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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.
+10
32
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
1,3
COMMENTS
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
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
EXAMPLE
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;
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 *)
PROG
(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
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).
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).
KEYWORD
nonn,tabl
AUTHOR
Boris Putievskiy, Aug 15 2013
EXTENSIONS
Cross-references corrected and extended by Philippe Deléham, Dec 27 2013
STATUS
approved
A triangle formed like Pascal's triangle, but with 3^n on the borders instead of 1.
+10
6
1, 3, 3, 9, 6, 9, 27, 15, 15, 27, 81, 42, 30, 42, 81, 243, 123, 72, 72, 123, 243, 729, 366, 195, 144, 195, 366, 729, 2187, 1095, 561, 339, 339, 561, 1095, 2187, 6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561, 19683, 9843, 4938, 2556, 1578, 1578, 2556, 4938
OFFSET
0,2
COMMENTS
All rows except the zeroth are divisible by 3. Is there a closed-form formula for these numbers, like for binomial coefficients?
Let b=3 and T(n,k) = A(n-k,k) be the associated reading of the symmetric array A by antidiagonals, then A(n,k) = sum_{r=1..n} b^r*A178300(n-r,k) + sum_{c=1..k} b^c*A178300(k-c,n). Similarly with b=4 and b=5 for A227074 and A227076. - R. J. Mathar, Aug 10 2013
EXAMPLE
Triangle:
1,
3, 3,
9, 6, 9,
27, 15, 15, 27,
81, 42, 30, 42, 81,
243, 123, 72, 72, 123, 243,
729, 366, 195, 144, 195, 366, 729,
2187, 1095, 561, 339, 339, 561, 1095, 2187,
6561, 3282, 1656, 900, 678, 900, 1656, 3282, 6561
MATHEMATICA
t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 3^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]
CROSSREFS
Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A166060 (row sums: 4*3^n - 3*2^n), A227074 (4^n edges), A227076 (5^n edges).
KEYWORD
nonn,tabl
AUTHOR
T. D. Noe, Aug 01 2013
STATUS
approved
A triangle formed like Pascal's triangle, but with (-1)^(n+1) on the borders instead of 1.
+10
6
-1, 1, 1, -1, 2, -1, 1, 1, 1, 1, -1, 2, 2, 2, -1, 1, 1, 4, 4, 1, 1, -1, 2, 5, 8, 5, 2, -1, 1, 1, 7, 13, 13, 7, 1, 1, -1, 2, 8, 20, 26, 20, 8, 2, -1, 1, 1, 10, 28, 46, 46, 28, 10, 1, 1, -1, 2, 11, 38, 74, 92, 74, 38, 11, 2, -1, 1, 1, 13, 49, 112, 166, 166, 112
OFFSET
0,5
COMMENTS
This sequence is almost the same as A026637.
T(n,k) = A026637(n-2,k-1) for n > 3, 1 < k < n-1. - Reinhard Zumkeller, Aug 08 2013
EXAMPLE
Example:
-1,
1, 1,
-1, 2, -1,
1, 1, 1, 1,
-1, 2, 2, 2, -1,
1, 1, 4, 4, 1, 1,
-1, 2, 5, 8, 5, 2, -1,
1, 1, 7, 13, 13, 7, 1, 1,
-1, 2, 8, 20, 26, 20, 8, 2, -1,
1, 1, 10, 28, 46, 46, 28, 10, 1, 1,
-1, 2, 11, 38, 74, 92, 74, 38, 11, 2, -1
MATHEMATICA
t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = (-1)^(n+1), m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]
PROG
(Haskell)
a228053 n k = a228053_tabl !! n !! k
a228053_row n = a228053_tabl !! n
a228053_tabl = iterate (\row@(i:_) -> zipWith (+)
([- i] ++ tail row ++ [0]) ([0] ++ init row ++ [- i])) [- 1]
-- Reinhard Zumkeller, Aug 08 2013
CROSSREFS
Cf. A007318 (Pascal's triangle), A026637 (many terms in common).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A097073 (row sums).
Cf. A227074 (4^n edges), A227075 (3^n edges), A227076 (5^n edges).
KEYWORD
sign,tabl
AUTHOR
T. D. Noe, Aug 07 2013
STATUS
approved
A triangle formed like Pascal's triangle, but with 4^n on the borders instead of 1.
+10
5
1, 4, 4, 16, 8, 16, 64, 24, 24, 64, 256, 88, 48, 88, 256, 1024, 344, 136, 136, 344, 1024, 4096, 1368, 480, 272, 480, 1368, 4096, 16384, 5464, 1848, 752, 752, 1848, 5464, 16384, 65536, 21848, 7312, 2600, 1504, 2600, 7312, 21848, 65536, 262144, 87384, 29160
OFFSET
0,2
COMMENTS
All rows except the zeroth are divisible by 4. Is there a closed-form formula for these numbers, like for binomial coefficients?
EXAMPLE
Example:
1,
4, 4,
16, 8, 16,
64, 24, 24, 64,
256, 88, 48, 88, 256,
1024, 344, 136, 136, 344, 1024,
4096, 1368, 480, 272, 480, 1368, 4096,
16384, 5464, 1848, 752, 752, 1848, 5464, 16384,
65536, 21848, 7312, 2600, 1504, 2600, 7312, 21848, 65536
MATHEMATICA
t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = 4^n, m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]
CROSSREFS
Cf. A007318 (Pascal's triangle), A228053 ((-1)^n on the borders).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A165665 (row sums: 3*4^n - 2*2^n), A227075 (3^n edges), A227076 (5^n edges).
KEYWORD
nonn,tabl
AUTHOR
T. D. Noe, Aug 06 2013
STATUS
approved

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