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A228196
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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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32
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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
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OFFSET
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1,3
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COMMENTS
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The third row is (n^4 - n^2 + 24*n + 24)/12.
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LINKS
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FORMULA
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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
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EXAMPLE
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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;
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MAPLE
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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:
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MATHEMATICA
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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 *)
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PROG
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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
(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
(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
(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
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CROSSREFS
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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).
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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