A305601 -id:A305601 - cp's OEIS frontend

A107877 Column 1 of triangle A107876.

1, 1, 2, 7, 37, 268, 2496, 28612, 391189, 6230646, 113521387, 2332049710, 53384167192, 1348601249480, 37291381915789, 1120914133433121, 36406578669907180, 1271084987848923282, 47487293697623885913, 1890771531272515677250, 79947079338974990793060

Offset: 0

Paul D. Hanna, Jun 04 2005, Apr 10 2007

nonn

Also number of subpartitions of partition consisting of first n-1 triangular numbers; e.g., a(4) = subp([1,3,6]) = 37. - Franklin T. Adams-Watters, Jun 26 2006
Number of length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(0)=0 and s(k) <= s(k-1)+k, see Fxtbook link and example. - Joerg Arndt, Apr 30 2011
Number of Dyck paths whose ascent lengths are exactly {1,2,...,n+1}; for example, the a(2) = 2 paths are uduuduuudddd and uduudduuuddd. - David Scambler, May 30 2012
Number of types of cells of a fine mixed subdivision of the Tesler flow polytope. - Alejandro H. Morales, Oct 11 2017

			1 = 1*(1-x)^1 + 1*x*(1-x)^2 + 2*x^2*(1-x)^4 + 7*x^3*(1-x)^7 + 37*x^4*(1-x)^11 + 268*x^5*(1-x)^16 + 2496*x^6*(1-x)^22 + ...
Also equals the final term in rows of the triangle where row n+1 equals the partial sums of row n with the final term repeated n+1 times, starting with a '1' in row 0, as illustrated by:
1;
1, 1;
1, 2,  2,  2;
1, 3,  5,  7,  7,  7,   7;
1, 4,  9, 16, 23, 30,  37,  37,  37,  37,  37;
1, 5, 14, 30, 53, 83, 120, 157, 194, 231, 268, 268, 268, 268, 268, 268; ...
Restricted growth strings: a(0)=1 corresponds to the empty string; a(1)=1 to [0];
a(2) = 2 to [00] and [01]; a(3)=7 to
  1:  [ 0 0 0 ],
  2:  [ 0 0 1 ],
  3:  [ 0 0 2 ],
  4:  [ 0 1 0 ],
  5:  [ 0 1 1 ],
  6:  [ 0 1 2 ],
  7:  [ 0 1 3 ].
[_Joerg Arndt_, Apr 30 2011]

R. P. Stanley, Enumerative Combinatorics volume 1, 2nd edition, Cambridge University Press, 2011, Ch. 3

Alois P. Heinz, Table of n, a(n) for n = 0..390
Joerg Arndt, Matters Computational (The Fxtbook), section 17.3.6, pp. 368-369
K. Mészáros, A. H. Morales, Volumes and Ehrhart polynomials of flow polytopes, arXiv:1710.00701 [math.CO], 2017, sections 6.1 and 7.

Cf. A107876, A107878, A107879, A115728, A115729.

Cf. A305605, A305601.

b:= proc(n, y) option remember; `if`(n=0, 1, add(
      b(n-1, y+i-n), i=max(1, n-y)..n*(n-1)/2+1-y))
    end:
a:= n-> b(n+1, 0):
seq(a(n), n=0..25);  # Alois P. Heinz, Nov 26 2016
# second Maple program:
a:= n-> LinearAlgebra:-Determinant(Matrix(n,(i,j)->
        binomial(binomial(n+1-i,2)+1,i-j+1))):
seq(a(n), n=0..25); # Alejandro H. Morales, Aug 31 2017

a[ n_, k_: 1, j_: 1] := If[ n < 2, Boole[n >= 0], a[n, k, j] = Sum[a[n - 1, i, j + 1], {i, k + j}]]; (* Michael Somos, Nov 26 2016 *)

{a(n)=polcoeff(1-sum(k=0,n-1,a(k)*x^k*(1-x+x*O(x^n))^(1+k*(k+1)/2)),n)}

G.f.: 1 = Sum_{k>=0} a(k)*x^k*(1-x)^(1 + k*(k+1)/2).
G.f.: 1 = Sum_{k>=0} a(k)*x^k/(1+x)^((k+1)*(k+2)/2).
From Benedict W. J. Irwin, Nov 26 2016: (Start)
Conjecture: a(n) can be expressed with a series of nested sums,
a(3) = Sum_{i=1..2} i+2,
a(4) = Sum_{i=1..2} Sum_{j=1..i+2} j+3,
a(5) = Sum_{i=1..2} Sum_{j=1..i+2} Sum_{k=1..j+3} k+4,
a(6) = Sum_{i=1..2} Sum_{j=1..i+2} Sum_{k=1..j+3} Sum_{l=1..k+4} l+5. (End)
Determinantal formula: a(n) = Det(A) where A is the n X n matrix with entries A(i,j) = binomial(binomial(n+1-i,2)+1,i-j+1). This follows by the formula by MacMahon (see EC1 Ex 3.63) for the number of such subpartitions. -  Alejandro H. Morales, Aug 31 2017

A127496 Triangle, read by rows of n*(n+1)/2 + 1 terms, generated by the following rule: start with a single '1' in row n=0; subsequently, row n+1 equals the partial sums of row n with the final term repeated n+1 more times at the end.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 1, 3, 5, 7, 7, 7, 7, 1, 4, 9, 16, 23, 30, 37, 37, 37, 37, 37, 1, 5, 14, 30, 53, 83, 120, 157, 194, 231, 268, 268, 268, 268, 268, 268, 1, 6, 20, 50, 103, 186, 306, 463, 657, 888, 1156, 1424, 1692, 1960, 2228, 2496, 2496, 2496, 2496, 2496, 2496, 2496

Offset: 0

Paul D. Hanna, Jan 16 2007

Last term in each row forms A107877, the number of subpartitions of the partition consisting of the triangular numbers.

			To obtain row 4 from row 3:
  [1, 3, _5, _7, _7, _7, __7];
take partial sums with final term '37' repeated 4 more times:
  [1, 4, _9, 16, 23, 30, _37, _37, _37, _37, _37].
To obtain row 5, take partial sums of row 4 with the final term '268' repeated 5 more times at the end:
  [1, 5, 14, 30, 53, 83, 120, 157, 194, 231, 268, 268,268,268,268,268].
Triangle begins:
  1;
  1, 1;
  1, 2, 2, 2;
  1, 3, 5, 7, 7, 7, 7;
  1, 4, 9, 16, 23, 30, 37, 37, 37, 37, 37;
  1, 5, 14, 30, 53, 83, 120, 157, 194, 231, 268, 268, 268, 268, 268, 268;
  1, 6, 20, 50, 103, 186, 306, 463, 657, 888, 1156, 1424, 1692, 1960, 2228, 2496, 2496, 2496, 2496, 2496, 2496, 2496;
Final term in rows forms A107877 which satisfies the g.f. 1/(1-x) = 1 + 1*x*(1-x) + 2*x^2*(1-x)^3 + ...

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A107877 (leading edge); diagonals: A127497, A127498.

Cf. A305605, A305601.

nxt[h_] :=Module[{c = Accumulate[h]}, Join[c, PadRight[{}, c[[2]], c[[-1]]]]]; Join[{1},Flatten[NestList[nxt,{1,1},5]]] (* Harvey P. Dale, Mar 10 2020 *)

T(n,k)=if(n<0 || k<0 || k>n*(n+1)/2,0,if(k==0,1, if(k<=n*(n-1)/2,T(n,k-1)+T(n-1,k),T(n,k-1))))

A305605 G.f. A(x) satisfies: [x^k] A(x) / (1-x)^n = 0 for k = n(n+1)/2 + 1 through k = (n+1)(n+2)/2 for n >= 0.

Original entry on oeis.org

1, 0, -1, 0, -2, 2, 0, -7, 14, -7, 0, -37, 111, -111, 37, 0, -268, 1072, -1608, 1072, -268, 0, -2496, 12480, -24960, 24960, -12480, 2496, 0, -28612, 171672, -429180, 572240, -429180, 171672, -28612, 0, -391189, 2738323, -8214969, 13691615, -13691615, 8214969, -2738323, 391189, 0, -6230646, 49845168, -174458088, 348916176, -436145220, 348916176, -174458088, 49845168, -6230646, 0

Offset: 0

Paul D. Hanna, Jun 14 2018

sign

			G.f.: A(x) = 1 - x^2 - 2*x^4 + 2*x^5 - 7*x^7 + 14*x^8 - 7*x^9 - 37*x^11 + 111*x^12 - 111*x^13 + 37*x^14 - 268*x^16 + 1072*x^17 - 1608*x^18 + 1072*x^19 - 268*x^20 - 2496*x^22 + 12480*x^23 - 24960*x^24 + 24960*x^25 - 12480*x^26 + 2496*x^27 - 28612*x^29 + 171672*x^30 + ...
The table of coefficients of x^k in A(x) / (1-x)^n, for n >= 0, begins:
[1, 0, -1,  0,  -2,   2,   0,   -7,  14,    -7,    0, -37, 111,-111,   37, ...];
[1, 1,  0,  0,  -2,   0,   0,   -7,   7,     0,    0, -37,  74, -37,    0, ...];
[1, 2,  2,  2,   0,   0,   0,   -7,   0,     0,    0, -37,  37,   0,    0, ...];
[1, 3,  5,  7,   7,   7,   7,    0,   0,     0,    0, -37,   0,   0,    0, ...];
[1, 4,  9, 16,  23,  30,  37,   37,  37,    37,   37,   0,   0,   0,    0, ...];
[1, 5, 14, 30,  53,  83, 120,  157,  194,  231,  268, 268, 268, 268,  268, ...];
[1, 6, 20, 50, 103, 186, 306,  463,  657,  888, 1156,1424,1692,1960, 2228, ...];
[1, 7, 27, 77, 180, 366, 672, 1135, 1792, 2680, 3836,5260,6952,8912,11140, ...]; ...
illustrating the occurrence of zeros.
Note that the initial terms of the rows in the above table forms the rows of irregular triangle A127496.
TRIANGULAR FORM.
This sequence may be arranged into a triangle like so:
1,
0, -1,
0, -2, 2,
0, -7, 14, -7,
0, -37, 111, -111, 37,
0, -268, 1072, -1608, 1072, -268,
0, -2496, 12480, -24960, 24960, -12480, 2496,
0, -28612, 171672, -429180, 572240, -429180, 171672, -28612,
...
in which the g.f. of the rows equal -x * A107877(n) * (1-x)^(n-1) for n > 0.

Paul D. Hanna, Table of n, a(n) for n = 0..2556

Cf. A305601, A107877, A127496.

/* Informal code to generate terms */
{A=[1, 0]; for(i=1, 465, A=concat(A, 0); m=floor(sqrt(2*#A-2) + 1/2); A[#A] = -polcoeff( Ser(A)/(1-x +x*O(x^#A))^(m-1), #A-1) ; print1(#A, ", ")); A}
/* Show that the definition is satisfied: */
for(n=0, sqrtint(2*#A)-1, print1(n": "); for(k=n*(n+1)/2+1, (n+1)*(n+2)/2, print1(polcoeff( Ser(A)/(1-x +x*O(x^#A))^n , k), ", ")); print(""))

G.f. A(x) = Sum_{n>=0} a(n) * x^n satisfies:
(1) A(x) = 1 - x*Sum_{n>=1} A107877(n) * x^(n*(n+1)/2) * (1-x)^(n-1).
(2) [x^k] A(x) / (1-x)^n = 0 for k = n*(n+1)/2 + 1 through (n+1)*(n+2)/2, n >= 0.
(3) [x^k] A(x) / (1-x)^n = A107877(n) for k = n*(n-1)/2 through n*(n+1)/2, n >= 0.
(4) [x^k] A(x) / (1-x)^n = A127496(n,k) for k = 0..n*(n+1)/2 for n >= 0.
(5) [x^n] A(x) / (1-x)^n = A127497(n) for n >= 0.
FORMULAS INVOLVING TERMS.
a(n*(n+1)/2) = 0 for n >= 1.
a(n*(n-1)/2) = (-1)^n * A107877(n) for n >= 0.
a(n*(n+1)/2 + 1) = -A107877(n) for n >= 0.
a(n) = [x^n] (1+x)^n * G(x) where G(x) is the g.f. of A305601, which is the inverse binomial transform of this sequence.

cp's OEIS Frontend

A107877 Column 1 of triangle A107876.

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A127496 Triangle, read by rows of n*(n+1)/2 + 1 terms, generated by the following rule: start with a single '1' in row n=0; subsequently, row n+1 equals the partial sums of row n with the final term repeated n+1 more times at the end.

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A305605 G.f. A(x) satisfies: [x^k] A(x) / (1-x)^n = 0 for k = n(n+1)/2 + 1 through k = (n+1)(n+2)/2 for n >= 0.

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cp's OEIS Frontend

A107877 Column 1 of triangle A107876.

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Examples

References

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Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A127496 Triangle, read by rows of n*(n+1)/2 + 1 terms, generated by the following rule: start with a single '1' in row n=0; subsequently, row n+1 equals the partial sums of row n with the final term repeated n+1 more times at the end.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A305605 G.f. A(x) satisfies: [x^k] A(x) / (1-x)^n = 0 for k = n*(n+1)/2 + 1 through k = (n+1)*(n+2)/2 for n >= 0.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

Formula

A305605 G.f. A(x) satisfies: [x^k] A(x) / (1-x)^n = 0 for k = n(n+1)/2 + 1 through k = (n+1)(n+2)/2 for n >= 0.