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%I #55 Mar 04 2023 22:31:41
%S 1,1,2,1,3,5,6,1,4,9,15,21,26,29,1,5,14,29,50,76,105,134,160,181,196,
%T 204,1,6,20,49,99,175,280,414,574,755,951,1155,1359,1554,1730,1876,
%U 1981,2036,1,7,27,76,175,350,630,1044,1618,2373,3324,4479,5838,7392,9122,10998,12979,15014,17044,19005,20832,22463,23842,24921,25662,26039
%N Triangle read by rows: T(n,k) is the number of n-tuples with sum k + n whose i-th element is a positive integer <= prime(i), 0 <= k < A070826(n).
%C Original name: Normal distributions from the primes. The contracted sequence of integers generated from the frequencies of the summation of elements of the subsets of the Cartesian product of the natural numbers of ascending prime cardinality. That is, given a number of sets of the natural numbers of ascending modulo P(n+1), the probabilities of generating a given number from the selection of one element from each set form the given sequence.
%C Although this sequence initially appears similar to A131791, its derivation is entirely different and it deviates quickly.
%C By sets of natural numbers of ascending prime cardinality, it is meant
%C N_1 = {1,2}, N_2 = {1,2,3}, N_3 = {1,2,3,4,5}, N_4 = {1,2,3,4,5,6,7}, ..., N_w = {1,2,3,...,p_w}, where the p_i are primes.
%C with Cartesian products
%C N_1 X N_2 = {1,2} X {1,2,3} = {(1,1),(1,2),(1,3),(2,1),(2,2),(2,3))}, etc.
%C and the sum of the elements of the product's subsets denoted
%C Sum[N_1 X N_2] = {(2),(3),(4),(3),(4),(5)}
%C whose elements have the frequencies [1,2,2,1] (with respect to magnitude). It is these frequencies that form the sequence, the symmetry allows for the omission of repeated terms, and hence the following contraction halves the data without loss of information:
%C [1,1] -> [1]
%C [1,2,2,1] -> [1,2]
%C [1,3,5,6,6,5,3,1] -> [1,3,5,6]
%C and so forth.
%C When arranged by row of the number of sets used, then P(S(u,r) = u + r - 1) = T(r,u)/prime(r)#, P(S = X) the probability that the sum S equals the value X, and prime(r)# is the product of the first r primes (A002110), then the structure and symmetry become more apparent.
%C Each row contains l(r) = (1/2)*(Sum p(r) - r + 1) terms and clearly the sum of each row must equal half the product of the primes used,
%C Sum_{u=1..l(r)} T(r,u) = (1/2)*prime(r)#,
%C and one can see that in general for all u, r:
%C P(S(u,r) = r) = P(S(u,r) = Sum p(r)) = 1/Product p(r),
%C P(S(u,r) = r + 1) = P(S(u,r) = Sum p(r) - 1) = r/prime(r)#,
%C P(S = r + i) = P(S = Sum p(r) - i) = T(r,u+i)/prime(r)#, [0 <= i <= l(r) - 1)],
%C S(u,r) ~ N(mu(r),sigma(r)^2).
%H Andrew Howroyd, <a href="/A239738/b239738.txt">Table of n, a(n) for n = 1..2005</a> (rows 1..20)
%H Mikhail Gaichenkov, <a href="https://math.stackexchange.com/questions/4628414/normal-distributions-from-the-primes-and-its-generation-function/4631059?noredirect=1#comment9775348_4631059">Normal distributions from the primes and its generation function</a>, Mathematics StackExchange, 2023.
%F T(n,k) = [x^k] (1/(x-1)^n) * Product_{i=1..n} (x^prime(i)-1). - _Steven Foster Clark_, Feb 05 2023
%F T(n,k) = [x^k] Product_{i=1..n} Sum_{j=0..prime(i)-1} x^j. - _Andrew Howroyd_, Feb 05 2023
%e Triangle T(n,k) begins: (n >= 1, k >= 0)
%e 1;
%e 1, 2;
%e 1, 3, 5, 6;
%e 1, 4, 9, 15, 21, 26, 29;
%e 1, 5, 14, 29, 50, 76, 105, 134, 160, 181, 196, 204;
%e 1, 6, 20, 49, 99, 175, 280, 414, 574, 755, 951, 1155, 1359, 1554, 1730, 1876, 1981, 2036;
%e ...
%e T(3, 2) = 5 because the following 3-tuples have sum 2 + 3 = 5: (1,1,3), (1,2,2), (1,3,1), (2,1,2), (2,2,1). The tuple (3,1,1) is excluded because the 1st term is required to be no greater than prime(1) = 2.
%t row[r_]:=Drop[#,-Length[#]/2]&[Transpose[Tally[Total[Tuples[Table[Range[1,Prime[k]],{k,1,r}]],{2}]]][[2]]] (* generates row r of the table *)
%t Grid@Table[row[r],{r,1,7}] (* generates the table *)
%t Flatten@Table[row[r],{r,1,7}] (* generates the sequence *) (* _Steven Foster Clark_, Feb 02 2023 *)
%t row[r_]:=Drop[#,-Length[#]/2]&[CoefficientList[1/(x-1)^r Product[(x^Prime[i]-1),{i,1,r}],x]] (* generates row r of the table *) (* _Steven Foster Clark_, Feb 07 2023 *)
%o (PARI) row(n)={my(v=Vecrev(prod(i=1, n, 1 - x^prime(i))/(1 - x)^n)); v[1..#v/2]} \\ _Andrew Howroyd_, Feb 06 2023
%Y Row sums are A070826.
%Y Cf. A005521 (row lengths).
%K nonn,tabf
%O 1,3
%A _Stuart Cooper_, Mar 26 2014
%E Name edited by _Andrew Howroyd_, Feb 05 2023
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