cp's OEIS Frontend

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.

Showing 1-2 of 2 results.

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

Original entry on oeis.org

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

Views

Author

Peter Bala, Mar 26 2020

Keywords

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.

A333562 a(n) = Sum_{j = 0..3*n} binomial(n+j-1,j)*2^j.

Original entry on oeis.org

1, 15, 769, 47103, 3080193, 208470015, 14413725697, 1011196362751, 71695889072129, 5124481173422079, 368599603785760769, 26648859989512290303, 1934777421539431153665, 140966705275001764839423, 10301634747725237826093057, 754776795329691207916847103
Offset: 0

Views

Author

Peter Bala, Mar 27 2020

Keywords

Comments

Column 3 of the square array A333560. Compare with A119259(n) = Sum_{j = 0..n} binomial(n+j-1,j)*2^j.
We conjecture that this sequence satisfies the congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Some examples are given below.

Examples

			Examples of congruences:
a(11) - a(1) = 26648859989512290303 - 15 = (2^4)*3*(11^3)*417118394526551 == 0 ( mod 11^3 ).
a(3*7) - a(3) = 121414496850169263529624169428526563327 - 47103 = (2^11)*(7^4)*24691554473186884926207539141513 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 3682696038139661781421472944275523824848470015 - 208470015 = (2^16)*(5^7)*71*1315737187*37481160881*205425986821331 == 0 ( mod 5^6 ).
		

Crossrefs

Programs

  • Maple
    seq(add( binomial(n+j-1,j)*2^j, j = 0..3*n), n = 0..25);
  • Mathematica
    Table[(-1)^n - 2^(3*n+1) * Binomial[4*n, 3*n+1] * Hypergeometric2F1[1, 4*n+1, 3*n+2, 2], {n, 0, 15}] (* Vaclav Kotesovec, Mar 28 2020 *)
  • PARI
    a(n) = sum(j = 0, 3*n, binomial(n+j-1,j)*2^j); \\ Michel Marcus, Mar 28 2020

Formula

Conjectural o.g.f.: 1/(1 + x) + 16*x*f'(8*x)/(2*f(8*x) - 1), where f(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + ... is the o.g.f. of A002293.
exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 15*x + 497*x^2 + 22031*x^3 + ... appears to be the o.g.f. of A062752.
a(n) ~ 2^(11*n + 3/2) / (5*sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
Showing 1-2 of 2 results.