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.

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

Original entry on oeis.org

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024
Offset: 1

Views

Author

Boris Putievskiy, Aug 15 2013

Keywords

Comments

The third row is (n^4 - n^2 + 24*n + 24)/12.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

Examples

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;

Links

Crossrefs

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).
Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

Programs

  • GAP
    T:= function(n,k)
    if k=0 then return n^2;
    elif k=n then return 2^n;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019
  • Maple
    T:= proc(n, k) option remember;
      if k=0 then n^2
    elif k=n then 2^k
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019
  • Mathematica
    T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)
  • PARI
    T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019
  • Python
    def funcL(n):
   q = n**2
   return q
def funcR(n):
   q = 2**n
   return q
for n in range (1,9871):
   t=int((math.sqrt(8*n-7) - 1)/ 2)
   i=n-t*(t+1)/2-1
   j=(t*t+3*t+4)/2-n-1
   sum1=0
   sum2=0
   for m1 in range (1,i+1):
      sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
   for m2 in range (1,j+1):
      sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
   sum=sum1+sum2
  • Sage
    @CachedFunction
def T(n, k):
    if (k==0): return n^2
    elif (k==n): return 2^n
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

Formula

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Extensions

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A164855 Generalized Lucas-Pascal triangle: (101*100^n,1).

Original entry on oeis.org

1, 101, 1, 10100, 102, 1, 1010000, 10202, 103, 1, 101000000, 1020202, 10305, 104, 1, 10100000000, 102020202, 1030507, 10409, 105, 1, 1010000000000, 10202020202, 103050709, 1040916, 10514, 106, 1, 101000000000000, 1020202020202
Offset: 0

Views

Author

Mark Dols, Aug 28 2009

Keywords

Examples

			Triangle begins:
1
101,1
10100,102,1
1010000,10202,103,1
101000000,1020202,10305,104,1
10100000000,102020202,1030507,10409,105,1
1010000000000,10202020202,103050709,1040916,10514,106,1
101000000000000,1020202020202,10305070911,104091625,1051430,10620,107,1

Crossrefs

Cf. A164847, A164851, A029635, A228196.

Programs

  • Maple
    A164855 := proc(n,k)
    option remember;
    if n=k then
        1;
    elif k>n or k<0 then
        0;
    elif k = 0 then
        101*100^(n-1) ;
    else
        procname(n-1,k-1)+procname(n-1,k) ;
    end if;
end proc:
seq(seq(A164855(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Nov 03 2016

Formula

T(0,0)=1, T(n+1,0)=101*100^n, T(n,n)=1, T(n,k)=T(n-1,k-1)+T(n-1,k) for 0Philippe Deléham, Dec 27 2013

Extensions

Initial 1 added by Philippe Deléham, Dec 27 2013

A164849 Diagonal sum of (100^n,1) Pascal triangle.

Original entry on oeis.org

1, 1, 101, 102, 10103, 10205, 1010308, 1020513, 101030821, 102051334, 10103082155, 10205133489, 1010308215644, 1020513349133, 101030821564777, 102051334913910, 10103082156478687, 10205133491392597, 1010308215647871284, 1020513349139263881, 101030821564787135165, 102051334913926399046
Offset: 1

Views

Author

Mark Dols, Aug 28 2009

Keywords

Links

Crossrefs

Cf. A164847.

Programs

  • PARI
    Vec(-x*(x-1)*(1+x) / ( (10*x+1)*(10*x-1)*(x^2+x-1) ) + O(x^25)) \\ Michel Marcus, Feb 02 2023

Formula

G.f.: -x*(x-1)*(1+x) / ( (10*x+1)*(10*x-1)*(x^2+x-1) ). - R. J. Mathar, Nov 03 2016

Extensions

More terms from Michel Marcus, Feb 02 2023

A164866 (101^n,1)-Pascal triangle.

Original entry on oeis.org

1, 1, 101, 1, 102, 10201, 1, 103, 10303, 1030301, 1, 104, 10406, 1040604, 104060401, 1, 105, 10510, 1051010, 105101005, 10510100501, 1, 106, 10615, 1061520, 106152015, 10615201506, 1061520150601
Offset: 1

Views

Author

Mark Dols, Aug 28 2009

Keywords

Examples

			Triangle begins:
1
1,101
1,102,10201
1,103,10303,1030301
1,104,10406,1040604,104060401
1,105,10510,1051010,105101005,10510100501
1,106,10615,1061520,106152015,10615201506,1061520150601

Crossrefs

Cf. A164847, A007318
Showing 1-4 of 4 results.

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