cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 21-26 of 26 results.

A111830 Triangle P, read by rows, that satisfies [P^7](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(7*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(0,k)=1 and P(k,k)=1 for all k>=0.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 154, 49, 1, 1, 16275, 8281, 343, 1, 1, 9106461, 6558209, 410914, 2401, 1, 1, 28543862991, 27307109501, 2298650515, 20170801, 16807, 1, 1, 521136519414483, 636922972420469, 67522139062441, 790856748801, 988621354
Offset: 0

Views

Author

Gottfried Helms and Paul D. Hanna, Aug 22 2005

Keywords

Comments

Also P(n,k) = partitions of (7^n - 7^(n-k)) into powers of 7 <= 7^(n-k).

Examples

			Let q=7; the g.f. of column k of matrix power P^m is:
1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
(m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
(m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
where L(x) satisfies:
x/(1-x) = L(x) + L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! + ...
and L(x) = x - 5/2!*x^2 + 83/3!*x^3 + 16110/4!*x^4 +... (A111834).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(7*x) + m^3/3!*L(x)*L(7*x)*L(7^2*x) +
m^4/4!*L(x)*L(7*x)*L(7^2*x)*L(7^3*x) + ...
Triangle P begins:
1;
1,1;
1,7,1;
1,154,49,1;
1,16275,8281,343,1;
1,9106461,6558209,410914,2401,1;
1,28543862991,27307109501,2298650515,20170801,16807,1; ...
where P^7 shifts columns left and up one place:
1;
7,1;
154,49,1;
16275,8281,343,1; ...
		

Crossrefs

Cf. A111831 (column 1), A111832 (row sums), A111833 (matrix log); triangles: A110503 (q=-1), A078121 (q=2), A078122 (q=3), A078536 (q=4), A111820 (q=5), A111825 (q=6), A111835 (q=8).

Programs

  • PARI
    P(n,k,q=7)=local(A=Mat(1),B);if(n
    				

Formula

Let q=7; the g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} L(q^j*x) where L(x) satisfies: x/(1-x) = Sum_{n>=1} Product_{j=0..n-1} L(q^j*x)/(j+1) and L(x) equals the g.f. of column 0 of the matrix log of P (A111834).

A111835 Triangle P, read by rows, that satisfies [P^8](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(8*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(0,k)=1 and P(k,k)=1 for all k>=0.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 232, 64, 1, 1, 36968, 16192, 512, 1, 1, 35593832, 21928768, 1047040, 4096, 1, 1, 219379963496, 178379459392, 11424946688, 67096576, 32768, 1, 1, 9003699178010216, 9288403489672000, 748093366229504, 5862250172416
Offset: 0

Views

Author

Gottfried Helms and Paul D. Hanna, Aug 22 2005

Keywords

Comments

Also P(n,k) = partitions of (8^n - 8^(n-k)) into powers of 8 <= 8^(n-k).

Examples

			Let q=8; the g.f. of column k of matrix power P^m is:
1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
(m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
(m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
where L(x) satisfies:
x/(1-x) = L(x) + L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! + ...
and L(x) = x - 6/2!*x^2 + 142/3!*x^3 + 31800/4!*x^4 +... (A111839).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(8*x) + m^3/3!*L(x)*L(8*x)*L(8^2*x) + m^4/4!*L(x)*L(8*x)*L(8^2*x)*L(8^3*x) + ...
Triangle P begins:
1;
1,1;
1,8,1;
1,232,64,1;
1,36968,16192,512,1;
1,35593832,21928768,1047040,4096,1;
1,219379963496,178379459392,11424946688,67096576,32768,1; ...
where P^8 shifts columns left and up one place:
1;
8,1;
232,64,1;
36968,16192,512,1; ...
		

Crossrefs

Cf. A111836 (column 1), A111837 (row sums), A111838 (matrix log); triangles: A110503 (q=-1), A078121 (q=2), A078122 (q=3), A078536 (q=4), A111820 (q=5), A111825 (q=6), A111830 (q=7).

Programs

  • PARI
    P(n,k,q=8)=local(A=Mat(1),B);if(n
    				

Formula

Let q=8; the g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} L(q^j*x) where L(x) satisfies: x/(1-x) = Sum_{n>=1} Product_{j=0..n-1} L(q^j*x)/(j+1) and L(x) equals the g.f. of column 0 of the matrix log of P (A111839).

A089177 Triangle read by rows: T(n,k) (n >= 0, 0 <= k <= 1+log_2(floor(n))) giving number of non-squashing partitions of n into k parts.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 4, 4, 1, 1, 5, 6, 2, 1, 6, 9, 4, 1, 7, 12, 6, 1, 8, 16, 10, 1, 1, 9, 20, 14, 2, 1, 10, 25, 20, 4, 1, 11, 30, 26, 6, 1, 12, 36, 35, 10, 1, 13, 42, 44, 14, 1, 14, 49, 56, 20, 1, 15, 56, 68, 26, 1, 16, 64, 84, 36, 1, 1, 17, 72, 100, 46, 2, 1, 18, 81, 120, 60, 4, 1
Offset: 0

Views

Author

N. J. A. Sloane, Dec 08 2003

Keywords

Comments

T(n,k) = A181322(n,k) - A181322(n,k-1) for n>0. - Alois P. Heinz, Jan 25 2014

Examples

			Triangle begins:
  1;
  1, 1;
  1, 2,  1;
  1, 3,  2;
  1, 4,  4,  1;
  1, 5,  6,  2;
  1, 6,  9,  4;
  1, 7, 12,  6;
  1, 8, 16, 10,  1;
		

Crossrefs

Cf. A078121, A089178. Columns give A002620, A008804, A088932, A088954. Row sums give A000123.

Programs

  • Maple
    T:= proc(n) option remember;
         `if`(n=0, 1, zip((x, y)-> x+y, [T(n-1)], [0, T(floor(n/2))], 0)[])
        end:
    seq(T(n), n=0..25);  # Alois P. Heinz, Apr 01 2012
  • Mathematica
    row[0] = {1}; row[1] = {1, 1}; row[n_] := row[n] = Plus @@ PadRight[ {row[n-1], Join[{0}, row[Floor[n/2]]]} ]; Table[row[n], {n, 0, 25}] // Flatten (* Jean-François Alcover, Jan 31 2014 *)

Formula

Row 0 = {1}, row 1 = {1 1}; for n >=2, row n = row n-1 + (row floor(n/2) shifted one place right).
G.f. for column k (k >= 2): x^(2^(k-2))/((1-x)*Product_{j=0..k-2} (1-x^(2^j))). [corrected by Jason Yuen, Jan 12 2025]
Conjecture: let R(n,x) be the n-th reversed row polynomial, then R(n,x) = Sum_{k=0..A000523(A053645(n)) + 1} T(A053645(n),k)*R(2^(A000523(n)-k),x) for n > 0, n != 2^m with R(0,x) = 1 where R(2^m,x) is the (m+1)-th row polynomial of A078121. - Mikhail Kurkov, Jun 28 2025

Extensions

More terms from Alford Arnold, May 22 2004

A078123 Square of infinite lower triangular matrix A078122.

Original entry on oeis.org

1, 2, 1, 5, 6, 1, 23, 51, 18, 1, 239, 861, 477, 54, 1, 5828, 32856, 25263, 4347, 162, 1, 342383, 3013980, 3016107, 699813, 39285, 486, 1, 50110484, 690729981, 865184724, 253656252, 19053063, 354051, 1458, 1, 18757984046, 406279238154
Offset: 0

Views

Author

Paul D. Hanna, Nov 18 2002

Keywords

Examples

			Square of A078122 = A078123 as can be seen by 4 X 4 submatrix:
[1,_0,_0,0]^2=[_1,_0,_0,_0]
[1,_1,_0,0]___[_2,_1,_0,_0]
[1,_3,_1,0]___[_5,_6,_1,_0]
[1,12,_9,1]___[23,51,18,_1]
		

Crossrefs

Programs

  • Maple
    S:= proc(i, j) option remember;
           add(M(i, k)*M(k, j), k=0..i)
        end:
    M:= proc(i, j) option remember; `if`(j=0 or i=j, 1,
           add(S(i-1, k)*M(k, j-1), k=0..i-1))
        end:
    seq(seq(S(n,k), k=0..n), n=0..10);  # Alois P. Heinz, Feb 27 2015
  • Mathematica
    S[i_, j_] := S[i, j] = Sum[M[i, k]*M[k, j], {k, 0, i}]; M[i_, j_] := M[i, j] = If[j == 0 || i == j, 1, Sum[S[i-1, k]*M[k, j-1], {k, 0, i-1}]]; Table[Table[S[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Mar 06 2015, after Alois P. Heinz *)

Formula

M(1, j) = A078125(j), M(j+1, j)=2*3^j.

A125799 Antidiagonal sums of table A125790.

Original entry on oeis.org

1, 2, 4, 9, 25, 94, 520, 4521, 64793, 1581010, 67106004, 5029631745, 673439168257, 162631617757086, 71416302988324776, 57430160224301687377, 85096038984339418975505, 233592305902515392375925762, 1193627868786115606927913952196, 11402285904243733254203516140245465
Offset: 0

Views

Author

Paul D. Hanna, Dec 10 2006

Keywords

Comments

Table A125790 is related to partitions into powers of 2, with A002577 in column 1 of A125790; further, column k of A125790 equals row sums of matrix power A078121^k, where triangle A078121 shifts left one column under matrix square.

Crossrefs

Programs

  • PARI
    a(n)=local(q=2,A=Mat(1), B); for(m=1, n+1, B=matrix(m, m); for(i=1, m, for(j=1, i, if(j==i || j==1, B[i, j]=1, B[i, j]=(A^q)[i-1, j-1]); )); A=B); return(sum(c=0,n,(A^(c+1))[n-c+1,1]))

A381810 Array read by downward antidiagonals: A(n,k) is a generalization of odd columns of A125790 defined in Comments for n > 0, k >= 0.

Original entry on oeis.org

2, 4, 4, 6, 16, 6, 8, 36, 20, 10, 10, 64, 42, 84, 14, 12, 100, 72, 286, 100, 20, 14, 144, 110, 680, 322, 120, 26, 16, 196, 156, 1330, 744, 364, 140, 36, 18, 256, 210, 2300, 1430, 816, 406, 656, 46, 20, 324, 272, 3654, 2444, 1540, 888, 3396, 740, 60, 22, 400, 342, 5456, 3850, 2600, 1650, 10816, 3682, 840, 74
Offset: 1

Views

Author

Mikhail Kurkov, May 05 2025

Keywords

Comments

This is generalization in the sense that first column of A125790 is A000123(2^(n-1)) while in this square array column zero is conjecturally A000123(n).
A(n,k) = v_{A001511(n)} where we start with vector v of fixed length L(n) = A070939(n) with elements v_i = A125790(i,2*k+1), pre-calculate A078121 up to L(n)-th row, reserve t as an empty vector of fixed length L(n) and for i=1..A119387(n+1), for j=1..L(n)-i+1 apply t := v (at the beginning of each cycle for i) and also apply v_j := Sum_{k=1..j+1} A078121(j,k-1)*t_k if R(n,L(n)-i) = 1, otherwise v_j := Sum_{k=1..j+1} A078121(j,k-1)*t_k*(-1)^(j+k+1). Here R(n,k) = floor(n/(2^k)) mod 2 is the (k+1)-th bit in the binary expansion of n.
Conjecture: sequence A(n,k) for fixed n is a polynomial of degree A070939(n).

Examples

			Array begins:
===========================================================
n\k|  0    1     2      3      4      5       6       7 ...
---+-------------------------------------------------------
1  |  2,   4,    6,     8,    10,    12,     14,     16 ...
2  |  4,  16,   36,    64,   100,   144,    196,    256 ...
3  |  6,  20,   42,    72,   110,   156,    210,    272 ...
4  | 10,  84,  286,   680,  1330,  2300,   3654,   5456 ...
5  | 14, 100,  322,   744,  1430,  2444,   3850,   5712 ...
6  | 20, 120,  364,   816,  1540,  2600,   4060,   5984 ...
7  | 26, 140,  406,   888,  1650,  2756,   4270,   6256 ...
8  | 36, 656, 3396, 10816, 26500, 55056, 102116, 174336 ...
  ...
		

Crossrefs

Programs

  • PARI
    upto1(n) = my(v1); v1 = vector(n+1, i, vector(i, j, j==1 || j==i)); for(i=2, n, for(j=1, i-1, v1[i+1][j+1] = sum(k=j-1, i-1, v1[i][k+1]*v1[k+1][j]))); v1
    A(n,m) = my(L = logint(n,2), A = valuation(n,2), B = logint(n>>A,2), v1, v2, v3); v1 = upto1(L+2); v2 = vector(L+2, i, vecsum(v1[i])); for(i=1, 2*m, v2 = vector(L+2, i, sum(j=1, i, v1[i][j]*v2[j]))); for(i=1, B, v3 = v2; for(j=1, L-i+1, v2[j+1] = sum(k=1, j+1, v1[j+1][k]*v3[k+1]*if(!bittest(n,L-i+1), (-1)^(j+k+1), 1)))); v2[A+2]
    
  • PARI
    upto1(n) = my(v1); v1 = vector(n+1, i, vector(i, j, j==1 || j==i)); for(i=2, n, for(j=1, i-1, v1[i+1][j+1] = sum(k=j-1, i-1, v1[i][k+1]*v1[k+1][j]))); v1
    upto2(n,m) = my(L = logint(n,2), A = valuation(n,2), B = logint(n>>A,2), v1, v2, v3, v4, v5); v1 = upto1(L+2); v2 = vector(L+2, i, 1); v3 = vector(m+1, i, 0); for(s=0, m, for(i=1, min(s+1,2), v2 = vector(L+2, i, sum(j=1, i, v1[i][j]*v2[j]))); v4 = v2; for(i=1, B, v5 = v4; for(j=1, L-i+1, v4[j+1] = sum(k=1, j+1, v1[j+1][k]*v5[k+1]*if(!bittest(n,L-i+1), (-1)^(j+k+1), 1)))); v3[s+1] = v4[A+2]); v3 \\ slightly modified version of the first program, some kind of memoization; generates A(n,k) for k=0..m

Formula

A(2^(n-1),k) = A125790(n,2*k+1) for n > 0, k >= 0.
Conjectured formulas: (Start)
A(n,0) = A000123(n) for n > 0.
A(n,k) = Sum_{j=0..k} A000123(A062383(n)*j+n)*A106400(k-j) for n > 0, k >= 0.
If we change v_i = A125790(i,2*k+1) to v_i = A125790(i,2*k) to get similar generalization of even columns, then for resulting array B(n,k) we have B(n,k) = Sum_{j=0..k} A000123(A062383(n)*j+A053645(n))*A106400(k-j) for n > 0, k >= 0.
2*(k+1) divides A(n,k) for n > 0 if (k+1) is a term of A236206.
G.f. for n-th row is f(A070939(n)+1,n) for n > 0 where f(n,k) = (Sum_{(c_0 + c_1 + ... + c_{n-1}) == 2*k (mod 2^n), 0 <= c_i < 2^n, 2^i divides c_i} x^((c_0 + c_1 + ... + c_{n-1} - 2*k)/2^n))/(1-x)^n for n > 0, k >= 0. Similarly, g.f. for n-th row of B(n,k) is f(A070939(n)+1,A053645(n)).
G.f. for n-th row is (Sum_{i=0..L(n)-1} x^i * Sum_{j=0..i} binomial(L(n)+1,j)*A(n,i-j)*(-1)^j)/(1-x)^(L(n)+1) for n > 0 where L(n) = A070939(n).
s(4*n+1) = 1 for n >= 0, s(4*n) = s(4*n+2) = 1 if A010060(n) = 1 for n > 0 where s(n) = A007814(Sum_{k=0..n-1} A(k+1,n-k-1)). (End)
Previous Showing 21-26 of 26 results.