Giuliano Cabrele - cp's OEIS frontend

User: Giuliano Cabrele

Giuliano Cabrele has authored 4 sequences.

A265644 Triangle read by rows: T(n,m) is the number of quaternary words of length n with m strictly increasing runs (0 <= m <= n).

1, 0, 4, 0, 6, 10, 0, 4, 40, 20, 0, 1, 65, 155, 35, 0, 0, 56, 456, 456, 56, 0, 0, 28, 728, 2128, 1128, 84, 0, 0, 8, 728, 5328, 7728, 2472, 120, 0, 0, 1, 486, 8451, 27876, 23607, 4950, 165

Offset: 0

Giuliano Cabrele, Dec 13 2015

In the following description the alphabet {0..r} is taken as a basis, with r = 3 in this case.
For example, the quaternary word 2|03|123|3 of length n=7, has m=4 strictly increasing runs.
The empty word has n = 0 and m = 0, and T(0, 0) = 1.
T(n, 0) = 0 for n >= 1.
T(n, m) <> 0 for m <= n <= m*(r+1). T(m*(r+1), m) = 1.
T(n,m) is a partition, based on m, of all the words of length n, so Sum_{k=0..n} T(n,k) = (r+1)^n.

			Triangle starts:
1;
0, 4;
0, 6, 10;
0, 4, 40,  20;
0, 1, 65, 155,  35;
0, 0, 56, 456, 456, 56;
.
T(3,2) = 40, which accounts for the following words:
[0 <= a <= 0, 1 |    0 <= b <= 1]  =   2
[0 <= a <= 1, 2 |    0 <= b <= 2]  =   6
[0 <= a <= 2, 3 |    0 <= b <= 3]  =  12
[0 <= a <= 3    | 0, 1 <= b <= 3]  =  12
[1 <= a <= 3    | 1, 2 <= b <= 3]  =   6
[2 <= a <= 3    | 2, 3 <= b <= 3]  =   2

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 24, p. 154.

MathPages, Balls In Bins With Limited Capacity

Cf. A119900 (r=1, binary words), A120987 (r=2, ternary words), A008287 (quadrinomial coefficients).
Row sums give A000302.
Cf. A000292.

T:=(n,m)->_plus((-1)^(m-j)*binomial(n+1, m-j)*binomial(4*j, n)$j=0..m):

T(n, k) = sum(j=0, k, (-1)^(k-j)*binomial(n+1, k-j)*binomial(4*j, n));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print()); \\ Michel Marcus, Feb 09 2016

Refer to comment to A120987 concerning formulas for general values of r and considerations.
Therefrom we get
T(n, m) = Qsc(3, n, m) =
Nb(4*m-n, 3, n+1) = Nb(4*(n-m)+3, 3, n+1) =
Sum_{j=0..n+1} (-1)^j*Cb(n+1, j)*Cb(4*(m-j), 4*(m-j)-n) =
Sum_{j=0..m} (-1)^(m-j)*Cb(n+1, m-j)*Cb(4*j, n) =
(in this last version Cb(n,m) can be replaced by binomial(n,m))
Sum_{j=0..m} (-1)^(m-j)*binomial(n+1, m-j)*binomial(4*j, n) = [z^n, t^m](1-t)/(1-t(1+(1-t)z)^4) where [x^n]F(x) denotes the coefficient of x^n in the formal power series expansion of F(x),
Nb(s,r,n) denotes the (r+1)-nomial coefficient [x^s](1+x+..+x^r)^n,(Nb(s,3,n) = A008287(n,s)).
Cb(x,m) denotes the binomial coefficient in its extended falling factorial notation (Cb(x,m)= x^_m/m! iff m is a nonnegative integer, 0 otherwise), as defined in the Graham et al. reference.
The diagonal T(n, n) = Nb(3, 3, n+1) = Sum_{j=0..n} (-1)^(n-j)*Cb(n+1, n-j)*Cb(4*j, n) = Cb(n+3, 3) = binomial(n+3, 3) = A000292(n+1).

A262706 Triangle: Newton expansion of C(n,m)^5, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 30, 1, 0, 150, 240, 1, 0, 240, 6810, 1020, 1, 0, 120, 63540, 94890, 3120, 1, 0, 0, 271170, 2615340, 740640, 7770, 1, 0, 0, 604800, 32186070, 47271840, 4029690, 16800, 1, 0, 0, 730800, 214628400, 1281612570, 518276640, 17075940, 32760, 1, 0, 0, 453600, 859992000, 18459063000, 26947757970, 4027831080, 60171300, 59040, 1

Offset: 0

Giuliano Cabrele, Sep 30 2015

Triangle here T_5(n,m) is such that C(n,m)^5 = Sum_{j=0..n} C(n,j)*T_5(j,m).
Equivalently, lower triangular matrix T_5 such that
|| C(n,m)^5 ||  =  P * T_5 = A007318 * T_5.
T_5(n,m) = 0 for n < m and for 5*m < n.
Refer to comment to A262704.
Example:
C(x,2)^5 = x^5*(x-1)^5/32 = 1*C(x,2) + 240*C(x,3) + 6810*C(x,4) + 63540*C(x,5) + 271170*C(x,6) + 604800*C(x,7) + 730800*C(x,8) + 453600*C(x,9) + 113400*C(x,10);
C(5,2)^5 = C(5,3)^5 = 100000 = 1*C(5,2) + 240*C(5,3) + 6810*C(5,4) + 63540*C(5,5) = 1*C(5,3) + 1020*C(5,4) + 94890*C(5,5).

			Triangle starts:
[1];
[0,   1];
[0,  30,      1];
[0, 150,    240,       1];
[0, 240,   6810,    1020,      1];
[0, 120,  63540,   94890,   3120,    1];
[0,   0, 271170, 2615340, 740640, 7770, 1];

P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.

Second diagonal (T_5(n+1,n)) is A061167(n+1).
Column T_5(n,2) is A122193(5,n).
Cf. A109983 (transpose of), A262704, A262705.
Cf. A078739, A078741.

[&+[(-1)^(n-j)*Binomial(n,j)*Binomial(j,m)^5: j in [0..n]]: m in [0..n], n in [0..10]]; // Bruno Berselli, Oct 01 2015

T5[n_, m_] := Sum[(-1)^(n - j) * Binomial[n, j] * Binomial[j, m]^5, {j, 0, n}]; Table[T5[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 01 2015 *)

// as a function
T_5:=(n,m)->_plus((-1)^(n-j)*binomial(n,j)*binomial(j,m)^5 $ j=0..n):
// as a matrix h x h
_P:=h->matrix([[binomial(n,m) $m=0..h]$n=0..h]):
_P_5:=h->matrix([[binomial(n,m)^5 $m=0..h]$n=0..h]):
_T_5:=h->_P(h)^-1*_P_5(h):

T_5(nmax) = {for(n=0, nmax, for(m=0, n, print1(sum(j=0, n, (-1)^(n-j)*binomial(n,j)*binomial(j,m)^5), ", ")); print())} \\ Colin Barker, Oct 01 2015

T_5(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^5.

Also, let S(r,s)(n,m) denote the Generalized Stirling2 numbers as defined in the link above, then T_5(n,m) = n! / (m!)^5 * S(m,m)(5,n).

A262705 Triangle: Newton expansion of C(n,m)^4, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 14, 1, 0, 36, 78, 1, 0, 24, 978, 252, 1, 0, 0, 4320, 8730, 620, 1, 0, 0, 8460, 103820, 46890, 1290, 1, 0, 0, 7560, 581700, 1159340, 185430, 2394, 1, 0, 0, 2520, 1767360, 13387570, 8314880, 595476, 4088, 1, 0, 0, 0, 3087000, 85806000, 170429490, 44341584, 1642788, 6552, 1

Offset: 0

Giuliano Cabrele, Sep 30 2015

Triangle here T_4(n,m) is such that C(n,m)^4 = Sum_{j=0..n} C(n,j)*T_4(j,m).
Equivalently, lower triangular matrix T_4 such that
|| C(n,m)^4 || = A202750 =  P * T_4 = A007318 * T_4.
T_4(n,m) = 0 for n < m and for 4*m < n.
Refer to comment to A262704.
Example:
C(x,2)^4 = x^4*(x-1)^4 /16 = 1*C(x,2) + 78*C(x,3) + 978*C(x,4) + 4320*C(x,5) + 8460*C(x,6) + 7560*C(x,7) + 2520*C(x,8);
C(5,2)^4 = C(5,3)^4 = 10000 = 1*C(5,2) + 78*C(5,3) + 978*C(5,4) + 4320*C(5,5) = 1*C(5,3) + 252*C(5,4) + 8730*C(5,5).

			Triangle starts:
[1];
[0,  1];
[0, 14,    1];
[0, 36,   78,      1];
[0, 24,  978,    252,     1];
[0,  0, 4320,   8730,   620,    1];
[0,  0, 8460, 103820, 46890, 1290, 1];

P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.

Row sums are, by definition, the inverse binomial transform of A005260.
Second diagonal (T_4(n+1,n)) is A058895(n+1).
Column T_4(n,2) is A122193(4,n).
Cf. A109983 (transpose of), A262704, A262706.
Cf. A078739, A078741.

[&+[(-1)^(n-j)*Binomial(n,j)*Binomial(j,m)^4: j in [0..n]]: m in [0..n], n in [0..10]]; // Bruno Berselli, Oct 01 2015

T4[n_, m_] := Sum[(-1)^(n - j) * Binomial[n, j] * Binomial[j, m]^4, {j, 0, n}]; Table[T4[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 01 2015 *)

// as a function
T_4:=(n,m)->_plus((-1)^(n-j)*binomial(n,j)*binomial(j,m)^4 $ j=0..n):
// as a matrix h x h
_P:=h->matrix([[binomial(n,m) $m=0..h]$n=0..h]):
_P_4:=h->matrix([[binomial(n,m)^4 $m=0..h]$n=0..h]):
_T_4:=h->_P(h)^-1*_P_4(h):

T_4(nmax) = {for(n=0, nmax, for(m=0, n, print1(sum(j=0, n, (-1)^(n-j)*binomial(n,j)*binomial(j,m)^4), ", ")); print())} \\ Colin Barker, Oct 01 2015

T_4(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^4.

Also, let S(r,s)(n,m) denote the Generalized Stirling2 numbers as defined in the link above, then T_4(n,m) = n! / (m!)^4 * S(m,m)(4,n).

A262704 Triangle: Newton expansion of C(n,m)^3, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 6, 1, 0, 6, 24, 1, 0, 0, 114, 60, 1, 0, 0, 180, 690, 120, 1, 0, 0, 90, 2940, 2640, 210, 1, 0, 0, 0, 5670, 21840, 7770, 336, 1, 0, 0, 0, 5040, 87570, 107520, 19236, 504, 1, 0, 0, 0, 1680, 189000, 735210, 407400, 42084, 720, 1, 0, 0, 0, 0, 224700, 2835756, 4280850, 1284360, 83880, 990, 1

Offset: 0

Giuliano Cabrele, Sep 27 2015

Triangle here T_3(n,m) is such that C(n,m)^3 = Sum_{j=0..n} C(n,j)*T_3(j,m).
Equivalently, lower triangular matrix T_3 such that
|| C(n,m)^3 || = A181583 =  P * T_3 = A007318 * T_3.
T_3(n,m) = 0 for n < m and for 3*m < n. In fact:
C(x,m)^q and C(x,m), with m nonnegative and q positive integer, are polynomial in x of degree m*q and m respectively, and C(x,m) is a divisor of C(x,m)^q.
Therefore the Newton series will give C(x,m)^q = T_q(m,m)*C(x,m) + T_q(m+1,m)*C(x,m+1) + ... + T_q(q*m,m)*C(x,q*m), where T_q(n,m) is the n-th forward finite difference of C(x,m)^q at x = 0.
Example:
C(x,2)^3 = x^3*(x-1)^3 / 8 = 1*C(x,2) + 24*C(x,3) + 114*C(x,4) + 180*C(x,5) + 90*C(x,6);
C(5,2)^3 = C(5,3)^3 = 1000 = 1*C(5,2) + 24*C(5,3) + 114*C(5,4) + 180*C(5,5) = 1*C(5,3) + 60*C(5,4) + 690*C(5,5).
So we get the expansion of the 3rd power of the binomial coefficient in terms of the binomial coefficients on the same row.
T_1 is the unitary matrix,
T_2 is the transpose of A109983,
T_3 is this sequence,
T_4, T_5 are A262705, A262706.

			Triangle starts:
n\m  [0]     [1]     [2]     [3]     [4]     [5]     [6]     [7]     [8]
[0]  1;
[1]  0,      1;
[2]  0,      6,      1;
[3]  0,      6,      24,     1;
[4]  0,      0,      114,    60,     1;
[5]  0,      0,      180,    690,    120,    1;
[6]  0,      0,      90,     2940,   2640,   210,    1;
[7]  0,      0,      0,      5670,   21840,  7770,   336,    1;
[8]  0,      0,      0,      5040,   87570,  107520, 19236,  504,    1;
[9]  ...

Gheorghe Coserea, Rows n = 0..200, flattened
P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.

Row sums are A172634, the inverse binomial transform of the Franel numbers (A000172).
Column sums are the A126086, per the comment given thereto by Brendan McKay.
Second diagonal (T_3(n+1,n)) is A007531 (n+2).
Column T_3(n,2) is A122193(3,n).
Cf. A109983 (transpose of), A262705, A262706.
Cf. A078739, A078741, A274786.

[&+[(-1)^(n-j)*Binomial(n,j)*Binomial(j,m)^3: j in [0..n]]: m in [0..n], n in [0..10]]; // Bruno Berselli, Oct 01 2015

T3[n_, m_] := Sum[(-1)^(n - j) * Binomial[n, j] * Binomial[j, m]^3, {j, 0, n}]; Table[T3[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 01 2015 *)

// as a function
T_3:=(n,m)->_plus((-1)^(n-j)*binomial(n,j)*binomial(j,m)^3 $ j=0..n):
// as a matrix h x h
_P:=h->matrix([[binomial(n,m) $m=0..h]$n=0..h]):
_P_3:=h->matrix([[binomial(n,m)^3 $m=0..h]$n=0..h]):
_T_3:=h->_P(h)^-1*_P_3(h):

T_3(nmax) = {for(n=0, nmax, for(m=0, n, print1(sum(j=0, n, (-1)^(n-j)*binomial(n,j)*binomial(j,m)^3), ", ")); print())} \\ Colin Barker, Oct 01 2015

t3(n,m) = sum(j=0, n,  (-1)^((n-j)%2)* binomial(n,j)*binomial(j,m)^3);
concat(vector(11, n, vector(n, k, t3(n-1,k-1)))) \\ Gheorghe Coserea, Jul 14 2016

T_3(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^3.

Also, let S(r,s)(n,m) denote the Generalized Stirling2 numbers as defined in the link above,then T_3(n,m) = n! / (m!)^3 * S(m,m)(3,n).

cp's OEIS Frontend

User: Giuliano Cabrele

A265644 Triangle read by rows: T(n,m) is the number of quaternary words of length n with m strictly increasing runs (0 <= m <= n).

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A262706 Triangle: Newton expansion of C(n,m)^5, read by rows.

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A262705 Triangle: Newton expansion of C(n,m)^4, read by rows.

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A262704 Triangle: Newton expansion of C(n,m)^3, read by rows.

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