A136654 G.f.: A(x) = (1/x)*Series_Reversion( x/Sum_{k=0..n} 2^[k(k-1)/2]*x^k ).

1, 1, 3, 15, 117, 1565, 41663, 2378147, 286991465, 71261033273, 35889915535835, 36421251158141399, 74222529448186797341, 303194457634544530959125, 2480120130065258782050157847, 40601998283406419045206334661611

Offset: 0

Author

Paul D. Hanna, Jan 15 2008

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 15*x^3 + 117*x^4 + 1565*x^5 + 41663*x^6 +...
Let F(x) = 1 + x + 2x^2 + 8x^3 + 64x^4 + 1024x^5 +...+ 2^(n(n-1)/2)*x^n +..
then A(x) = F(x*A(x)), A(x/F(x)) = F(x).
a(n) = coefficient of x^n in F(x)^(n+1)/(n+1),
as can be seen by the main diagonal in the array of
coefficients in the initial powers of F(x):
F^1: [(1), 1, 2, 8, 64, 1024, 32768, 2097152, 268435456,...;
F^2: [1, (2), 5, 20, 148, 2208, 67904, 4264960, 541216768,...;
F^3: [1, 3, (9), 37, 258, 3588, 105704, 6507552, 818458752,...;
F^4: [1, 4, 14, (60), 401, 5208, 146520, 8829536, 1100282640,...;
F^5: [1, 5, 20, 90, (585), 7121, 190770, 11236080, 1386816800,...;
F^6: [1, 6, 27, 128, 819, (9390), 238949, 13733004, 1678197564,...;
F^7: [1, 7, 35, 175, 1113, 12089, (291641), 16326885, 1974570178,...;
F^8: [1, 8, 44, 232, 1478, 15304, 349532, (19025176), 2276089889,...;
F^9: [1, 9, 54, 300, 1926, 19134, 413424, 21836340, (2582923185),...;
dividing each diagonal term in row n by (n+1) gives a(n) for n>=0.
The diagonal above the main diagonal gives coefficients of l.g.f.:
log(A(x)) = x + 5*x^2/2 + 37*x^3/3 + 401*x^4/4 + 7121*x^5/5 +...
		

Crossrefs

Cf. A136653.

Programs

Mathematica

max = 15; s = x/Sum[2^(k*(k-1)/2)*x^k, {k, 0, max}] + O[x]^(max+2);(1/x)*InverseSeries[s] + O[x]^(max+1) // CoefficientList[#, x]& (* Jean-François Alcover, Sep 03 2017 *)

PARI

a(n)=polcoeff(1/x*serreverse(x/sum(k=0,n,2^(k*(k-1)/2)*x^k +x*O(x^n))),n)

Formula

a(n) = coefficient of x^n in [Sum_{k=0..n} 2^(k(k-1)/2)*x^k]^(n+1)/(n+1).