A241199 -id:A241199 - cp's OEIS frontend

A241200 For the n in A241199, the index of the first of 4 terms in binomial(n,k) that satisfy a quadratic relation.

2, 4, 9, 12, 19, 23, 32, 37, 48, 54, 67, 74, 89, 97, 114, 123, 142, 152, 173, 184, 207, 219, 244, 257, 284, 298, 327, 342, 373, 389, 422, 439, 474, 492, 529, 548, 587, 607, 648, 669, 712, 734, 779, 802, 849, 873, 922, 947, 998, 1024, 1077, 1104, 1159, 1187

Offset: 1

T. D. Noe, Apr 17 2014

This value of k appears to approach n/2 as n grows larger.

			Binomial(14,k) = (1, 14, 91, 364, 1001, 2002, 3003, 3432) for k = 0..7. The 4 quadratic terms begin at k = 2.

Colin Barker, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A008865 (binomial(n,k) has 3 consecutive terms in a linear relation).
Cf. A062730 (3 terms in arithmetic progression in Pascal's triangle).
Cf. A241199 (the values of n).

t = {}; Do[b = Binomial[n, Range[0, n/2]]; d = Differences[b, 3]; If[MemberQ[d, 0], AppendTo[t, Position[d, 0, 1, 1][[1, 1]] - 1]], {n, 3000}]; t
LinearRecurrence[{1,2,-2,-1,1},{2,4,9,12,19},60] (* Harvey P. Dale, Dec 18 2022 *)

Vec(x*(x^2-2)*(x^2+x+1)/((x-1)^3*(x+1)^2) + O(x^100)) \\ Colin Barker, Apr 29 2015

a(n) = (-11-5*(-1)^n-2*(-15+(-1)^n)*n+6*n^2)/16. G.f.: x*(x^2-2)*(x^2+x+1) / ((x-1)^3*(x+1)^2). - Colin Barker, Apr 18 2014 and Apr 29 2015

The terms appear to satisfy a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5), with initial terms 2, 4, 9, 12, 19. - T. D. Noe, Apr 18 2014

A200182 Number of -n..n arrays x(0..3) of 4 elements with zero sum and no two consecutive declines, no adjacent equal elements, and no element more than one greater than the previous (random base sawtooth pattern).

Original entry on oeis.org

3, 6, 11, 14, 19, 26, 31, 38, 47, 54, 63, 74, 83, 94, 107, 118, 131, 146, 159, 174, 191, 206, 223, 242, 259, 278, 299, 318, 339, 362, 383, 406, 431, 454, 479, 506, 531, 558, 587, 614, 643, 674, 703, 734, 767, 798, 831, 866, 899, 934, 971, 1006, 1043, 1082, 1119, 1158

Offset: 1

R. H. Hardin, Nov 13 2011

nonn

			Some solutions for n=6:
..3....4....2....6....5....2....0....6....1....0....0....5....6....1....4....3
.-2....0....1...-2....6....3...-1...-1....2....1....1....0...-3....0...-1....1
.-1....1....2...-1...-6...-3....0....0....3....2...-1....1...-2....1....0....2
..0...-5...-5...-3...-5...-2....1...-5...-6...-3....0...-6...-1...-2...-3...-6

R. H. Hardin, Table of n, a(n) for n = 1..200

Row 4 of A200181.

A014206 is a related sequence.

Empirical: a(n) = 2*a(n-1) -a(n-2) +a(n-3) -2*a(n-4) +a(n-5)
     a(3*k-2) = ((3*k+1)^2)/3 - 7/3.
     a(3*k-1) = ((3*k+2)^2)/3 - 7/3.
     a(3*k)   = ((3*k+3)^2)/3 -  1  = 3*(k+1)^2 - 1.
     a(3*k+1) = ((3*k+4)^2)/3 - 7/3.
     a(3*k+2) = ((3*k+5)^2)/3 - 7/3 ... and so on.
The terms a(3*k-1) and a(3*k+1) seem to be terms of A241199: numbers n such that 4 consecutive terms of binomial(n,k) satisfy a quadratic relation for 0 <= k <= n/2. - Avi Friedlich, Apr 28 2015
Empirical g.f.: -x*(2*x^4-5*x^3+2*x^2+3) / ((x-1)^3*(x^2+x+1)). - Colin Barker, Apr 28 2015

A241201 a(n) is the least r such that there are n+2 consecutive increasing terms in the r-th row of Pascal's triangle (binomial(r,*)) which satisfy a polynomial of degree n.

Original entry on oeis.org

7, 14, 62, 31, 339, 1022

Offset: 1

T. D. Noe, Apr 21 2014

Old definition: "Numbers k such that n+2 consecutive terms of binomial(n,k) satisfy a polynomial relation of degree n for some k in the range 0 <= k <= n/2.".

Is this sequence finite?

			a(1) = 7 because the 3 terms 7, 21, 35 are linear.

Cf. A008865 (binomial(n,k) has 3 consecutive terms in a linear relation).
Cf. A062730 (3 terms in arithmetic progression in Pascal's triangle).
Cf. A241199, A241200 (similar, but quadratic).
Cf. A241202 (position of the first of terms).

t = Table[k = 1; While[b = Binomial[k, Range[0, k/2]]; d = Differences[b, n + 1]; ! MemberQ[d, 0], k++]; {k, Position[d, 0, 1, 1][[1, 1]] - 1}, {n, 6}]; Transpose[t][[1]]

Definition clarified by Don Reble, Dec 14 2020

A241202 Beginning of a polynomial relation of degree n in n+2 terms in the first half of Pascal's triangle. See A241201.

Original entry on oeis.org

1, 2, 26, 9, 149, 489

Offset: 1

T. D. Noe, Apr 21 2014

Is this sequence finite?

Cf. A008865 (binomial(n,k) has 3 consecutive terms in a linear relation).
Cf. A062730 (3 terms in arithmetic progression in Pascal's triangle).
Cf. A241199, A241200 (similar, but quadratic).

t = Table[k = 1; While[b = Binomial[k, Range[0, k/2]]; d = Differences[b, n + 1]; ! MemberQ[d, 0], k++]; {k, Position[d, 0, 1, 1][[1, 1]] - 1}, {n, 6}]; Transpose[t][[2]]

cp's OEIS Frontend

A241200 For the n in A241199, the index of the first of 4 terms in binomial(n,k) that satisfy a quadratic relation.

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A200182 Number of -n..n arrays x(0..3) of 4 elements with zero sum and no two consecutive declines, no adjacent equal elements, and no element more than one greater than the previous (random base sawtooth pattern).

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A241201 a(n) is the least r such that there are n+2 consecutive increasing terms in the r-th row of Pascal's triangle (binomial(r,*)) which satisfy a polynomial of degree n.

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A241202 Beginning of a polynomial relation of degree n in n+2 terms in the first half of Pascal's triangle. See A241201.

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