A172171 -id:A172171 - cp's OEIS frontend

A172172 Sums of NW-SE diagonals of triangle A172171.

0, 1, 10, 20, 39, 68, 116, 193, 318, 520, 847, 1376, 2232, 3617, 5858, 9484, 15351, 24844, 40204, 65057, 105270, 170336, 275615, 445960, 721584, 1167553, 1889146, 3056708, 4945863, 8002580, 12948452, 20951041, 33899502, 54850552, 88750063, 143600624, 232350696

Offset: 0

Mark Dols, Jan 28 2010

This is the sequence A(0,1;1,1;9) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010

Colin Barker, Table of n, a(n) for n = 0..1000
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

Cf. A000032, A000045, A172171.

[Lucas(n+2) +6*Fibonacci(n+1) -9: n in [0..50]]; // G. C. Greubel, Apr 25 2022

CoefficientList[Series[x*(1+8*x)/((1-x)*(1-x-x^2)), {x,0,50}], x] (* G. C. Greubel, Jul 13 2017 *)

concat(0, Vec(x*(1+8*x)/((1-x)*(1-x-x^2)) + O(x^50))) \\ Colin Barker, Jul 13 2017

[fibonacci(n+3) +7*fibonacci(n+1) -9 for n in (0..50)] # G. C. Greubel, Apr 25 2022

a(n) = a(n-1) + a(n-2) + 9 with a(0)=0 and a(1)=1.
From Wolfdieter Lang, Oct 18 2010: (Start)
O.g.f.: x*(1+8*x)/((1-x)*(1-x-x^2)).
a(n) = 2*a(n-1) - a(n-3), a(0)=0, a(1)=1, a(2)=10 (Observation by G. Detlefs).
(End)
a(n+1) - a(n) = A022099(n). - R. J. Mathar, Apr 22 2013
a(n) = -9 + ( (11 + 9*sqrt(5))*(1 + sqrt(5))^n - (11 - 9*sqrt(5))*(1 - sqrt(5))^n )/(2^(n+1)*sqrt(5)). - Colin Barker, Jul 13 2017
a(n) = Fibonacci(n+3) + 7*Fibonacci(n+1) - 9. - G. C. Greubel, Apr 25 2022

Wrong offset 1 changed into 0 Wolfdieter Lang, Oct 18 2010

A172173 Sums of NE-SW diagonals of triangle A172171.

Original entry on oeis.org

0, 1, 1, 11, 12, 32, 44, 85, 129, 223, 352, 584, 936, 1529, 2465, 4003, 6468, 10480, 16948, 27437, 44385, 71831, 116216, 188056, 304272, 492337, 796609, 1288955, 2085564, 3374528, 5460092, 8834629, 14294721, 23129359, 37424080, 60553448, 97977528, 158530985

Offset: 0

Mark Dols, Jan 28 2010

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1).

Cf. A000032, A000045, A172171, A172172.

[Lucas(n) +7*Fibonacci(n-1) -9*((n+1) mod 2): n in [0..50]]; // G. C. Greubel, Apr 25 2022

CoefficientList[Series[x*(1+8*x^2)/((1-x^2)*(1-x-x^2)), {x,0,50}], x] (* G. C. Greubel, Jul 13 2017 *)

concat(0, Vec(x*(1+8*x^2)/((1-x)*(1+x)*(1-x-x^2)) + O(x^50))) \\ Colin Barker, Jul 13 2017

[fibonacci(n+1) +8*fibonacci(n-1) -9*((n+1)%2) for n in (0..50)] # G. C. Greubel, Apr 25 2022

For n=even: a(n) = a(n-1) + a(n-2); for n=odd: a(n) = a(n-1) + a(n-2) + 9 ; with a(0) = 0 and a(1) = 1.
From Colin Barker, Feb 18 2013: (Start)
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4) for n>3.
G.f.: x*(1+8*x^2) / ((1-x)*(1+x)*(1-x-x^2)).
(End)
a(n) = (2^(-1-n)*(-45*((-2)^n+2^n) + (45-7*sqrt(5))*(1+sqrt(5))^n + (1-sqrt(5))^n*(45+7*sqrt(5)))) / 5. - Colin Barker, Jul 13 2017
a(n) = Fibonacci(n+1) + 8*Fibonacci(n-1) - 9*((1+(-1)^n)/2). - G. C. Greubel, Apr 25 2022

Offset corrected by Colin Barker, Feb 18 2013

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

Boris Putievskiy, Aug 15 2013

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

			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;

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Rely Pellicer and David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4
Index entries for triangles and arrays related to Pascal's triangle

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).

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

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

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 *)

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

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

@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

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

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

A172179 (1,[99n+1]) Pascal Triangle.

Original entry on oeis.org

1, 1, 100, 1, 101, 199, 1, 102, 300, 298, 1, 103, 402, 598, 397, 1, 104, 505, 1000, 995, 496, 1, 105, 609, 1505, 1995, 1491, 595, 1, 106, 714, 2114, 3500, 3486, 2086, 694, 1, 107, 820, 2828, 5614, 6986, 5572, 2780, 793, 1, 108, 927, 3648, 8442, 12600, 12558

Offset: 1

Mark Dols, Jan 28 2010

			Triangle begins as:
  1;
  1, 100;
  1, 101, 199;
  1, 102, 300,  298;
  1, 103, 402,  598,  397;
  1, 104, 505, 1000,  995,   496;
  1, 105, 609, 1505, 1995,  1491,   595;
  1, 106, 714, 2114, 3500,  3486,  2086,  694;
  1, 107, 820, 2828, 5614,  6986,  5572, 2780,  793;
  1, 108, 927, 3648, 8442, 12600, 12558, 8352, 3573, 892;

G. C. Greubel, Rows n = 1..50 of the triangle, flattened

Cf. A000225, A007318, A172171, A172178.

Table[99*Binomial[n-1, k-2] + Binomial[n-1, k-1], {n,12}, {k,n}]//Flatten (* G. C. Greubel, Apr 27 2022 *)

flatten([[98*binomial(n-1,k-2) + binomial(n,k-1) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Apr 27 2022

T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-1) - T(n-2,k-2), T(1,0) = T(2,0) = 1, T(2,1) = 100, T(n,k)=0 if k<0 or if k>=n. - Philippe Deléham, Dec 26 2013
From G. C. Greubel, Apr 27 2022: (Start)
T(n, k) = 99*binomial(n-1, k-2) + binomial(n-1, k-1).
T(n, n) = A172178(n-1).
Sum_{k=1..n} T(n, k) = 100*A000225(n-1) + 1. (End)

cp's OEIS Frontend

A172172 Sums of NW-SE diagonals of triangle A172171.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

Extensions

A172173 Sums of NE-SW diagonals of triangle A172171.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

Extensions

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Maple

Mathematica

PARI

Python

Sage

Formula

Extensions

A172179 (1,[99n+1]) Pascal Triangle.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

SageMath

Formula