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
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
- 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
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).
Programs
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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
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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
Comments