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.

A185778 Second weight array of Pascal's triangle (formatted as a rectangle), by antidiagonals.

Original entry on oeis.org

1, -1, -1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 0, 1, 4, 6, 4, 1, 0, 0, 0, 0, 1, 5, 10, 10, 5, 1, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 1, 0, 0, 0, 0, 1, 7, 21, 35, 35, 21, 7, 1, 0, 0, 0, 0, 1, 8, 28, 56, 70, 56, 28, 8, 1, 0, 0, 0, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 0, 0
Offset: 1

Views

Author

Clark Kimberling, Feb 03 2011

Keywords

Comments

Using "->" to mean "is the weight array of" as defined at A144112:
A185779->A144225->A007318->A014430->A077023->A185779, where each of these is formatted as a rectangle (e.g., A007318 is Pascal's triangle). Read in reverse order, each is the accumulation array of the preceding array. It appears that successive weight arrays of A185779 contain Pascal's triangle except for initial terms.

Examples

			Northwest corner:
1....-1....0....0....0....0....0,...0
-1....2....0....0....0....0....0....0
0.....0....0....1....1....1....1....1
0.....0....1....2....3....4....5....6
0.....0....1....3....6....10...15...21
0.....0....1....4....10...20...35...56
		

Crossrefs

Programs

  • Mathematica
    (* This code produces three arrays: A144225, A007318, A185778. *)
    f[n_,0]:=0;f[0,k_]:=0;  (* Used to make the weight array *)
    f[1,1]:=1;f[n_,1]:=0;f[1,k_]:=0
    f[n_,2]:=1;f[2,k_]:=1;
    f[n_,k_]:=-1+(n+k-4)!/((n-2)!*(k-2)!)/;k>1&&n>1;
    TableForm[Table[f[n,k],{n,1,10},{k,1,15}]] (* A144225 *)
    s[n_,k_]:=Sum[f[i,j],{i,1,n},{j,1,k}]; (* accumulation array of {f(n,k)} *)
    TableForm[Table[s[n,k],{n,1,10},{k,1,15}]] (* A007318, Pascal's triangle formatted as a rectangle *)
    w[m_,n_]:=f[m,n]+f[m-1,n-1]-f[m,n-1]-f[m-1,n]/;Or[m>0,n>0];
    TableForm[Table[w[n,k],{n,1,10},{k,1,15}]] (* A185778 *)
    Table[w[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten

Formula

(See the Mathematica code.)

A185779 Third accumulation array of Pascal's triangle (as a rectangle), by antidiagonals.

Original entry on oeis.org

1, 4, 4, 10, 17, 10, 20, 45, 45, 20, 35, 95, 126, 95, 35, 56, 175, 281, 281, 175, 56, 84, 294, 546, 662, 546, 294, 84, 120, 462, 966, 1358, 1358, 966, 462, 120, 165, 690, 1596, 2534, 2941, 2534, 1596, 690, 165, 220, 990, 2502, 4410, 5790, 5790, 4410, 2502, 990, 220, 286, 1375, 3762, 7272, 10620, 12021, 10620, 7272, 3762, 1375, 286, 364, 1859, 5467, 11484, 18432, 23229, 23229, 18432, 11484, 5467, 1859, 364, 455
Offset: 1

Views

Author

Clark Kimberling, Feb 03 2011

Keywords

Comments

Using "Axxxxxx < Ayyyyyy" to mean that Ayyyyyy is the accumulation array of Axxxxxx, as defined at A144112:
A185779 < A144225 < A007318 < A014430 < A077023 < A185779, where each of these is formatted as a rectangle (e.g., A007318 is Pascal's triangle). See A185778.
row 1: A000292
row 2: A095667

Examples

			Northwest corner:
1....4...10...20...35
4....17..45...95...175
10...45..126..281..546
20...95..281..662..1358
		

Crossrefs

Programs

  • Mathematica
    f[n_, k_] := Binomial[n + k + 4, n + 2] - (k + 3)*(k + 4)/2 - (k + 2)* n*(k*n + n + 3*k + 7)/4; TableForm[Table[f[n, k], {n, 1, 5}, {k, 1, 5}]]
    Table[f[n - k + 1, k], {n, 10}, {k, n, 1, -1}] // Flatten (* G. C. Greubel, Jul 12 2017 *)

Formula

T(n,k) = C(n+k+4,n+2) - (k+3)*(k+4)/2 - (k+2)*n*(k*n+n+3*k+7)/4, for k>=1, n>=1.

A077022 Values of k such that sum of first k primes squared is divisible by k-th prime.

Original entry on oeis.org

1, 6, 17, 11156, 16548
Offset: 1

Views

Author

Randy L. Ekl and Zak Seidov, Oct 17 2002

Keywords

Comments

Numbers k such that A072004(k) = 0.
a(6) > 10^11. - Giovanni Resta, Mar 07 2013

Examples

			6 is a term because A024450(6)/prime(6) = 29;
17 is a term because A024450(17)/prime(17) = 284;
11156 is a term because A024450(11156)/prime(11156) = 410066261;
16548 is a term because A024450(16548)/prime(16548) = 941945317.
		

Crossrefs

Programs

  • Mathematica
    Module[{nn=17000,prs},prs=Accumulate[Prime[Range[nn]]^2];Select[Range[ nn],Divisible[prs[[#]],Prime[#]]&]] (* Harvey P. Dale, Nov 23 2013 *)
Showing 1-3 of 3 results.