0,3

Row n starts with n!, after which the following pattern holds. When terms of row n are divided by a list of factorials, with (n-j-1)! repeated A002865(j+1) times in the list as j=1..n-1, the result is the initial terms of A110142. E.g. row 6 is: {720,48,18,16,8,6,5,48,8,18,6}; divide by respective factorials: {6!,4!,3!,2!,2!,1!,1!,0!,0!,0!,0!} with {4!,3!,2!,1!,0!} respectively occurring {1,1,2,2,4} times (A002865), yields the initial terms of A110142: {1,2,3,8,4,6,5,48,8,18,6}.

Number of terms in row n is A000041(n) (partition numbers). The unit fractions of each row sum to unity: Sum_{k=1..A000041(n)} 1/T(n, k) = 1.

Coefficients [x^n] exp(c1*x + c2/2*x^2 + c3/3*x^3 +...) begin:

[x^0]: 1;

[x^1]: 1*c1;

[x^2]: 1/2*c1^2 + 1/2*c2;

[x^3]: 1/6*c1^3 + 1/2*c1*c2 + 1/3*c3;

[x^4]: 1/24*c1^4 + 1/4*c1^2*c2 + 1/3*c1*c3 + 1/8*c2^2 + 1/4*c4;

[x^5]: 1/120*c1^5 + 1/12*c1^3*c2 + 1/6*c1^2*c3 + 1/8*c1*c2^2 + 1/4*c1*c4 + 1/6*c2*c3 + 1/5*c5;

[x^6]: 1/720*c1^6 + 1/48*c1^4*c2 + 1/18*c1^3*c3 + 1/16*c1^2*c2^2 + 1/8*c1^2*c4 + 1/6*c1*c2*c3 + 1/5*c1*c5 + 1/48*c2^3 + 1/8*c2*c4 + 1/18*c3^2 + 1/6*c6;

forming this triangle of unit fraction coefficients:

1;

1;

2,2;

6,2,3;

24,4,3,8,4;

120,12,6,8,4,6,5;

720,48,18,16,8,6,5,48,8,18,6;

5040,240,72,48,24,12,10,48,8,18,6,24,10,12,7;

40320,1440,360,192,96,36,30,96,16,36,12,24,10,12,7,384,32,36,12,15,32,8;

362880,10080,2160,960,480,144,120,288,48,108,36,48,20,24,14,384,32,36,12,15,32,8,144,40,24,14,162,18,20,9; ...

Cf. A110142, A000041, A002865, A110143 (row sums).

Sequence in context: A046110 A126889 A134339 this_sequence A129750 A068976 A124859

Adjacent sequences: A110138 A110139 A110140 this_sequence A110142 A110143 A110144

nonn,tabl,frac

Paul D. Hanna (pauldhanna(AT)juno.com), Jul 13 2005