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 1-4 of 4 results.

A241199 Numbers n such that 4 consecutive terms of binomial(n,k) satisfy a quadratic relation for 0 <= k <= n/2.

Original entry on oeis.org

14, 19, 31, 38, 54, 63, 83, 94, 118, 131, 159, 174, 206, 223, 259, 278, 318, 339, 383, 406, 454, 479, 531, 558, 614, 643, 703, 734, 798, 831, 899, 934, 1006, 1043, 1119, 1158, 1238, 1279, 1363, 1406, 1494, 1539, 1631, 1678, 1774, 1823, 1923, 1974, 2078, 2131
Offset: 1

Views

Author

T. D. Noe, Apr 17 2014

Keywords

Comments

From Robert Israel, Apr 28 2015: (Start)
Numbers n >= 14 such that 3*n + 7 is a square.
This is because
C(n,i+3) - 3*C(n,i+2) + 3*C(n,i+1) - C(n,i) = n!/((n-i)!*(i+3)!) * g(n,i)
where g(n,i) = (n-3-2*i) * ((n-3-2*i)^2 - 3*n - 7). (End)

Examples

			Binomial(14,k) = (1, 14, 91, 364, 1001, 2002, 3003, 3432) for k = 0..7. The 4 terms beginning with 91 equal 182 - 273*x + 182*x^2 for x = 1..4.

Links

Crossrefs

Sequence A241200 gives the position of the first of the 4 terms. Sequence A008865 gives the terms greater than 2 for which 3 consecutive terms satisfy a linear relation.
A014206 is a related sequence. - Avi Friedlich, Apr 28 2015
Cf. A062730 (3 terms in arithmetic progression in Pascal's triangle).

Programs

  • Maple
    map(k -> (3*k^2+8*k+3,3*k^2+10*k+6),[$1..100]); # Robert Israel, Apr 28 2015
  • Mathematica
    Select[Range[2500], MemberQ[Differences[Binomial[#, Range[0, #/2]], 3], 0] &]
LinearRecurrence[{1,2,-2,-1,1},{14,19,31,38,54},50] (* Harvey P. Dale, Oct 29 2017 *)
  • PARI
    Vec(-x*(6*x^4-3*x^3-16*x^2+5*x+14)/((x-1)^3*(x+1)^2) + O(x^100)) \\ Colin Barker, Apr 29 2015
  • PARI
    a(n)=(6*n^2+42*n+55-(-1)^n*(2*n+7))/8 \\ Charles R Greathouse IV, Apr 15 2016

Formula

a(n) = (55-7*(-1)^n-2*(-21+(-1)^n)*n+6*n^2)/8. G.f.: -x*(6*x^4-3*x^3-16*x^2+5*x+14) / ((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 14, 19, 31, 38, 54. - T. D. Noe, Apr 18 2014
Numbers are of the form A200182(3n+1) and A200182(3n-1). - Avi Friedlich, Apr 25 2015
a(2*k-1) = 3*k^2 + 8*k + 3, a(2*k) = 3*k^2 + 10*k + 6. - Robert Israel, Apr 28 2015

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

Original entry on oeis.org

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

Views

Author

T. D. Noe, Apr 17 2014

Keywords

Comments

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

Examples

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

Links

Crossrefs

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).

Programs

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

Formula

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

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

Views

Author

T. D. Noe, Apr 21 2014

Keywords

Comments

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?

Examples

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

Crossrefs

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).

Programs

  • Mathematica
    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]]

Extensions

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

Views

Author

T. D. Noe, Apr 21 2014

Keywords

Comments

Is this sequence finite?

Crossrefs

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).

Programs

  • Mathematica
    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]]
Showing 1-4 of 4 results.

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