A118191 -id:A118191 - cp's OEIS frontend

A118195 Self-convolution square-root of A118191, where A118191 is column 0 of the matrix square of triangle A118190 with A118190(n,k) = (5^k)^(n-k).

1, 1, 3, 23, 411, 15771, 1353045, 252512065, 106798723795, 99080638950595, 208993838938550873, 968425792397232696773, 10208662119796586878979989, 236472963735267887311598074949, 12462692176683507314938059670486683

Offset: 0

Paul D. Hanna, Apr 15 2006

nonn

In general, sqrt( Sum_{n>=0} x^n/(1 - q^n*x) ) is an integer series whenever q == 1 (mod 4).

			A(x) = 1 + x + 3*x^2 + 23*x^3 + 411*x^4 + 15771*x^5 + ...
A(x)^2 = 1 + 2*x + 7*x^2 + 52*x^3 + 877*x^4 + 32502*x^5 + ...
= 1/(1-x) + x/(1-5x) + x^2/(1-25x) + x^3/(1-125x) + ...

G. C. Greubel, Table of n, a(n) for n = 0..75

Cf. A118190, A118191.

m:=30;
R:=PowerSeriesRing(Rationals(), m);
Coefficients(R!( Sqrt( (&+[x^j/(1-5^j*x): j in [0..m+2]]) ) )); // G. C. Greubel, Jun 30 2021

With[{m = 30}, CoefficientList[Series[Sqrt[Sum[x^j/(1 - 5^j*x), {j, 0, m + 2}]], {x, 0, m}], x]] (* G. C. Greubel, Jun 30 2021 *)

a(n)=polcoeff(sqrt(sum(k=0,n,sum(j=0, k, (5^j)^(k-j) )*x^k+x*O(x^n))),n)

m=30;
def A118195_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( sqrt(sum( x^j/(1-5^j*x) for j in (0..m+2))) ).list()
A118195_list(m) # G. C. Greubel, Jun 30 2021

G.f.: A(x) = sqrt( Sum_{n>=0} x^n/(1-5^n*x) ).

a(n) ~ A118191(n)/2. - Vaclav Kotesovec, Aug 20 2025

A118190 Triangle T(n,k) = 5^(k*(n-k)) for n >= k >= 0, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 25, 25, 1, 1, 125, 625, 125, 1, 1, 625, 15625, 15625, 625, 1, 1, 3125, 390625, 1953125, 390625, 3125, 1, 1, 15625, 9765625, 244140625, 244140625, 9765625, 15625, 1, 1, 78125, 244140625, 30517578125, 152587890625, 30517578125, 244140625, 78125, 1

Offset: 0

Paul D. Hanna, Apr 15 2006

Matrix power T^m satisfies: [T^m](n,k) = [T^m](n-k,0)*T(n,k) for all m and so the triangle has an invariant character. For example, the matrix inverse is defined by [T^-1](n,k) = A118193(n-k)*T(n,k); also, the matrix log is given by [log(T)](n,k) = A118194(n-k)*T(n,k).

For any column vector C, the matrix product of T*C transforms the g.f. of C: Sum_{n>=0} c(n)*x^n into the g.f.: Sum_{n>=0} c(n)*x^n/(1-5^n*x).

			A(x,y) = 1/(1-x*y) + x/(1-5*x*y) + x^2/(1-25*x*y) + x^3/(1-125*x*y) + ...
Triangle begins:
  1;
  1,     1;
  1,     5,       1;
  1,    25,      25,         1;
  1,   125,     625,       125,         1;
  1,   625,   15625,     15625,       625,       1;
  1,  3125,  390625,   1953125,    390625,    3125,     1;
  1, 15625, 9765625, 244140625, 244140625, 9765625, 15625, 1; ...
The matrix inverse T^-1 starts:
         1;
        -1,       1;
         4,      -5,        1;
       -76,     100,      -25,     1;
      7124,   -9500,     2500,  -125,    1;
  -3326876, 4452500, -1187500, 62500, -625, 1; ...
where [T^-1](n,k) = A118193(n-k)*(5^k)^(n-k).

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A118191 (row sums), A118192 (antidiagonal sums), A118193, A118194.

Cf. A117401 (m=0), A118180 (m=1), A118185 (m=2), this sequence (m=3), A158116 (m=4), A176642 (m=6), A158117 (m=8), A176627 (m=10), A176639 (m=13), A156581 (m=15).

[5^(k*(n-k)): k in [0..n], n in [0..12]]; // G. C. Greubel, Jun 29 2021

With[{m=3}, Table[(m+2)^(k*(n-k)), {n,0,12}, {k,0,n}]//Flatten] (* G. C. Greubel, Jun 29 2021 *)

PARI
```
T(n, k)=if(n
				
```

flatten([[5^(k*(n-k)) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 29 2021

G.f.: A(x,y) = Sum_{n>=0} x^n/(1-5^n*x*y).
G.f. satisfies: A(x,y) = 1/(1-x*y) + x*A(x,5*y).
T(n,k) = (1/n)*( 5^(n-k)*k*T(n-1,k-1) + 5^k*(n-k)*T(n-1,k) ), where T(i,j)=0 if j>i. - Tom Edgar, Feb 21 2014
T(n, k, m) = (m+2)^(k*(n-k)) with m = 3. - G. C. Greubel, Jun 29 2021

A118192 Antidiagonal sums of triangle A118190: a(n) = Sum_{k=0..floor(n/2)} 5^(k(n-2k)) for n>=0.

Original entry on oeis.org

1, 1, 2, 6, 27, 151, 1252, 18876, 421877, 11797501, 489062502, 36867190626, 4119892578127, 576049853531251, 119400024902343752, 45003894807128984376, 25145828723919677734377, 17579646409034759521875001

Offset: 0

Paul D. Hanna, Apr 15 2006

nonn

			A(x) = 1/(1-x^2) + x/(1-5*x^2) + x^2/(1-25*x^2) + x^3/(1-125*x^2) + ...
  = 1 + x + 2*x^2 + 6*x^3 + 27*x^4 + 151*x^5 + ...

G. C. Greubel, Table of n, a(n) for n = 0..100

Cf. A118190 (triangle), A118191 (row sums).

[(&+[5^(k*(n-2*k)): k in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, Jun 29 2021

Table[Sum[5^(k*(n-2*k)), {k,0,Floor[n/2]}], {n,0,30}] (* G. C. Greubel, Jun 29 2021 *)

PARI
```
a(n)=sum(k=0, n\2, (5^k)^(n-2*k) )
```

[sum(5^(k*(n-2*k)) for k in (0..n//2)) for n in (0..30)] # G. C. Greubel, Jun 29 2021

G.f.: A(x) = Sum_{n>=0} x^n/(1-5^n*x^2).
a(2*n) = Sum_{k=0..n} 5^(2*k*(n-k)).
a(2*n+1) = Sum_{k=0..n} 5^(k*(2*(n-k)+1)).

cp's OEIS Frontend

A118195 Self-convolution square-root of A118191, where A118191 is column 0 of the matrix square of triangle A118190 with A118190(n,k) = (5^k)^(n-k).

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A118190 Triangle T(n,k) = 5^(k*(n-k)) for n >= k >= 0, read by rows.

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Magma

Mathematica

PARI

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A118192 Antidiagonal sums of triangle A118190: a(n) = Sum_{k=0..floor(n/2)} 5^(k*(n-2*k)) for n>=0.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

A118192 Antidiagonal sums of triangle A118190: a(n) = Sum_{k=0..floor(n/2)} 5^(k(n-2k)) for n>=0.