OFFSET
0,3
COMMENTS
What is the limit of a(n)^(1/n) as n grows? For example, a(520)^(1/520) = 3.38918103746...
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..520
EXAMPLE
G.f.: A(x) = 1 + x + 4*x^2 + 17*x^3 + 44*x^4 + 219*x^5 + 798*x^6 + 1770*x^7 + 7584*x^8 + 12029*x^9 + 75410*x^10 + 299250*x^11 + 959820*x^12 + ...
ILLUSTRATION OF DEFINITION.
Set a(0) = 1; for n > 0, term a(n) equals the sum, from k = 0 to n-1, of 2^(n-k-1) times the residue of the coefficient of x^k in A(x)^n modulo 2^n, as illustrated below.
The table of coefficients of x^k in A(x)^n starts
n=1: [1, 1, 4, 17, 44, 219, 798, 1770, 7584, ...];
n=2: [1, 2, 9, 42, 138, 662, 2675, 8384, 34474, ...];
n=3: [1, 3, 15, 76, 294, 1428, 6235, 23238, 98388, ...];
n=4: [1, 4, 22, 120, 525, 2632, 12246, 50976, 225286, ...];
n=5: [1, 5, 30, 175, 845, 4406, 21660, 97735, 451590, ...];
n=6: [1, 6, 39, 242, 1269, 6900, 35634, 171414, 826746, ...];
n=7: [1, 7, 49, 322, 1813, 10283, 55552, 281968, 1416156, ...];
n=8: [1, 8, 60, 416, 2494, 14744, 83048, 441728, 2304509, ...];
n=9: [1, 9, 72, 525, 3330, 20493, 120030, 665748, 3599541, ...];
...
from which we take the residue of the coefficient of x^k in A(x)^n modulo 2^n for k = 0..n-1 to form the n-th row of a triangle that begins:
A^1 (mod 2): [1];
A^2 (mod 4): [1, 2];
A^3 (mod 8): [1, 3, 7];
A^4 (mod 16): [1, 4, 6, 8];
A^5 (mod 32): [1, 5, 30, 15, 13];
A^6 (mod 64): [1, 6, 39, 50, 53, 52];
A^7 (mod 128): [1, 7, 49, 66, 21, 43, 0];
A^8 (mod 256): [1, 8, 60, 160, 190, 152, 104, 128];
A^9 (mod 512): [1, 9, 72, 13, 258, 13, 222, 148, 181];
...
From this triangle, we see that a(n) may be obtained in the following manner.
a(1) = 1;
a(2) = 1*2 + 2 = 4;
a(3) = 1*2^2 + 3*2 + 7 = 17;
a(4) = 1*2^3 + 4*2^2 + 6*2 + 8 = 44;
a(5) = 1*2^4 + 5*2^3 + 30*2^2 + 15*2 + 13 = 219;
a(6) = 1*2^5 + 6*2^4 + 39*2^3 + 50*2^2 + 53*2 + 52 = 798;
a(7) = 1*2^6 + 7*2^5 + 49*2^4 + 66*2^3 + 21*2^2 + 43*2 + 0 = 1770;
a(8) = 1*2^7 + 8*2^6 + 60*2^5 + 160*2^4 + 190*2^3 + 152*2^2 + 104*2 + 128 = 7584;
a(9) = 1*2^8 + 9*2^7 + 72*2^6 + 13*2^5 + 258*2^4 + 13*2^3 + 222*2^2 + 148*2 + 181 = 12029;
...
Continuing in this way, we may generate all the terms of this sequence.
SPECIFIC VALUES.
A(1/4) = 3.0899296035448858291992137...
Summing by columns in the above triangle, A(x) at x = 1/4 equals the series
A(1/4) = 1 + 2/4 + 6/4^2 + 30/4^3 + 72/4^4 + 190/4^5 + 532/4^6 + 876/4^7 + 2768/4^8 + 6054/4^9 + 13672/4^10 + 35908/4^11 + 68326/4^12 + 151164/4^13 + 335368/4^14 + 795092/4^15 + 1692334/4^16 + 3193270/4^17 + 6213312/4^18 + 15607836/4^19 + 29314746/4^20 + 77483276/4^21 + 146205638/4^22 + 274169006/4^23 + 605329890/4^24 + 1178955004/4^25 + 2563244588/4^26 + 4920278244/4^27 + 12067394712/4^28 + 19529893388/4^29 + 50954511294/4^30 + 105054406670/4^31 + 201512427802/4^32 + 425611605014/4^33 + 846996442308/4^34 + 1816352319384/4^35 + 3855385008770/4^36 + 8364698676580/4^37 + 17699377790690/4^38 + 36199808127270/4^39 + 70964452308758/4^40 + ...
PROG
(PARI) {a(n) = my(A=[1]);
for(m=1, n, A = concat(A, sum(k=0, m-1, 2^(m-k-1)*(Vec(Ser(A)^m)[k+1]%2^m))) ); A[n+1]}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 16 2024
STATUS
approved