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Irregular triangle read by rows: the n-th row gives the exponents of the powers of x corresponding to the maximal coefficient of the product x^2*(x^2 + x^3)*(x^2 + x^3 + x^5)*...*(x^2 + x^3 + x^5 + ... + x^prime(n)).
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%I #11 Dec 31 2022 15:18:15

%S 0,2,4,5,7,12,16,17,22,24,32,42,53,65,79,96,114,134,155,180,205,233,

%T 263,294,329,364,403,442,485,529,576,625,676,729,785,842,902,964,1029,

%U 1097,1167,1238,1313,1390,1469,1552,1636,1723,1813,1904,1999,2096,2195,2298

%N Irregular triangle read by rows: the n-th row gives the exponents of the powers of x corresponding to the maximal coefficient of the product x^2*(x^2 + x^3)*(x^2 + x^3 + x^5)*...*(x^2 + x^3 + x^5 + ... + x^prime(n)).

%C Conjecture: except for n = 2, 5, and 6, the rows have length equal to 1.

%e The irregular triangle begins:

%e 0;

%e 2;

%e 4, 5;

%e 7;

%e 12;

%e 16, 17;

%e 22, 24;

%e 32;

%e 42;

%e 53;

%e 65;

%e ...

%t b[n_]:=CoefficientList[Product[Sum[x^Prime[i],{i,k}],{k,n}],x]; Table[Position[b[n],Max[b[n]]]-1,{n,0,50}]//Flatten

%Y Cf. A000040, A359328.

%Y Cf. A359338 (minimal exponent), A359339 (maximal exponent).

%K nonn,tabf

%O 0,2

%A _Stefano Spezia_, Dec 27 2022