A153056 -id:A153056 - cp's OEIS frontend

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.

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

A056520 a(n) = (n + 2)(2n^2 - n + 3)/6.

Original entry on oeis.org

1, 2, 6, 15, 31, 56, 92, 141, 205, 286, 386, 507, 651, 820, 1016, 1241, 1497, 1786, 2110, 2471, 2871, 3312, 3796, 4325, 4901, 5526, 6202, 6931, 7715, 8556, 9456, 10417, 11441, 12530, 13686, 14911, 16207, 17576, 19020, 20541, 22141, 23822

Offset: 0

N. J. A. Sloane, Laura Kasavan (maui12129(AT)cswebmail.com), Aug 26 2000

Hankel transform of A030238. - Paul Barry, Oct 16 2007
Equals (1, 2, 3, 4, 5, ...) convolved with (1, 0, 3, 5, 7, 9, ...). - Gary W. Adamson, Jul 31 2010
a(n) equals n!^2 times the determinant of the n X n matrix whose (i,j)-entry is 1 + KroneckerDelta[i, j] (-1 + (1 + i^2)/i^2). - John M. Campbell, May 20 2011
Positions of ones in A253903 (with offset 1). - Harvey P. Dale, Mar 05 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Kassie Archer, Ethan Borsh, Jensen Bridges, Christina Graves, and Millie Jeske, Cyclic permutations avoiding patterns in both one-line and cycle forms, arXiv:2312.05145 [math.CO], 2023. See p. 2.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Amit Kumar Singh, Akash Kumar and Thambipillai Srikanthan, Accelerating Throughput-aware Run-time Mapping for Heterogeneous MPSoCs, ACM Transactions on Design Automation of Electronic Systems, 2012. - From _N. J. A. Sloane_, Dec 25 2012
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000330, A153056, A153057, A153058, A179904.

Cf. A030238, A253903.

[(n+2)*(2*n^2-n+3)/6: n in [0..40]]; // Vincenzo Librandi, May 24 2011

a[n_] := (n+2)*(2*n^2-n+3)/6; Table[a[n], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Dec 17 2008 *)
s = 1; lst = {s}; Do[s += n^2; AppendTo[lst, s], {n, 1, 41, 1}]; lst (* Zerinvary Lajos, Jul 12 2009 *)
Table[n!^2*Det[Array[KroneckerDelta[#1,#2](((#1^2+1)/(#1^2))-1)+1&,{n,n}]],{n,1,20}] (* John M. Campbell, May 20 2011 *)
FoldList[#1 + #2^2 &, 1, Range@ 40] (* Robert G. Wilson v, Oct 28 2011 *)

a(n)=(n+2)*(2*n^2-n+3)/6 \\ Charles R Greathouse IV, Jul 02 2013

a(n) = a(n-1) + n^2.
a(n) = A000330(n) + 1.
G.f.: (1 - 2*x + 4*x^2 - x^3)/(1 - x)^4. - Paul Barry, Apr 14 2010
Let b(0) = b(1) = 1, b(n) = max(b(n-1) + (n - 1)^2, b(n-2) + (n - 2)^2) for n >= 2; then a(n) = b(n+1). - Yalcin Aktar, Jul 28 2011

A153058 a(0)=4; a(n)=n^2+a(n-1) for n>0.

Original entry on oeis.org

4, 5, 9, 18, 34, 59, 95, 144, 208, 289, 389, 510, 654, 823, 1019, 1244, 1500, 1789, 2113, 2474, 2874, 3315, 3799, 4328, 4904, 5529, 6205, 6934, 7718, 8559, 9459, 10420, 11444, 12533, 13689, 14914, 16210, 17579, 19023, 20544, 22144, 23825, 25589

Offset: 0

Vladimir Joseph Stephan Orlovsky, Dec 17 2008

Indranil Ghosh, Table of n, a(n) for n = 0..64950
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

Cf. A000330, A056520, A153056, A153057, A179904.

a=4;lst={};Do[a=n^2+a;AppendTo[lst,a],{n,0,5!}];lst
RecurrenceTable[{a[0]==4,a[n]==n^2+a[n-1]},a,{n,50}] (* Harvey P. Dale, Apr 27 2012 *)

G.f.: (4-11x+13x^2-4x^3)/(1-x)^4. a(n)=4+A000330(n). - R. J. Mathar, Jan 17 2009

Added indices to definition and corrected offset. - R. J. Mathar, Jan 17 2009

A153057 a(0)=3; a(n) = n^2 + a(n-1) for n>0.

Original entry on oeis.org

3, 4, 8, 17, 33, 58, 94, 143, 207, 288, 388, 509, 653, 822, 1018, 1243, 1499, 1788, 2112, 2473, 2873, 3314, 3798, 4327, 4903, 5528, 6204, 6933, 7717, 8558, 9458, 10419, 11443, 12532, 13688, 14913, 16209, 17578, 19022, 20543, 22143, 23824, 25588

Offset: 0

Vladimir Joseph Stephan Orlovsky, Dec 17 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000330, A056520, A153056, A153058, A179904.

I:=[3,4,8,17]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..60]]; // Vincenzo Librandi, May 09 2017

a=3;lst={};Do[a=n^2+a;AppendTo[lst,a],{n,0,5!}];lst
CoefficientList[Series[(3 - 8 x + 10 x^2 - 3 x^3) / (1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, May 09 2017 *)

From R. J. Mathar, Jan 17 2009: (Start)
G.f.: (3-8*x + 10*x^2 - 3*x^3)/(1 - x)^4.
a(n) = 3+A000330(n). (End)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, May 09 2017

Added indices to definition. Corrected offset R. J. Mathar, Jan 17 2009

A179904 a(n) = A056520(n)+1 for n>0, a(0)=1.

Original entry on oeis.org

1, 3, 7, 16, 32, 57, 93, 142, 206, 287, 387, 508, 652, 821, 1017, 1242, 1498, 1787, 2111, 2472, 2872, 3313, 3797, 4326, 4902, 5527, 6203, 6932, 7716, 8557, 9457, 10418, 11442, 12531, 13687, 14912, 16208, 17577, 19021, 20542, 22142, 23823, 25587

Offset: 0

Gary W. Adamson, Jul 31 2010

Original name: (1,3,5,7,9,..) = A005408 convolved with (1,0,2,3,4,..) = 1 followed by A087156.

			a(3) = 16 = 1 + A056520(3) = (1 + 15).
a(4) = 32 = (9, 7, 5, 3, 1) dot (1, 0, 2, 3, 4) = (9 + 0 + 10 + 9 + 4).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000330, A056520, A153056, A153057, A153058.

LinearRecurrence[{4,-6,4,-1},{1,3,7,16,32},50] (* Harvey P. Dale, Apr 25 2020 *)

From Bruno Berselli, Aug 26 2011: (Start)
G.f.: (1 + x)*(1 - 2*x + 3*x^2 - x^3)/(1 - x)^4.
a(n) = (1/6)*(2*n^3 + 3*n^2 + n + 12) for n>0, a(0)=1. (End)
a(n) = A153056(n) for n > 0. - Georg Fischer, Oct 24 2018

More terms and a(20) added by Bruno Berselli, Aug 26 2011

cp's OEIS Frontend

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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A056520 a(n) = (n + 2)*(2*n^2 - n + 3)/6.

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A153058 a(0)=4; a(n)=n^2+a(n-1) for n>0.

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A153057 a(0)=3; a(n) = n^2 + a(n-1) for n>0.

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A179904 a(n) = A056520(n)+1 for n>0, a(0)=1.

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A056520 a(n) = (n + 2)(2n^2 - n + 3)/6.