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.

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A062751 Coefficient array for certain polynomials N(4; k,x) (rising powers in x).

Original entry on oeis.org

1, 4, -6, 4, -1, 22, -80, 139, -140, 84, -28, 4, 140, -851, 2500, -4536, 5516, -4616, 2640, -990, 220, -22, 969, -8420, 35504, -94584, 175564, -237600, 239250, -179960, 100078, -40040, 10920, -1820, 140, 7084, -80776, 448056
Offset: 0

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Author

Wolfdieter Lang, Jul 12 2001

Keywords

Comments

The g.f. for the sequence of column r=3*k+j, k >= 0, j=1,2,3, of the staircase array A062750(n,r) is N(4; k,x)*(x^(k+1))/(1-x)^(3*k+1+j) with N(4; k,x) := sum(a(k,p)*x^p,p=0..3*k).
The m=0 column gives: A002293(n+1). The row sums give A000012 (powers of 1) and (unsigned) A062752.
The sequence of step width of this staircase array is [1,3,3,3,...], i.e. the degree of the row polynomials is [0,3,6,9,...]= A008585.

Examples

			{1}; {4,-6,4,-1}; {22,-80,139,-140,84,-28,4}; ...; N(4; 1,x)= 4-6*x+4*x^2-x^3 =(2-x)*(2-2*x+x^2).
		

Formula

a(k, p) := [x^p]N(4; k, x) with N(4; k, x)=(N(4; k-1, x)-A002293(k)*(1-x)^(3*k+1))/x, N(4; 0, x) := 1.
a(n, k)= a(n-1, k+1)+((-1)^k)*binomial(3*n+1, k+1)*binomial(4*n+1, n)/(4*n+1) if k=0, .., (3*n-4); a(n, k)= ((-1)^k)*binomial(3*n+1, k+1)*binomial(4*n+1, n)/(4*n+1) if k=(3*n-3), ..., 3*n; else 0.

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.