A164845 -id:A164845 - cp's OEIS frontend

A128134 A128132 * A007318.

1, 1, 2, 2, 5, 3, 3, 10, 11, 4, 4, 17, 27, 19, 5, 5, 26, 54, 56, 29, 6, 6, 37, 95, 130, 100, 41, 7, 7, 50, 153, 260, 265, 162, 55, 8, 8, 65, 231, 469, 595, 483, 245, 71, 9, 9, 82, 332, 784, 1190, 1204, 812, 352, 89, 10

Offset: 1

Gary W. Adamson, Feb 15 2007

A007318 * A128132 = A128133. Row sums = A128135: (1, 3, 10, 28, 72, 176, ...).

			Triangle T(n,k) (with rows n >= 1 and columns k = 1..n) begins:
  1;
  1,  2;
  2,  5,  3;
  3, 10, 11,   4;
  4, 17, 27,  19,   5;
  5, 26, 54,  56,  29,  6;
  6, 37, 95, 130, 100, 41, 7;
  ...

Cf. A002522, A007318, A028387, A128132, A128133, A128135, A164845, A332697.

A128132 * A007318 as infinite lower triangular matrices (assuming the top of the Pascal triangle A007318 is shifted from (0,0) to (1,1)).
From Petros Hadjicostas, Jul 26 2020: (Start)
T(n,k) = n*binomial(n-1, k-1) - binomial(n-2, k-1)*[n <> k] for 1 <= k <= n, where [ ] is the Iverson bracket.
Bivariate o.g.f.:  x*y*(1 - x + x^2*(1 + y))/(1 - x*(1 + y))^2.
T(n,k) = T(n-1,k) + T(n-1,k-1) + binomial(n-1,k-1) for 2 <= k <= n with (n,k) <> (2,2).
T(n,k) = 2*T(n-1,k) - T(n-2,k) - T(n-2,k-2) + 2*T(n-1,k-1) - 2*T(n-2,k-1) for 3 <= k <= n with (n,k) <> (3,3).
T(n,1) = n - 1 for n >= 2.
T(n,2) = A002522(n-1) for n >= 2.
T(n,3) = A164845(n-3) for n >= 3.
T(n,4) = A332697(n-3) for n >= 4.
T(n,n) = n for n >= 1.
T(n,n-1) = A028387(n-2) for n >= 2. (End)

A164900 a(2n) = 4n(n+1) + 3; a(2n+1) = 2n(n+2) + 3.

Original entry on oeis.org

3, 3, 11, 9, 27, 19, 51, 33, 83, 51, 123, 73, 171, 99, 227, 129, 291, 163, 363, 201, 443, 243, 531, 289, 627, 339, 731, 393, 843, 451, 963, 513, 1091, 579, 1227, 649, 1371, 723, 1523, 801, 1683, 883, 1851, 969, 2027, 1059, 2211, 1153

Offset: 0

Paul Curtz, Aug 30 2009

a(n) = largest odd divisor of A059100(n+1). Proof: Observe that a(2n) = A059100(2n+1) and a(2n+1) = (A059100(2n+2))/2 and note that (A059100(m))/2 is odd for even m. - Jeremy Gardiner, Aug 25 2013

a(n) is also the denominator of the (n+1)-st largest circle in a special case of the Pappus chain inspired by the Yin-Yang symbol. See illustration in the links. - Kival Ngaokrajang, Jun 20 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Kival Ngaokrajang, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A026741, A058331, A059100, A164845, A164897.

[((-1)^n+3)*(n^2+2*n+3)/4: n in [0..50]]; // Vincenzo Librandi, Aug 07 2011

LinearRecurrence[{0, 3, 0, -3, 0, 1}, {3, 3, 11, 9, 27, 19}, 50] (* Amiram Eldar, Aug 09 2022 *)

vector(100,n,n--;(1/4)*((-1)^n+3)*(n^2+2*n+3)) \\ Derek Orr, Jun 27 2015

a(2n) = A164897(n); a(2n+1) = A058331(n+1).
a(n) = A164845(n-1)/A026741(n), n>0.
G.f.: ( -3-3*x-2*x^2-3*x^4-x^5 ) / ( (x-1)^3*(1+x)^3 ). - R. J. Mathar, Jan 21 2011
a(n) = ((-1)^n+3)*(n^2+2*n+3)/4. - Bruno Berselli, Jan 21 2011
From Amiram Eldar, Aug 09 2022: (Start)
a(n) = numerator(((n+1)^2 + 2)/2).
Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(2))*Pi/sqrt(2) + tanh(Pi/sqrt(2))*Pi/(2*sqrt(2)) - 1)/2. (End)
E.g.f.: ((6 + 3*x + 2*x^2)*cosh(x) + (3 + 6*x + x^2)*sinh(x))/2. - Stefano Spezia, Oct 19 2024

A125027 Binomial transform of the "1,2,3,..." triangle.

Original entry on oeis.org

1, 3, 3, 9, 11, 6, 26, 32, 27, 10, 72, 86, 85, 54, 15, 192, 222, 233, 189, 95, 21, 496, 558, 597, 549, 371, 153, 28, 1248, 1374, 1473, 1446, 1160, 664, 231, 36, 3072, 3326, 3549, 3600, 3203, 2246, 1107, 332, 45, 7424, 7934, 8409, 8659, 8201, 6567, 4051, 1745, 459, 55, 17664, 18686, 19669, 20367, 20015, 17503, 12597, 6893, 2629, 615, 66

Offset: 1

Gary W. Adamson, Nov 15 2006

			First few rows of the triangle:
   1;
   3,  3;
   9, 11,  6;
  26, 32, 27, 10;
  72, 86, 85, 54, 15;
  ...

Cf. A072863 (first column), A000217 (diagonal), A164845 (subdiagonal).

A27 := proc(n,k)
    option remember;
    if k>= 0 and k <=n then
        if k = 0 then
            1+procname(n-1,n-1) ;
        else
            procname(n,0)+k ;
        end if;
    else
        0;
    end if;
end proc:
A125027 := proc(n,k)
    add( binomial(n,j)*A27(j,k),j=k..n) ;
end proc: # R. J. Mathar, May 21 2018

A27[n_, k_] := A27[n, k] = If[k >= 0 && k <= n, If[k == 0, 1+A27[n-1, n-1], A27[n, 0]+k], 0];
A125027[n_, k_] := Sum[Binomial[n, j]*A27[j, k], {j, k, n}];
Table[A125027[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 27 2024, after R. J. Mathar *)

Given the triangle (natural numbers in succession: 1; 2,3; 4,5,6; ...) as an infinite matrix M and P = Pascal's triangle as a lower triangular matrix, perform P*M, deleting the zeros.

The row sums s(n) = 1, 6, 26, 95, 312, 952, ... obey (-3*n+2)*s(n) +(9*n+7)*s(n-1) + 2*(-3*n-2)*s(n-2) = 0. - R. J. Mathar, May 21 2018

cp's OEIS Frontend

A128134 A128132 * A007318.

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Formula

A164900 a(2n) = 4n(n+1) + 3; a(2n+1) = 2n(n+2) + 3.

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Magma

Mathematica

PARI

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A125027 Binomial transform of the "1,2,3,..." triangle.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

cp's OEIS Frontend

A128134 A128132 * A007318.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A164900 a(2n) = 4*n*(n+1) + 3; a(2n+1) = 2*n*(n+2) + 3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A125027 Binomial transform of the "1,2,3,..." triangle.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A164900 a(2n) = 4n(n+1) + 3; a(2n+1) = 2n(n+2) + 3.