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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A339019 Square table read by upwards antidiagonals: T(m,n) = A103438(2*m-1,n)/A103438(1,n) for m>=1, n>=1.

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%I A339019 #42 Dec 22 2024 14:28:50
%S A339019 1,1,1,1,3,1,1,11,6,1,1,43,46,10,1,1,171,386,130,15,1,1,683,3366,1870,
%T A339019 295,21,1,1,2731,29866,28234,6455,581,28,1,1,10923,267086,437350,
%U A339019 149031,17941,1036,36,1
%N A339019 Square table read by upwards antidiagonals: T(m,n) = A103438(2*m-1,n)/A103438(1,n) for m>=1, n>=1.
%F A339019 Let a(i,m) = ((-2)^i)*Sum_{j=0..i} C(2*m,i-j)*C(i+j,j)*((i-j)/(i+j))*B(2*m-i+j), B(s) = A027641(s)/A027642(s) the Bernoulli numbers and N = n*(n+1)/2, then T(m,n) = (1/(2*m))*Sum_{i=2..m} a(i,m)*N^(i-1).
%e A339019 T(3,4) = A103438(2*3-1,4)/A103438(1,4) = 1300/10 = 130.
%e A339019 By formula: a(2,3) = 4*15*1*1*B(4) = -2 and a(3,3) = (-8)*15*4*(2/4)*B(4) = 8 yields T(3,n) = (-N+4*N^2)/3. Since N = 4*5/2 = 10, T(3,4) = (4*10^2-10)/3 = 130.
%e A339019 Table begins:
%e A339019 m\n| 1    2      3       4        5         6          7
%e A339019 ---+-----------------------------------------------------
%e A339019 1  | 1    1      1       1        1         1          1
%e A339019 2  | 1    3      6      10       15        21         28
%e A339019 3  | 1   11     46     130      295       581       1036
%e A339019 4  | 1   43    386    1870     6455     17941      42868
%e A339019 5  | 1  171   3366   28234   149031    586341    1880956
%e A339019 6  | 1  683  29866  437350  3546775  19809461   85475908
%e A339019 7  | 1 2731 267086 6871138 85960967 683338501 3972825676
%Y A339019 Cf. A103438.
%K A339019 nonn,tabl
%O A339019 1,5
%A A339019 _Franz Vrabec_, Dec 24 2020