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G.f. A(x) = Sum_{n>=0} a(n)*x^n where a(n) = Sum_{k=0..n-1} ( ([x^k] A(x)^n) (mod 2^n) ) for n > 0, with a(0) = 1.
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%I #17 Sep 17 2024 08:27:28

%S 1,1,3,8,15,49,136,446,815,2045,3833,8717,14272,39260,72840,210072,

%T 340127,791503,1695607,2740397,7948517,14258853,31589187,62871303,

%U 172680120,259553374,491879422,1390281126,3193983644,5509013560,11070386930,27804975886,57398653055,108014137383

%N G.f. A(x) = Sum_{n>=0} a(n)*x^n where a(n) = Sum_{k=0..n-1} ( ([x^k] A(x)^n) (mod 2^n) ) for n > 0, with a(0) = 1.

%C What is the limit of a(n)/(n*2^n) as n grows? For example, a(520)/(520*2^520) = 0.37178628499...

%H Paul D. Hanna, <a href="/A376231/b376231.txt">Table of n, a(n) for n = 0..520</a>

%e 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 + ...

%e ILLUSTRATION OF DEFINITION.

%e 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.

%e The table of coefficients of x^k in A(x)^n starts

%e n=1: [1, 1, 3, 8, 15, 49, 136, 446, 815, ...];

%e n=2: [1, 2, 7, 22, 55, 176, 524, 1698, 4347, ...];

%e n=3: [1, 3, 12, 43, 129, 432, 1380, 4581, 13587, ...];

%e n=4: [1, 4, 18, 72, 247, 880, 3006, 10376, 33591, ...];

%e n=5: [1, 5, 25, 110, 420, 1596, 5805, 20950, 72305, ...];

%e n=6: [1, 6, 33, 158, 660, 2670, 10297, 38904, 141573, ...];

%e n=7: [1, 7, 42, 217, 980, 4207, 17136, 67740, 258349, ...];

%e n=8: [1, 8, 52, 288, 1394, 6328, 27128, 112048, 446135, ...];

%e n=9: [1, 9, 63, 372, 1917, 9171, 41250, 177714, 736668, ...];

%e ...

%e 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:

%e A^1 (mod 2): [1];

%e A^2 (mod 4): [1, 2];

%e A^3 (mod 8): [1, 3, 4];

%e A^4 (mod 16): [1, 4, 2, 8];

%e A^5 (mod 32): [1, 5, 25, 14, 4];

%e A^6 (mod 64): [1, 6, 33, 30, 20, 46];

%e A^7 (mod 128): [1, 7, 42, 89, 84, 111, 112];

%e A^8 (mod 256): [1, 8, 52, 32, 114, 184, 248, 176];

%e A^9 (mod 512): [1, 9, 63, 372, 381, 467, 290, 50, 412];

%e ...

%e The row sums of the above triangle generate all the terms of this sequence.

%e SPECIFIC VALUES.

%e A(1/3) = 3.19165654953338695440681583335928571940026307605013927...

%e Summing by columns in the above triangle, A(x) at x = 1/3 equals the series

%e 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 + ...

%e A(1/4) = 1.75741768996442511768326999491510596752804202708999837...

%e Summing by columns in the above triangle, A(x) at x = 1/4 equals the series

%e 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 + ...

%o (PARI) {a(n) = my(A=[1]);

%o for(m=1,n, A = concat(A, sum(k=0,m-1,Vec(Ser(A)^m)[k+1]%2^m)) );A[n+1]}

%o for(n=0,30,print1(a(n),", "))

%Y Cf. A376232.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Sep 16 2024