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A333560 Square array read by antidiagonals: T(n,k) = Sum_{j = 0..n*k} binomial(n+j-1,j)*2^j; n,k >= 0.

%I A333560 #15 May 01 2023 09:28:41
%S A333560 1,1,1,1,3,1,1,17,7,1,1,111,129,15,1,1,769,2815,769,31,1,1,5503,65537,
%T A333560 47103,4097,63,1,1,40193,1579007,3080193,647167,20481,127,1,1,297727,
%U A333560 38862849,208470015,109051905,7929855,98305,255,1,1,2228225,970522623,14413725697,19012780031,3271557121,90177535,458753,511,1
%N A333560 Square array read by antidiagonals: T(n,k) = Sum_{j = 0..n*k} binomial(n+j-1,j)*2^j; n,k >= 0.
%C A333560 We conjecture that each column sequence satisfies the following supercongruences:
%C A333560 Column k: T(n*p^j, k) == T(n*p^(j-1),k) ( mod p^(3*j) ) for prime p >= 5 and positive integers n and j. Some examples are given below.
%F A333560 T(n,k) = Sum_{j = 0..n*k} binomial(n+j-1,j)*2^j.
%F A333560 Conjectural o.g.f. for column k: 2^(k+1)*x*f'(k,(2^k)*x)/(2*f(k,(2^k)*x) - 1) + 1/(1 + x), where f(k,x) = Sum_{n >= 0} 1/((k+1)*n+1)*C((k+1)*n+1,n)* x^n.
%e A333560 Square array begins
%e A333560       |k=0    k=1       k=2           k=3             k=4
%e A333560   - - - - - - - - - - - - - - - - - - - - - - - - - - - -
%e A333560   n=0 | 1      1         1             1               1
%e A333560   n=1 | 1      3         7            15              31
%e A333560   n=2 | 1     17       129           769            4097
%e A333560   n=3 | 1    111      2815         47103          647167
%e A333560   n=4 | 1    769     65537       3080193       109051905
%e A333560   n=5 | 1   5503   1579007     208470015     19012780031
%e A333560   n=6 | 1  40193  38862849   14413725697   3385776406529
%e A333560   n=7 | 1 297727 970522623 1011196362751 611732191969279
%e A333560   ...
%e A333560 Examples of congruences for column k = 1:
%e A333560 T(5,1) - T(1,1) = 5503 - 3 = (2^2)*(5^3)*11 == 0 ( mod 5^3 ).
%e A333560 T(7,1) - T(1,1) = 297727 - 3 = (2^2)*(7^4)*31 == 0 ( mod 7^3 ).
%e A333560 T(2*11,1) - T(2,1) = 5913649000782757889 - 17 = (2^4)*(3^2)*(11^3)*107*288357478039 == 0 ( mod 11^3 ).
%e A333560 T(5^2,1) - T(5,1) = 2840491845703386005503 - 5503 = (2^7)*(3^3)*(5^6)*7*19*1123*352183001 == 0 ( mod 5^6 ).
%p A333560 T := (n, k) -> add(binomial(n+j-1, j)*2^j, j = 0..n*k):
%p A333560 T_col := k -> seq(T(n, k), n = 0..7):
%p A333560 seq(print(T_col(k)), k = 0..10);
%Y A333560 A119259 (column 1), A333561 (column 2), A333562 (column 3). Cf. A333563.
%K A333560 nonn,easy,tabf
%O A333560 0,5
%A A333560 _Peter Bala_, Mar 26 2020