A164844 -id:A164844 - cp's OEIS frontend

A180175 Diagonal sums of A164844.

1, 10, 101, 1011, 10112, 101123, 1011235, 10112358, 101123593, 1011235951, 10112359544, 101123595495, 1011235955039, 10112359550534, 101123595505573, 1011235955056107, 10112359550561680, 101123595505617787, 1011235955056179467, 10112359550561797254

Offset: 1

Mark Dols, Aug 15 2010

Sums are built along inclined lines through the triangle with (1,2)-steps in the (row,column) indices. - R. J. Mathar, Aug 19 2010

			From _R. J. Mathar_, Aug 19 2010: (Start)
One example is a(5), the sum of numbers in parentheses:
  1;
  1, 10;
  (1), 11, 100;
  1, 12, (111) ; 1000;;
  1, 13, 123 ; 1111, (10000); (End)

Index entries for linear recurrences with constant coefficients, signature (11,-9,-10).

Cf. A022100, A164844.

From R. J. Mathar, Aug 19 2010: (Start)
G.f.: ( 1-x ) / ( (10*x-1)*(x^2+x-1) ).
a(n) = +11*a(n-1) -9*a(n-2) -10*a(n-3).
a(n) = (90*10^n -A022100(n))/89. (End)

More terms from R. J. Mathar, Aug 19 2010

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

A108561 Triangle read by rows: T(n,0)=1, T(n,n)=(-1)^n, T(n+1,k)=T(n,k-1)+T(n,k) for 0 < k < n.

Original entry on oeis.org

1, 1, -1, 1, 0, 1, 1, 1, 1, -1, 1, 2, 2, 0, 1, 1, 3, 4, 2, 1, -1, 1, 4, 7, 6, 3, 0, 1, 1, 5, 11, 13, 9, 3, 1, -1, 1, 6, 16, 24, 22, 12, 4, 0, 1, 1, 7, 22, 40, 46, 34, 16, 4, 1, -1, 1, 8, 29, 62, 86, 80, 50, 20, 5, 0, 1, 1, 9, 37, 91, 148, 166, 130, 70, 25, 5, 1, -1, 1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6, 0, 1, 1, 11, 56, 174, 367

Offset: 0

Reinhard Zumkeller, Jun 10 2005

Sum_{k=0..n} T(n,k) = A078008(n);
Sum_{k=0..n} abs(T(n,k)) = A052953(n-1) for n > 0;
T(n,1) = n - 2 for n > 1;
T(n,2) = A000124(n-3) for n > 2;
T(n,3) = A003600(n-4) for n > 4;
T(n,n-6) = A001753(n-6) for n > 6;
T(n,n-5) = A001752(n-5) for n > 5;
T(n,n-4) = A002623(n-4) for n > 4;
T(n,n-3) = A002620(n-1) for n > 3;
T(n,n-2) = A008619(n-2) for n > 2;
T(n,n-1) = n mod 2 for n > 0;
T(2*n,n) = A072547(n+1).
Sum_{k=0..n} T(n,k)*x^k = A232015(n), A078008(n), A000012(n), A040000(n), A001045(n+2), A140725(n+1) for x = 2, 1, 0, -1, -2, -3 respectively. - Philippe Deléham, Nov 17 2013, Nov 19 2013
(1,a^n) Pascal triangle with a = -1. - Philippe Deléham, Dec 27 2013
T(n,k) = A112465(n,n-k). - Reinhard Zumkeller, Jan 03 2014

			From _Philippe Deléham_, Nov 17 2013: (Start)
Triangle begins:
  1;
  1, -1;
  1,  0,  1;
  1,  1,  1, -1;
  1,  2,  2,  0,  1;
  1,  3,  4,  2,  1, -1;
  1,  4,  7,  6,  3,  0,  1; (End)

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
C. Merino, S. D. Noble, M. Ramirez-Ibanez, R. Villarroel-Flores, On the structure of the h-vector of a paving matroid, Eur. J. Comb. 33, No. 8, 1787-1799 (2012).
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318 (a=1), A008949(a=2), A164844(a=10).
Similar to the triangles A035317, A059259, A080242, A112555.
Cf. A228196
Cf. A072547 (central terms).

Flat(List([0..13],n->List([0..n],k->Sum([0..k],i->Binomial(n,i)*(-2)^(k-i))))); # Muniru A Asiru, Feb 19 2018

a108561 n k = a108561_tabl !! n !! k
a108561_row n = a108561_tabl !! n
a108561_tabl = map reverse a112465_tabl
-- Reinhard Zumkeller, Jan 03 2014

A108561 := (n, k) -> add(binomial(n, i)*(-2)^(k-i), i = 0..k):
seq(seq(A108561(n,k), k = 0..n), n = 0..12); # Peter Bala, Feb 18 2018

Clear[t]; t[n_, 0] = 1; t[n_, n_] := t[n, n] = (-1)^Mod[n, 2]; t[n_, k_] := t[n, k] = t[n-1, k] + t[n-1, k-1]; Table[t[n, k], {n, 0, 13}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)

def A108561_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
    return [(-1)^k*prec(n, k) for k in (1..n-1)]+[(-1)^(n+1)]
for n in (1..12): print(A108561_row(n)) # Peter Luschny, Mar 16 2016

G.f.: (1-y*x)/(1-x-(y+y^2)*x). - Philippe Deléham, Nov 17 2013
T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 17 2013
From Peter Bala, Feb 18 2018: (Start)
T(n,k) = Sum_{i = 0..k} binomial(n,i)*(-2)^(k-i), 0 <= k <= n.
The n-th row polynomial is the n-th degree Taylor polynomial of the rational function (1 + x)^n/(1 + 2*x) about 0. For example, for n = 4, (1 + x)^4/(1 + 2*x) = 1 + 2*x + 2*x^2 + x^4 + O(x^5). (End)

Definition corrected by Philippe Deléham, Dec 26 2013

A164851 Generalized Lucas-Pascal triangle; (11*10^n, 1).

Original entry on oeis.org

1, 11, 1, 110, 12, 1, 1100, 122, 13, 1, 11000, 1222, 135, 14, 1, 110000, 12222, 1357, 149, 15, 1, 1100000, 122222, 13579, 1506, 164, 16, 1, 11000000, 1222222, 135801, 15085, 1670, 180, 17, 1

Offset: 0

Mark Dols, Aug 28 2009

			Triangle begins:
         1;
        11,      1;
       110,     12,      1;
      1100,    122,     13,     1;
     11000,   1222,    135,    14,    1;
    110000,  12222,   1357,   149,   15,   1;
   1100000, 122222,  13579,  1506,  164,  16,  1;
  11000000,1222222, 135801, 15085, 1670, 180, 17, 1;
  ...

Robert Israel, Table of n, a(n) for n = 0..10010(rows 0 to 140, flattened)

Cf. A029635, A164844, A228196.

G[0]:= 1;
G[1]:= 11+x;
G[2]:= 110+12*x+x^2;
for nn from 3 to 20 do
  G[nn]:= expand((x+11)*G[nn-1]-10*(x+1)*G[nn-2]);
od:
seq(seq(coeff(G[n],x,j),j=0..n),n=0..20); # Robert Israel, Jul 17 2017

T[0, 0] := 1; T[n_, n_] := 1; T[n_, 0] := 11*10^(n - 1); T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k];  Table[T[n, k], {n, 0, 10}, {k, 0, n}] //Flatten (* G. C. Greubel, Dec 22 2017 *)

T(0,0)=1, T(n+1,0)=11*10^n, T(n,n)=1, T(n,k)=T(n-1,k-1)+T(n-1,k) for 0Philippe Deléham, Dec 27 2013

G.f. as triangle: (1-x^2)/((1-10*x)*(1-x-x*y)). - Robert Israel, Jul 17 2017

Initial 1 added by Philippe Deléham, Dec 27 2013

A164899 Binomial matrix (1,10^n) read by antidiagonals.

Original entry on oeis.org

1, 1, 10, 1, 11, 100, 1, 12, 110, 1000, 1, 13, 121, 1100, 10000, 1, 14, 133, 1210, 11000, 100000, 1, 15, 146, 1331, 12100, 110000, 1000000, 1, 16, 160, 1464, 13310, 121000, 1100000, 10000000, 1, 17, 175, 1610, 14641, 133100, 1210000, 11000000, 100000000

Offset: 1

Mark Dols, Aug 30 2009

			Matrix array, A(n, k), begins:
  1, 10, 100, 1000, ...
  1, 11, 110, 1100, ...
  1, 12, 121, 1210, ...
  1, 13, 133, 1331, ...
  1, 14, 146, 1464, ...
  1, 15, 160, 1610, ...
Antidiagonal triangle, T(n, k), begins as:
  1;
  1, 10;
  1, 11, 100;
  1, 12, 110, 1000;
  1, 13, 121, 1100, 10000;
  1, 14, 133, 1210, 11000, 100000;
  1, 15, 146, 1331, 12100, 110000, 1000000;

G. C. Greubel, Antidiagonals n = 1..50, flattened

Cf. A001020, A007318, A021093, A164844.

Cf. A094704 (antidiagonal row sums).

function T(n,k) // T = A164899
  if k eq n then return 10^(n-1);
  elif k eq 1 then return 1;
  else return T(n-1,k) + T(n-2,k-1);
  end if; return T;
end function;
[T(n,k): k in [1..n], n in [1..15]]; // G. C. Greubel, Feb 10 2023

T[n_, k_]:= T[n,k]= If[k==n, 10^(n-1), If[k==1, 1, T[n-1,k] +T[n-2, k-1]]];
Table[T[n, k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Feb 10 2023 *)

def T(n,k): # T = A164899
    if (k==n): return 10^(n-1)
    elif (k==1): return 1
    else: return T(n-1,k) + T(n-2,k-1)
flatten([[T(n,k) for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Feb 10 2023

Sum_{k=1..n} T(n, k) = A094704(n).
As a triangle T(n,k) read by rows, T(n,1) = 1, T(n,n) = 10^(n-1), and T(n,k) = T(n-1, k) + T(n-2, k-1) otherwise. - Joerg Arndt, Dec 10 2016
From G. C. Greubel, Feb 10 2023: (Start)
A(n, k) = A(n-1, k) + A(n-1, k-1), with A(n, 1) = 1 and A(1, k) = 10^(k-1) (array).
T(n, k) = A(n-k+1, k) (antidiagonal triangle). (End)

A164854 Diagonal sum of generalized Pascal triangle; (10^n,1).

Original entry on oeis.org

1, 1, 11, 12, 113, 125, 1138, 1263, 11401, 12664, 114065, 126729, 1140794, 1267523, 11408317, 12675840, 114084157, 126759997, 1140844154, 1267604151, 11408448305, 12676052456, 114084500761, 126760553217, 1140845053978, 1267605607195, 11408450661173

Offset: 0

Mark Dols, Aug 28 2009

Robert Israel, Table of n, a(n) for n = 0..1990
Index entries for linear recurrences with constant coefficients, signature (1,11,-10,-10).

Cf. A164844, A164852, A000045.

f:= gfun:-rectoproc({10*a(n-4)+10*a(n-3)-11*a(n-2)-a(n-1)+a(n),
a(0)=1,a(1)=1,a(2)=11,a(3)=12},a(n),remember):
map(f, [$0..30]); # Robert Israel, Jul 01 2016

LinearRecurrence[{1,11,-10,-10},{1,1,11,12},30] (* Harvey P. Dale, Apr 07 2022 *)

From Robert Israel, Jul 01 2016: (Start)
G.f.: (1-x^2)/((1-10*x^2)*(1-x-x^2)).
a(n) = (171-9*(-1)^n)*10^floor(n/2)/142 + (A000045(n)-10*A000045(n+2))/71. (End)
a(n) = a(n-1)+11*a(n-2)-10*a(n-3)-10*a(n-4). - Wesley Ivan Hurt, Apr 21 2021

More terms from Harvey P. Dale, Apr 07 2022

cp's OEIS Frontend

A164881 Inverse of A164844.

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A168647 Reverse (palindrome) of A164844.

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A180175 Diagonal sums of A164844.

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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.

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A108561 Triangle read by rows: T(n,0)=1, T(n,n)=(-1)^n, T(n+1,k)=T(n,k-1)+T(n,k) for 0 < k < n.

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A164851 Generalized Lucas-Pascal triangle; (11*10^n, 1).

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Maple

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A164899 Binomial matrix (1,10^n) read by antidiagonals.

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