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A227532
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Logarithmic derivative, wrt x, of triangle A227543, as read by terms k=0..n*(n-1)/2 in rows n>=1.
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3
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1, 1, 2, 1, 3, 3, 3, 1, 4, 6, 8, 8, 4, 4, 1, 5, 10, 15, 20, 20, 20, 15, 10, 5, 5, 1, 6, 15, 26, 39, 48, 57, 60, 54, 48, 36, 30, 18, 12, 6, 6, 1, 7, 21, 42, 70, 98, 126, 154, 168, 175, 168, 154, 133, 112, 84, 70, 49, 35, 21, 14, 7, 7, 1, 8, 28, 64, 118, 184, 256, 336, 408, 472, 516, 536, 532, 504, 464, 408, 360, 296, 248, 192, 152, 112, 88, 56, 40, 24, 16, 8, 8
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OFFSET
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1,3
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LINKS
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FORMULA
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L.g.f.: Sum_{k=0..n*(n-1)/2, n>=1} T(n,k)*x^n*q^k/n = Log(G(x,q)) where G(x,q) = 1 + x*G(q*x,q)*G(x,q) is the g.f. of triangle A227543.
Row sums form A001700, the logarithmic derivative of the Catalan numbers.
Sum_{k=0..n*(n-1)/2} T(n,k) = binomial(2*n-1, n-1), for n>=1.
Sum_{k=0..n*(n-1)/2} T(n,k)*(-1)^k = (-1)^[n/2] * binomial(n-1, [(n-1)/2]).
Sum_{k=0..n*(n-1)/2} k*T(n,k) = n*2^(2*n-2) - (2*n-1)*binomial(2*n-2,n-1) = A153338(n), for n>=1.
Sum_{k=0..n*(n-1)/2} T(n,k)*exp(2*Pi*I*k/n) = (-1)^(n-1) for n>=1; i.e., the n-th row sum at q = exp(2Pi*I/n), the n-th root of unity, equals -(-1)^n for n>=1.
Sum_{k=0..[n/2]} T(n, n*k) = A145855(n), the number of n-member subsets of 1..2n-1 whose elements sum to a multiple of n.
L.g.f. satisfies: L'(x,q) = P'(x,q)/P(x,q) - Q'(x,q)/Q(x,q), where
P(x,q) = Sum_{n>=0} q^(n^2) * (-x)^n / Product_{k=1..n} (1-q^k),
Q(x,q) = Sum_{n>=0} q^(n*(n-1)) * (-x)^n / Product_{k=1..n} (1-q^k),
due to Ramanujan's continued fraction identity. - Paul D. Hanna, Dec 28 2016
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EXAMPLE
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L.g.f.: L(x,q) = x*(1) + x^2*(1 + 2*q)/2 + x^3*(1 + 3*q + 3*q^2 + 3*q^3)/3
+ x^4*(1 + 4*q + 6*q^2 + 8*q^3 + 8*q^4 + 4*q^5 + 4*q^6)/4
+ x^5*(1 + 5*q + 10*q^2 + 15*q^3 + 20*q^4 + 20*q^5 + 20*q^6 + 15*q^7 + 10*q^8 + 5*q^9 + 5*q^10)/5
+ x^6*(1 + 6*q + 15*q^2 + 26*q^3 + 39*q^4 + 48*q^5 + 57*q^6 + 60*q^7 + 54*q^8 + 48*q^9 + 36*q^10 + 30*q^11 + 18*q^12 + 12*q^13 + 6*q^14 + 6*q^15)/6 +...
where exponentiation yields the g.f. of triangle A227543:
exp(L(x,q)) = 1 + x*(1) + x^2*(1 + q) + x^3*(1 + 2*q + q^2 + q^3)
+ x^4*(1 + 3*q + 3*q^2 + 3*q^3 + 2*q^4 + q^5 + q^6)
+ x^5*(1 + 4*q + 6*q^2 + 7*q^3 + 7*q^4 + 5*q^5 + 5*q^6 + 3*q^7 + 2*q^8 + q^9 + q^10)
+ x^6*(1 + 5*q + 10*q^2 + 14*q^3 + 17*q^4 + 16*q^5 + 16*q^6 + 14*q^7 + 11*q^8 + 9*q^9 + 7*q^10 + 5*q^11 + 3*q^12 + 2*q^13 + q^14 + q^15) +...
This triangle begins:
1;
1, 2;
1, 3, 3, 3;
1, 4, 6, 8, 8, 4, 4;
1, 5, 10, 15, 20, 20, 20, 15, 10, 5, 5;
1, 6, 15, 26, 39, 48, 57, 60, 54, 48, 36, 30, 18, 12, 6, 6;
1, 7, 21, 42, 70, 98, 126, 154, 168, 175, 168, 154, 133, 112, 84, 70, 49, 35, 21, 14, 7, 7;
1, 8, 28, 64, 118, 184, 256, 336, 408, 472, 516, 536, 532, 504, 464, 408, 360, 296, 248, 192, 152, 112, 88, 56, 40, 24, 16, 8, 8;
1, 9, 36, 93, 189, 324, 489, 684, 891, 1101, 1305, 1476, 1611, 1683, 1701, 1665, 1593, 1476, 1350, 1197, 1053, 900, 765, 630, 522, 405, 324, 243, 189, 135, 99, 63, 45, 27, 18, 9, 9; ...
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PROG
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(PARI) {T(n, k)=local(A=1); for(i=1, n, A=1+x*subst(A, x, q*x)*A +x*O(x^n)); n*polcoeff(polcoeff(log(A), n, x), k, q)}
for(n=1, 10, for(k=0, n*(n-1)/2, print1(T(n, k), ", ")); print(""))
(PARI) /* By Ramanujan's continued fraction identity: */
{T(n, k)=local(P=1, Q=1);
P=sum(m=0, n+1, q^(m^2)*(-x)^m/prod(k=1, m, 1-q^k) +O(x^(n+2)));
Q=sum(m=0, n+1, q^(m*(m-1))*(-x)^m/prod(k=1, m, 1-q^k) +O(x^(n+2)));
polcoeff(polcoeff(P'/P - Q'/Q, n-1, x), k, q)}
for(n=1, 10, for(k=0, n*(n-1)/2, print1(T(n, k), ", ")); print("")) \\ Paul D. Hanna, Dec 28 2016
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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