OFFSET
0,3
COMMENTS
What is the limit of a(n)/(n*2^n) as n grows? For example, a(520)/(520*2^520) = 0.37178628499...
LINKS
Paul D. Hanna, Table of n, a(n) for n = 0..520
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 8*x^3 + 15*x^4 + 49*x^5 + 136*x^6 + 446*x^7 + 815*x^8 + 2045*x^9 + 3833*x^10 + 8717*x^11 + 14272*x^12 + ...
ILLUSTRATION OF DEFINITION.
Set a(0) = 1; for n > 0, term a(n) equals the sum of the residues of the first n coefficients 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, 3, 8, 15, 49, 136, 446, 815, ...];
n=2: [1, 2, 7, 22, 55, 176, 524, 1698, 4347, ...];
n=3: [1, 3, 12, 43, 129, 432, 1380, 4581, 13587, ...];
n=4: [1, 4, 18, 72, 247, 880, 3006, 10376, 33591, ...];
n=5: [1, 5, 25, 110, 420, 1596, 5805, 20950, 72305, ...];
n=6: [1, 6, 33, 158, 660, 2670, 10297, 38904, 141573, ...];
n=7: [1, 7, 42, 217, 980, 4207, 17136, 67740, 258349, ...];
n=8: [1, 8, 52, 288, 1394, 6328, 27128, 112048, 446135, ...];
n=9: [1, 9, 63, 372, 1917, 9171, 41250, 177714, 736668, ...];
...
from which we take the residues of the coefficient of x^k in A(x)^n modulo 2^n for k = 0..n-1 to form rows of a triangle that begins:
A^1 (mod 2): [1];
A^2 (mod 4): [1, 2];
A^3 (mod 8): [1, 3, 4];
A^4 (mod 16): [1, 4, 2, 8];
A^5 (mod 32): [1, 5, 25, 14, 4];
A^6 (mod 64): [1, 6, 33, 30, 20, 46];
A^7 (mod 128): [1, 7, 42, 89, 84, 111, 112];
A^8 (mod 256): [1, 8, 52, 32, 114, 184, 248, 176];
A^9 (mod 512): [1, 9, 63, 372, 381, 467, 290, 50, 412];
...
The row sums of the above triangle generate all the terms of this sequence.
SPECIFIC VALUES.
A(1/3) = 3.19165654953338695440681583335928571940026307605013927...
Summing by columns in the above triangle, A(x) at x = 1/3 equals the series
A(1/3) = 1 + 81/162 + 10935/162^2 + 1502469/162^3 + 189737559/162^4 + 15771994461/162^5 + 3592080833679/162^6 + 366658848770517/162^7 + 26092620616963383/162^8 + 3647452065131814669/162^9 + 266473731108842221599/162^10 + 38235892646929415624325/162^11 + 4603705873782929233296327/162^12 + 382148211586497427156392765/162^13 + 54436746171829224258109609263/162^14 + 3004165735518365517122973022005/162^15 + 776638170361137221049441318486615/162^16 + ...
A(1/4) = 1.75741768996442511768326999491510596752804202708999837...
Summing by columns in the above triangle, A(x) at x = 1/4 equals the series
A(1/4) = 1 + 16/48 + 448/48^2 + 11776/48^3 + 319744/48^4 + 4597504/48^5 + 291495424/48^6 + 7282842112/48^7 + 112783994368/48^8 + 3511778478592/48^9 + 54943154970112/48^10 + 1875309044480512/48^11 + 53126556811013632/48^12 + 701084699667237376/48^13 + 28392513835468185088/48^14 + 317012819269037278720/48^15 + 19479870906603281961472/48^16 + 439076694241734531053056/48^17 + 4950820105940741443838464/48^18 + 168210496994570296771552768/48^19 + 1244871541635655380054438400/48^20 + ...
PROG
(PARI) {a(n) = my(A=[1]);
for(m=1, n, A = concat(A, sum(k=0, m-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