A304403 G.f. A(x) satisfies: [x^n] A(x) * (1+x)^(n*(n-1)/2) = [x^n] (1+x)^(n*(n+1)/2) for n >= 0.

1, 1, 2, 10, 85, 1001, 15036, 273932, 5858560, 143735650, 3976623010, 122427035732, 4150101179665, 153570442058684, 6158890134498661, 266074435570524219, 12318188650382356297, 608375312208623381681, 31927089332971578025902, 1774159611825531210120848, 104068978022940760659203857, 6425920321260029677988702979, 416624218261246444665784515673

Offset

0,3

Comments

Limit ( a(n) / n! )^(1/n) seems to exist and is near 3.1...

Links

Paul D. Hanna, Table of n, a(n) for n = 0..520

Formula

a(n) = binomial(n*(n+1)/2, n) - Sum_{k=0..n-1} a(k) * binomial(n*(n-1)/2, n-k), for n >= 0.

Example

G.f.: A(x) =  1 + x + 2*x^2 + 10*x^3 + 85*x^4 + 1001*x^5 + 15036*x^6 + 273932*x^7 + 5858560*x^8 + 143735650*x^9 + 3976623010*x^10 + 122427035732*x^11 + ...
such that
binomial(n*(n+1)/2, n) = Sum_{k=0..n} a(k) * binomial(n*(n-1)/2, n-k) for n >= 0.
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x) * (1+x)^(n*(n-1)/2) begins:
n=0: [1, 1, 2, 10, 85, 1001, 15036, 273932, 5858560, ...];
n=1: [1, 1, 2, 10, 85, 1001, 15036, 273932, 5858560, ...];
n=2: [1, 2, 3, 12, 95, 1086, 16037, 288968, 6132492, ...];
n=3: [1, 4, 8, 20, 122, 1288, 18304, 322128, 6726465, ...];
n=4: [1, 7, 23, 57, 210, 1722, 22554, 381026, 7749049, ...];
n=5: [1, 11, 57, 195, 605, 3003, 30953, 482471, 9415575, ...];
n=6: [1, 16, 122, 600, 2265, 8604, 54264, 674348, 12170710, ...];
n=7: [1, 22, 233, 1592, 8030, 33880, 153790, 1184040, 17240535, ...];
n=8: [1, 29, 408, 3720, 24872, 132468, 623924, 3313868, 30260340, ...]; ...
in which the main diagonal equals A014068:
[1, 1, 3, 20, 210, 3003, 54264, 1184040, ..., binomial(n*(n+1)/2, n), ...]
illustrating [x^n] A(x) * (1+x)^(n*(n-1)/2) = [x^n] (1+x)^(n*(n+1)/2) for n >= 0.
ILLUSTRATION OF RECURRENCE.
The table of coefficients of x^k in (1+x)^(n*(n-1)/2) begins:
n=1: [1, 0, 0, 0, 0, 0, 0, 0, 0, ...];
n=2: [1, 1, 0, 0, 0, 0, 0, 0, 0, ...];
n=3: [1, 3, 3, 1, 0, 0, 0, 0, 0, ...];
n=4: [1, 6, 15, 20, 15, 6, 1, 0, 0, ...];
n=5: [1, 10, 45, 120, 210, 252, 210, 120, 45, ...];
n=6: [1, 15, 105, 455, 1365, 3003, 5005, 6435, 6435, ...];
n=7: [1, 21, 210, 1330, 5985, 20349, 54264, 116280, 203490, ...];
n=8: [1, 28, 378, 3276, 20475, 98280, 376740, 1184040, 3108105, ...];
n=9: [1, 36, 630, 7140, 58905, 376992, 1947792, 8347680, 30260340, ...]; ...
The recurrence uses the above coefficients like so:
a(0) = 1;
a(1) = 1 - (0*a(0) = 1;
a(2) = 3 - (0*a(0) + 1*a(1)) = 2;
a(3) = 20 - (1*a(0) + 3*a(1) + 3*a(2)) = 10;
a(4) = 210 - (15*a(0) + 20*a(1) + 15*a(2) + 6*a(3)) = 85;
a(5) = 3003 - (252*a(0) + 210*a(1) + 120*a(2) + 45*a(3) + 10*a(4)) = 1001; ...
illustrating a(n) = C(n*(n+1)/2, n) - Sum_{k=0..n-1} C(n*(n-1)/2, n-k) * a(k), for n >= 0.

Prog

(PARI) {a(n) = my(A=[1]); for(i=1, n, A=concat(A, 0); m=#A-1; A[#A] = binomial(m*(m+1)/2, m) - polcoeff( Ser(A)*(1+x +x*O(x^m))^(m*(m-1)/2) , m) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))
(PARI) /* Recurrence: */
{a(n) = if(n==0, 1, binomial(n*(n+1)/2, n) - sum(k=0, n-1, a(k) * binomial(n*(n-1)/2, n-k) ) )}
for(n=0, 30, print1(a(n), ", "))

Crossrefs

Sequence in context: A302935 A332655 A156466 * A132397 A202745 A364396

Adjacent sequences: A304400 A304401 A304402 * A304404 A304405 A304406

Keyword

nonn

Author

Paul D. Hanna, May 18 2018

Status

approved