x^N*( sum(k=-N-2, N+2, x^k * (1 - x^k +x*O(x^N))^(n+k) ) + O(x) )}
\\ print this triangle
{T(n, k) = polcoef(R(n), ((n+1)\2)^2 - k)}
{for(n=1, 12, for(k=0, ((n+1)\2)^2, print1(T(n, k), ", ")); print(""))}
\\ faster method is to invoke R(n) only once for each row
{for(n=1, 12, V = Vec(truncate(R(n))); for(k=0, ((n+1)\2)^2, print1(V[k+1], ", ")); print(""))}
\\ demonstrate row sums are zero
{for(n=1, 30, V = Vec(truncate(R(n))); print1(V*vector(#V, k, 1)~, ", "))}
\\ demonstrate Sum_{k=0..floor((n+1)/2)^2} k * T(n, k) = 1
{for(n=1, 30, V = Vec(truncate(R(n))); print1(V*vector(#V, k, k-1)~, ", "))}
\\ demonstrate Sum_{k=0..floor((n+1)/2)^2} -(-1)^k * T(n, k) = 2*round(2^n/3)
{for(n=1, 30, V = Vec(truncate(R(n))); print1(V*vector(#V, k, (-1)^k)~, ", "))}
CROSSREFS
KEYWORD
sign,tabf
AUTHOR
Paul D. Hanna, Dec 21 2025
STATUS
approved