Date: Tue, 29 Sep 2009 14:34:24 +1300
From: "Sean A. Irvine" <sairvin(AT)xtra.co.nz>

I have computed the first 55 terms of A000135. The terms seem to have good
agreement with exp(x^(3/4)) as shown in the graph.  Perhaps this is a
known result, I don't have access to the Agarwala & Auluck reference so
cannot check. (Please note, the values of a(n) below could be slightly in
error for larger values of n, since I used ordinary double precision
arithmetic for computing the sums).

n a(n) exp(n^(3/4))
1 1 2.71828
2 2 5.37518
3 6 9.77186
4 13 16.9188
5 24 28.3238
6 42 46.2314
7 73 73.9595
8 125 116.376
9 204 180.576
10 324 276.833
11 511 419.937
12 801 631.072
13 1228 940.419
14 1856 1390.79
15 2780 2042.63
16 4135 2980.96
17 6084 4324.9
18 8873 6240.73
19 12847 8959.76
20 18481 12802.8
21 26416 18213.1
22 37473 25802
23 52871 36409.2
24 74216 51186.1
25 103596 71706.7
26 143841 100118
27 198839 139339
28 273654 193336
29 374987 267478
30 511735 369022
31 695559 507757
32 941932 696862
33 1271139 954042
34 1709474 1.30304e+06
35 2291195 1.77565e+06
36 3061385 2.41434e+06
37 4078152 3.2758e+06
38 5416322 4.43551e+06
39 7173114 5.99386e+06
40 9473896 8.08414e+06
41 12479712 1.08831e+07
42 16396918 1.46246e+07
43 21490368 1.9618e+07
44 28098433 2.62716e+07
45 36652969 3.51236e+07
46 47704268 4.68827e+07
47 61951295 6.24808e+07
48 80282074 8.31417e+07
49 103822515 1.10471e+08
50 133995350 1.4657e+08
51 172598882 1.94193e+08
52 221901148 2.56934e+08
53 284757153 3.39491e+08
54 364757460 4.47987e+08
55 466411124 5.90399e+08