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

1, 1, 1, 1, 3, 1, 1, 17, 7, 1, 1, 111, 129, 15, 1, 1, 769, 2815, 769, 31, 1, 1, 5503, 65537, 47103, 4097, 63, 1, 1, 40193, 1579007, 3080193, 647167, 20481, 127, 1, 1, 297727, 38862849, 208470015, 109051905, 7929855, 98305, 255, 1, 1, 2228225, 970522623, 14413725697, 19012780031, 3271557121, 90177535, 458753, 511, 1

Offset: 0

Author

Peter Bala, Mar 26 2020

Keywords

• nonn
• easy
• tabf

Comments

We conjecture that each column sequence satisfies the following supercongruences:

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.

Examples

			Square array begins
      |k=0    k=1       k=2           k=3             k=4
  - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  n=0 | 1      1         1             1               1
  n=1 | 1      3         7            15              31
  n=2 | 1     17       129           769            4097
  n=3 | 1    111      2815         47103          647167
  n=4 | 1    769     65537       3080193       109051905
  n=5 | 1   5503   1579007     208470015     19012780031
  n=6 | 1  40193  38862849   14413725697   3385776406529
  n=7 | 1 297727 970522623 1011196362751 611732191969279
  ...
Examples of congruences for column k = 1:
T(5,1) - T(1,1) = 5503 - 3 = (2^2)*(5^3)*11 == 0 ( mod 5^3 ).
T(7,1) - T(1,1) = 297727 - 3 = (2^2)*(7^4)*31 == 0 ( mod 7^3 ).
T(2*11,1) - T(2,1) = 5913649000782757889 - 17 = (2^4)*(3^2)*(11^3)*107*288357478039 == 0 ( mod 11^3 ).
T(5^2,1) - T(5,1) = 2840491845703386005503 - 5503 = (2^7)*(3^3)*(5^6)*7*19*1123*352183001 == 0 ( mod 5^6 ).
		

Crossrefs

A119259 (column 1), A333561 (column 2), A333562 (column 3). Cf. A333563.

Programs

Maple

T := (n, k) -> add(binomial(n+j-1, j)*2^j, j = 0..n*k):
T_col := k -> seq(T(n, k), n = 0..7):
seq(print(T_col(k)), k = 0..10);

Formula

T(n,k) = Sum_{j = 0..n*k} binomial(n+j-1,j)*2^j.

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.