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 1-3 of 3 results.

A189391 The minimum possible value for the apex of a triangle of numbers whose base consists of a permutation of the numbers 0 to n, and each number in a higher row is the sum of the two numbers directly below it.

Original entry on oeis.org

0, 1, 3, 8, 19, 44, 98, 216, 467, 1004, 2134, 4520, 9502, 19928, 41572, 86576, 179587, 372044, 768398, 1585416, 3263210, 6711176, 13775068, 28255568, 57863214, 118430584, 242061468, 494523536, 1009105372, 2058327344, 4194213448
Offset: 0

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Author

Nathaniel Johnston, Apr 20 2011

Keywords

Comments

This is the Riordan transform of A000217 (triangular numbers) with the Riordan matrix (of the Bell type) A053121 (inverse of the Chebyshev S Bell matrix). See the resulting formulae below. - Wolfdieter Lang, Feb 18 2017.
Conjecture: a(n) is also half the sum of the "cuts-resistance" (see A319416, A319420, A319421) of all binary vectors of length n (see Lenormand, page 4). - N. J. A. Sloane, Sep 20 2018

Examples

			For n = 4 consider the triangle:
....19
...8  11
..5  3  8
.4  1 2  6
3  1 0 2  4
This triangle has 19 at its apex and no other such triangle with the numbers 0 - 4 on its base has a smaller apex value, so a(4) = 19.
		

Crossrefs

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); [0] cat Coefficients(R!((2*x+Sqrt(1-4*x^2)-1)/(2*(2*x-1)^2))); // G. C. Greubel, Aug 24 2018
  • Maple
    a:=proc(n)return add((2*n-4*k-1)*binomial(n,k),k=0..floor((n-1)/2)): end:
    seq(a(n),n=0..50);
  • Mathematica
    CoefficientList[Series[(2*x+Sqrt[1-4*x^2]-1) / (2*(2*x-1)^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 16 2014 *)
  • PARI
    A189391(n)=sum(i=0,(n-1)\2,(2*n-4*i-1)*binomial(n,i))  \\ M. F. Hasler, Jan 24 2012
    

Formula

If n even, a(n) = (n+1/2)*binomial(n,n/2) - 2^(n-1); if n odd, a(n) = ((n+1)/2)*binomial(n+1,(n+1)/2) - 2^(n-1). - N. J. A. Sloane, Nov 01 2018
a(n) = Sum_{k=0..floor((n-1)/2)} (2*n-4*k-1)*binomial(n,k).
G.f.: (2*x+sqrt(1-4*x^2)-1) / (2*(2*x-1)^2). - Alois P. Heinz, Feb 09 2012
a(n) ~ 2^n * (sqrt(2n/Pi)- 1/2). - Vaclav Kotesovec, Mar 16 2014 (formula simplified by Lewis Chen, May 25 2017)
D-finite with recurrence n*a(n) + (n-5)*a(n-1) + 2*(-5*n+6)*a(n-2) + 4*(-n+8)*a(n-3) + 24*(n-3)*a(n-4) = 0. - R. J. Mathar, Jan 04 2017
From Wolfdieter Lang, Feb 18 2017:(Start)
a(n) = Sum_{m=0..n} A053121(n, m)*A000217(m), n >= 0.
G.f.: c(x^2)*Tri(x*c(x^2)), with c and Tri the g.f. of A000108 and A000217, respectively. See the explicit form of the g.f. given above by Alois P. Heinz.
(End)
2*a(n) = A152548(n)-2^n. - R. J. Mathar, Jun 17 2021

A162958 Equals A162956 convolved with (1, 3, 3, 3, ...).

Original entry on oeis.org

1, 4, 10, 19, 25, 40, 67, 94, 100, 115, 142, 175, 208, 280, 388, 469, 475, 490, 517, 550, 583, 655, 763, 850, 883, 955, 1069, 1201, 1372, 1696, 2101, 2344, 2350, 2365, 2392, 2425, 2458, 2530, 2638, 2725, 2758, 2830, 2944, 3076, 3247, 3571, 3976, 4225, 4258
Offset: 1

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Author

Gary W. Adamson, Jul 18 2009

Keywords

Comments

Can be considered a toothpick sequence for N=3, following rules analogous to those in A160552 (= special case of "A"), A151548 = special case "B", and the toothpick sequence A139250 (N=2) = special case "C".
To obtain the infinite set of toothpick sequences, (N = 2, 3, 4, ...), replace the multiplier "2" in A160552 with any N, getting a triangle with 2^n terms. Convolve this A sequence with (1, N, 0, 0, 0, ...) = B such that row terms of A triangles converge to B.
Then generalized toothpick sequences (C) = A convolved with (1, N, N, N, ...).
Examples: A160552 * (1, 2, 0, 0, 0,...) = a B-type sequence A151548.
A160552 * (1, 2, 2, 2, 2,...) = the toothpick sequence A139250 for N=2.
A162956 is analogous to A160552 but replaces "2" with the multiplier "3".
Row terms of A162956 tend to A162957 = (1, 3, 0, 0, 0, ...) * A162956.
Toothpick sequence for N = 3 = A162958 = A162956 * (1, 3, 3, 3, ...).
Row sums of "A"-type triangles = powers of (N+2); since row sums of A160552 = (1, 4, 16, 64, ...), while row sums of A162956 = (1, 5, 25, 125, ...).
Is there an illustration of this sequence using toothpicks? - Omar E. Pol, Dec 13 2016

Crossrefs

Third diagonal of A163311.

Programs

  • Maple
    b:= proc(n) option remember; `if`(n<2, n,
          (j-> 3*b(j)+b(j+1))(n-2^ilog2(n)))
        end:
    a:= proc(n) option remember;
          `if`(n=0, 0, a(n-1)+2*b(n-1)+b(n))
        end:
    seq(a(n), n=1..100);  # Alois P. Heinz, Jan 28 2017
  • Mathematica
    b[n_] := b[n] = If[n<2, n, Function[j, 3*b[j]+b[j+1]][n-2^Floor[Log[2, n]] ]];
    a[n_] := a[n] = If[n == 0, 0, a[n-1] + 2*b[n-1] + b[n]];
    Array[a, 100] (* Jean-François Alcover, Jun 11 2018, after Alois P. Heinz *)

Extensions

Clarified definition by Omar E. Pol, Feb 06 2017

A152547 Triangle, read by rows, derived from Pascal's triangle (see g.f. and example for generating methods).

Original entry on oeis.org

1, 2, 3, 1, 4, 2, 2, 5, 3, 3, 3, 1, 1, 6, 4, 4, 4, 4, 2, 2, 2, 2, 2, 7, 5, 5, 5, 5, 5, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 8, 6, 6, 6, 6, 6, 6, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 9, 7, 7, 7, 7, 7, 7, 7, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
Offset: 0

Views

Author

Paul D. Hanna, Dec 14 2008

Keywords

Examples

			The number of terms in row n is C(n,[n/2]).
Triangle begins:
[1],
[2],
[3,1],
[4,2,2],
[5,3,3,3,1,1],
[6,4,4,4,4,2,2,2,2,2],
[7,5,5,5,5,5,3,3,3,3,3,3,3,3,3,1,1,1,1,1],
[8,6,6,6,6,6,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4,2,2,2,2,2,2,2,2,2,2,2,2,2,2],
[9,7,7,7,7,7,7,7,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
...
ILLUSTRATION OF GENERATING METHOD.
Row n is derived from the binomial coefficients in the following way.
Place markers in an array so that the number of contiguous markers
in row k is C(n,k) and then count the markers along columns.
For example, row 6 of this triangle is generated from C(6,k) like so:
------------------------------------------
1: o - - - - - - - - - - - - - - - - - - -
6: o o o o o o - - - - - - - - - - - - - -
15:o o o o o o o o o o o o o o o - - - - -
20:o o o o o o o o o o o o o o o o o o o o
15:o o o o o o o o o o o o o o o - - - - -
6: o o o o o o - - - - - - - - - - - - - -
1: o - - - - - - - - - - - - - - - - - - -
------------------------------------------
Counting the markers along the columns gives row 6 of this triangle:
[7,5,5,5,5,5,3,3,3,3,3,3,3,3,3,1,1,1,1,1].
Continuing in this way generates all the rows of this triangle.
...
Number of repeated terms in each row of this triangle forms A008315:
1;
1;
1, 1;
1, 2;
1, 3, 2;
1, 4, 5;
1, 5, 9, 5;
1, 6, 14, 14;
1, 7, 20, 28, 14;...
		

Crossrefs

Cf. A152548 (row squared sums), A008315; A152545.

Programs

  • PARI
    {T(n,k)=polcoeff(sum(j=0,n,(x^binomial(n,j) - 1)/(x-1)),k)}
    for(n=0,10, for(k=0, binomial(n,n\2)-1, print1(T(n,k),","));print(""))

Formula

G.f. of row n: Sum_{k=0..n} (x^binomial(n,k) - 1)/(x-1) = Sum_{k=0..binomial(n,n\2)-1} T(n,k)*x^k.
A152548(n) = Sum_{k=0..C(n,[n/2])-1} T(n,k)^2 = Sum_{k=0..[(n+1)/2]} C(n+1, k)*(n+1-2k)^3/(n+1).
Showing 1-3 of 3 results.