A096956 -id:A096956 - cp's OEIS frontend

A096957 Fourth column (m=3) of (1,6)-Pascal triangle A096956.

6, 19, 40, 70, 110, 161, 224, 300, 390, 495, 616, 754, 910, 1085, 1280, 1496, 1734, 1995, 2280, 2590, 2926, 3289, 3680, 4100, 4550, 5031, 5544, 6090, 6670, 7285, 7936, 8624, 9350, 10115, 10920, 11766, 12654, 13585, 14560, 15580, 16646, 17759, 18920

Offset: 0

Wolfdieter Lang, Aug 13 2004

If Y is a 6-subset of an n-set X then, for n>=8, a(n-8) is the number of 3-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 16 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000292, A096956.

Cf. other columns: A096958 (m = 4), A096959 (m = 5), A097297 (m = 6), A097298 (m = 7), A097299 (m = 8), A097300 (m = 9).

I:=[6,19,40,70]; [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..50]]; // Vincenzo Librandi, Apr 19 2017

CoefficientList[Series[(6 - 5*x)/(1 - x)^4, {x, 0, 40}], x] (* Wesley Ivan Hurt, Apr 18 2017 *)
LinearRecurrence[{4, -6, 4, -1}, {6, 19, 40, 70}, 50] (* Vincenzo Librandi, Apr 19 2017 *)

a(n) = A096956(n+3, 3) = 6*b(n) - 5*b(n-1) = (n+18)*binomial(n+2, 2)/3, with b(n) = A000292(n) = binomial(n+3, 3).
G.f.: (6-5*x)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - Vincenzo Librandi, Apr 19 2017
E.g.f.: exp(x)*(36 + 78*x + 24*x^2 + x^3)/6. - Stefano Spezia, May 02 2025

A097297 Seventh column (m=6) of (1,6)-Pascal triangle A096956.

Original entry on oeis.org

6, 37, 133, 364, 840, 1722, 3234, 5676, 9438, 15015, 23023, 34216, 49504, 69972, 96900, 131784, 176358, 232617, 302841, 389620, 495880, 624910, 780390, 966420, 1187550, 1448811, 1755747, 2114448, 2531584, 3014440, 3570952, 4209744

Offset: 0

Wolfdieter Lang, Aug 13 2004

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000579, A096956.

Cf. other columns: A096957 (m = 3), A096958 (m = 4), A096959 (m = 5), A097298 (m = 7), A097299 (m = 8), A097300 (m = 9).

A097297[n_] := (n + 36)*Binomial[n + 5, 5]/6;
Array[A097297, 50, 0] (* Paolo Xausa, May 02 2025 *)

a(n) = A096956(n+6, 6) = 6*b(n) - 5*b(n-1) = (n+36)*binomial(n+5, 5)/6, with b(n) = A000579(n+6) = binomial(n+6, 6).

G.f.: (6-5*x)/(1-x)^7.

A096958 Fifth column (m=4) of (1,6)-Pascal triangle A096956.

Original entry on oeis.org

6, 25, 65, 135, 245, 406, 630, 930, 1320, 1815, 2431, 3185, 4095, 5180, 6460, 7956, 9690, 11685, 13965, 16555, 19481, 22770, 26450, 30550, 35100, 40131, 45675, 51765, 58435, 65720, 73656, 82280, 91630, 101745, 112665, 124431, 137085, 150670

Offset: 0

Wolfdieter Lang, Aug 13 2004

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000332, A096956.

Cf. other columns: A096957 (m = 3), A096959 (m = 5), A097297 (m = 6), A097298 (m = 7), A097299 (m = 8), A097300 (m = 9).

[(n+24)*Binomial(n+3, 3) div 4: n in [0..40]]; // Vincenzo Librandi, Oct 01 2013

Table[(n + 24) Binomial[n+3, 3]/4, {n, 0, 50}] (* Vincenzo Librandi, Oct 01 2013 *)

a(n) = A096956(n+4, 4) = 6*b(n) - 5*b(n-1) = (n+24)*binomial(n+3, 3)/4, with b(n) = A000332(n) = binomial(n+4, 4).
G.f.: (6-5*x)/(1-x)^5.
a(n) = sum_{k=1..n+1} ( sum_{i=1..k} i*(n-k+7) ). - Wesley Ivan Hurt, Sep 26 2013

A096959 Sixth column (m=5) of (1,6)-Pascal triangle A096956.

Original entry on oeis.org

6, 31, 96, 231, 476, 882, 1512, 2442, 3762, 5577, 8008, 11193, 15288, 20468, 26928, 34884, 44574, 56259, 70224, 86779, 106260, 129030, 155480, 186030, 221130, 261261, 306936, 358701, 417136, 482856, 556512, 638792, 730422, 832167, 944832

Offset: 0

Wolfdieter Lang, Aug 13 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A096958 (fifth column), A097297 (seventh column).

[(n+30)*Binomial(n+4, 4)/5: n in [0..30]]; // G. C. Greubel, Nov 24 2017

Table[(n + 30)*Binomial[n + 4, 4]/5, {n, 0, 50}] (* G. C. Greubel, Nov 24 2017 *)

for(n=0,30, print1((n+30)*binomial(n+4, 4)/5, ", ")) \\ G. C. Greubel, Nov 24 2017

a(n) = A096956(n+5, 5).
a(n) = 6*b(n) - 5*b(n-1), with b(n) = A000389(n+5) = binomial(n+5, 5).
a(n) = (n+30)*binomial(n+4, 4)/5.
G.f.: (6-5*x)/(1-x)^6.
E.g.f.: x*(720 + 1140*x + 420*x^2 + 45*x^3 + x^4)*exp(x)/120. - G. C. Greubel, Nov 24 2017

A097300 Tenth column (m=9) of (1,6)-Pascal triangle A096956.

Original entry on oeis.org

6, 55, 280, 1045, 3190, 8437, 20020, 43615, 88660, 170170, 311168, 545870, 923780, 1514870, 2416040, 3759074, 5720330, 8532425, 12498200, 18007275, 25555530, 35767875, 49424700, 67492425, 91158600, 121872036, 161388480, 211822380

Offset: 0

Wolfdieter Lang, Aug 13 2004

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A000582, A096956.

Cf. other columns: A096957 (m = 3), A096958 (m = 4), A096959 (m = 5), A097297 (m = 6), A097298 (m = 7), A097299 (m = 8).

A097300[n_] := (n + 54)*Binomial[n + 8, 8]/9;
Array[A097300, 50, 0] (* Paolo Xausa, May 02 2025 *)

a(n) = A096956(n+9, 9) = 6*b(n) - 5*b(n-1) = (n+54)*binomial(n+8, 8)/9, with b(n) = A000582(n+9) = binomial(n+9, 9).

G.f.: (6-5*x)/(1-x)^10.

A097298 Eighth column (m=7) of (1,6)-Pascal triangle A096956.

Original entry on oeis.org

6, 43, 176, 540, 1380, 3102, 6336, 12012, 21450, 36465, 59488, 93704, 143208, 213180, 310080, 441864, 618222, 850839, 1153680, 1543300, 2039180, 2664090, 3444480, 4410900, 5598450, 7047261, 8803008, 10917456, 13449040, 16463480

Offset: 0

Wolfdieter Lang, Aug 13 2004

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A000580, A096956.

Cf. other columns: A096957 (m = 3), A096958 (m = 4), A096959 (m = 5), A097297 (m = 6), A097299 (m = 8), A097300 (m = 9).

A097298[n_] := (n + 42)*Binomial[n + 6, 6]/7;
Array[A097298, 50, 0] (* Paolo Xausa, May 02 2025 *)

a(n) = A096956(n+7, 7) = 6*b(n) - 5*b(n-1) = (n+42)*binomial(n+6, 6)/7, with b(n) = A000580(n+7) = binomial(n+7, 7).

G.f.: (6-5*x)/(1-x)^8.

A097299 Ninth column (m=8) of (1,6)-Pascal triangle A096956.

Original entry on oeis.org

6, 49, 225, 765, 2145, 5247, 11583, 23595, 45045, 81510, 140998, 234702, 377910, 591090, 901170, 1343034, 1961256, 2812095, 3965775, 5509075, 7548255, 10212345, 13656825, 18067725, 23666175, 30713436, 39516444, 50433900, 63882940

Offset: 0

Wolfdieter Lang, Aug 13 2004

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000581, A096956.

Cf. other columns: A096957 (m = 3), A096958 (m = 4), A096959 (m = 5), A097297 (m = 6), A097298 (m = 7), A097300 (m = 9).

A097299[n_] := (n + 48)*Binomial[n + 7, 7]/8;
Array[A097299, 50, 0] (* Paolo Xausa, May 02 2025 *)

a(n) = A096956(n+8, 8) = 6*b(n) - 5*b(n-1) = (n+48)*binomial(n+7, 7)/8, with b(n) = A000581(n+8) = binomial(n+8, 8).

G.f.: (6-5*x)/(1-x)^9.

A005009 a(n) = 7*2^n.

Original entry on oeis.org

7, 14, 28, 56, 112, 224, 448, 896, 1792, 3584, 7168, 14336, 28672, 57344, 114688, 229376, 458752, 917504, 1835008, 3670016, 7340032, 14680064, 29360128, 58720256, 117440512, 234881024, 469762048, 939524096, 1879048192, 3758096384

Offset: 0

N. J. A. Sloane

The first differences are the sequence itself. - Alexandre Wajnberg & Eric Angelini, Sep 07 2005

Vincenzo Librandi, Table of n, a(n) for n = 0..3000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2).

Sequences of the form (2*m+1)*2^n: A000079 (m=0), A007283 (m=1), A020714 (m=2), this sequence (m=3), A005010 (m=4), A005015 (m=5), A005029 (m=6), A110286 (m=7), A110287 (m=8), A110288 (m=9), A175805 (m=10), A248646 (m=11), A164161 (m=12), A175806 (m=13), A257548 (m=15).
Row sums of (6, 1)-Pascal triangle A093563 and of (1, 6)-Pascal triangle A096956, n>=1.
Cf. A000120, A014311, A118416, A173787, A212191.

a005009 = (* 7) . (2 ^) -- Reinhard Zumkeller, May 03 2012

[7*2^n:n in [0..50]]; // Vincenzo Librandi, Sep 20 2011

7*2^Range[0,50] (* Vladimir Joseph Stephan Orlovsky, Mar 14 2011 *)
NestList[2#&,7,30] (* Harvey P. Dale, Aug 10 2024 *)

a(n)=7<Charles R Greathouse IV, Dec 22 2011

[7*2^n for n in range(51)] # G. C. Greubel, Jan 05 2023

G.f.: 7/(1-2*x).
a(n) = A118416(n+1,4) for n > 3. - Reinhard Zumkeller, Apr 27 2006
a(n) = 2*a(n-1), for n > 0, with a(0)=7 . - Philippe Deléham, Nov 23 2008
a(n) = 7 * A000079(n). - Omar E. Pol, Dec 16 2008
a(n) = A173787(n+3,n). - Reinhard Zumkeller, Feb 28 2010
Intersection of A014311 and A212191: all terms and their squares are the sum of exactly three distinct powers of 2, A000120(a(n)) = A000120(a(n)^2) = 3. - Reinhard Zumkeller, May 03 2012
G.f.: 2/x/G(0) - 1/x + 9, where G(k)= 1 + 1/(1 - x*(7*k+2)/(x*(7*k+9) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 03 2013
E.g.f.: 7*exp(2*x). - Stefano Spezia, May 15 2021

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

A056115 a(n) = n*(n+11)/2.

Original entry on oeis.org

0, 6, 13, 21, 30, 40, 51, 63, 76, 90, 105, 121, 138, 156, 175, 195, 216, 238, 261, 285, 310, 336, 363, 391, 420, 450, 481, 513, 546, 580, 615, 651, 688, 726, 765, 805, 846, 888, 931, 975, 1020, 1066, 1113, 1161, 1210, 1260, 1311, 1363, 1416, 1470, 1525

Offset: 0

Barry E. Williams, Jul 04 2000

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A001477, A055999, A056000, A056119.

Third column of Pascal (1, 6) triangle A096956.

List([0..50], n-> n*(n+11)/2 ); # G. C. Greubel, Jan 18 2020

[n*(n+11)/2: n in [0..50]]; // G. C. Greubel, Jan 18 2020

((2*Range[0,50]+11)^2 -11^2)/8 (* G. C. Greubel, Jan 18 2020 *)

a(n)=n*(n+11)/2; \\ Joerg Arndt, Oct 25 2014

[n*(n+11)/2 for n in (0..50)] # G. C. Greubel, Jan 18 2020

G.f.: x*(6-5*x)/(1-x)^3.
a(n) = A000096(n) + 4*A001477(n) = A056000(n) + A001477(n) = A056119(n) - A001477(n). - Zerinvary Lajos, Oct 01 2006
a(n) = A126890(n,5) for n>4. - Reinhard Zumkeller, Dec 30 2006
Equals A119412/2. - Zerinvary Lajos, Feb 12 2007
If we define f(n,i,a) = Sum_{k=0..n-i} ( binomial(n,k)*stirling1(n-k,i) *Product_{j=0..k-1} (-a-j) ), then a(n) = -f(n,n-1,6), for n>=1. - Milan Janjic, Dec 20 2008
a(n) = a(n-1) + n + 5 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
Sum_{n>=1} 1/a(n) = 83711/152460. - R. J. Mathar, Jul 14 2012
a(n) = 6*n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
E.g.f.: x*(12 + x)*exp(x)/2. - G. C. Greubel, Jan 18 2020
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/11 - 20417/152460. - Amiram Eldar, Jan 10 2021

cp's OEIS Frontend

A096957 Fourth column (m=3) of (1,6)-Pascal triangle A096956.

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A097297 Seventh column (m=6) of (1,6)-Pascal triangle A096956.

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A096958 Fifth column (m=4) of (1,6)-Pascal triangle A096956.

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A096959 Sixth column (m=5) of (1,6)-Pascal triangle A096956.

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A097300 Tenth column (m=9) of (1,6)-Pascal triangle A096956.

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A097298 Eighth column (m=7) of (1,6)-Pascal triangle A096956.

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A097299 Ninth column (m=8) of (1,6)-Pascal triangle A096956.

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A005009 a(n) = 7*2^n.

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