A078125
Number of partitions of 3^n into powers of 3.
Original entry on oeis.org
1, 2, 5, 23, 239, 5828, 342383, 50110484, 18757984046, 18318289003448, 47398244089264547, 329030840161393127681, 6190927493941741957366100, 318447442589056401640929570896, 45106654667152833836835578059359839
Offset: 0
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]
To obtain t_3(5,2) we use the table T, defined as T[i,j]= t_3(i,j), for i=1,2,...,5(=n), and j= 0,1,2,...,162(= k.m^{n-1}). It is: 1,2,3,4,5,6,7,8,...,162; 1,5,12,22,35,51,...,4510; (this row contains the first 55 members of A000326 - the pentagonal numbers) 1,23,93,238,485,...,29773; 1,239,1632,5827,15200,32856,62629; 1,5828,68457; Column 1 contains the first 5 members of this sequence. - _Valentin Bakoev_, Feb 22 2009
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import Data.MemoCombinators (memo2, list, integral)
a078125 n = a078125_list !! n
a078125_list = f [1] where
f xs = (p' xs $ last xs) : f (1 : map (* 3) xs)
p' = memo2 (list integral) integral p
p 0 = 1; p [] = 0
p ks'@(k:ks) m = if m < k then 0 else p' ks' (m - k) + p' ks m
-- Reinhard Zumkeller, Nov 27 2015
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m[i_, j_] := m[i, j]=If[j==0||i==j, 1, m3[i-1, j-1]]; m2[i_, j_] := m2[i, j]=Sum[m[i, k]m[k, j], {k, j, i}]; m3[i_, j_] := m3[i, j]=Sum[m[i, k]m2[k, j], {k, j, i}]; a[n_] := m2[n, 0]
A078122
Infinite lower triangular matrix, M, that satisfies [M^3](i,j) = M(i+1,j+1) for all i,j>=0 where [M^n](i,j) denotes the element at row i, column j, of the n-th power of matrix M, with M(0,k)=1 and M(k,k)=1 for all k>=0.
Original entry on oeis.org
1, 1, 1, 1, 3, 1, 1, 12, 9, 1, 1, 93, 117, 27, 1, 1, 1632, 3033, 1080, 81, 1, 1, 68457, 177507, 86373, 9801, 243, 1, 1, 7112055, 24975171, 15562314, 2371761, 88452, 729, 1, 1, 1879090014, 8786827629, 6734916423, 1291958181, 64392813, 796797, 2187, 1
Offset: 0
The cube of the matrix is the same matrix excluding the first row and column:
[1, 0, 0, 0]^3 = [ 1, 0, 0, 0]
[1, 1, 0, 0] [ 3, 1, 0, 0]
[1, 3, 1, 0] [12, 9, 1, 0]
[1,12, 9, 1] [93,117,27, 1]
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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(M(n,k), k=0..n), n=0..10); # Alois P. Heinz, Feb 27 2015
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m[i_, j_] := m[i, j]=If[j==0||i==j, 1, m3[i-1, j-1]]; m2[i_, j_] := m2[i, j]=Sum[m[i, k]m[k, j], {k, j, i}]; m3[i_, j_] := m3[i, j]=Sum[m[i, k]m2[k, j], {k, j, i}]; Flatten[Table[m[i, j], {i, 0, 8}, {j, 0, i}]]
A078124
Second column, M(n+1,1) for n>=0, of infinite lower triangular matrix M defined in A078122.
Original entry on oeis.org
1, 3, 12, 93, 1632, 68457, 7112055, 1879090014, 1287814075131, 2325758241901161, 11213788533232011006, 145939965725683888932081, 5174322925070232320838406581, 503750821963423009552527526376232
Offset: 0
a(1)=3 since the coefficient of x^6 in 1/Product_{j=0..inf}(1-x^(3^j)) = 1 + x + x^2 + 2x^3 + 2x^4 + 2x^5 + 3x^6 + ... is 3.
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m[i_, j_] := m[i, j]=If[j==0||i==j, 1, m3[i-1, j-1]]; m2[i_, j_] := m2[i, j]=Sum[m[i, k]m[k, j], {k, j, i}]; m3[i_, j_] := m3[i, j]=Sum[m[i, k]m2[k, j], {k, j, i}]; a[n_] := m[n+1, 1]
A125791
a(n) = 2^(n*(n-1)*(n-2)/6) for n>=1.
Original entry on oeis.org
1, 1, 2, 16, 1024, 1048576, 34359738368, 72057594037927936, 19342813113834066795298816, 1329227995784915872903807060280344576, 46768052394588893382517914646921056628989841375232
Offset: 1
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seq(2^(binomial(n, n-3)), n=1..10); # Zerinvary Lajos, Jun 16 2007 [modified by Georg Fischer, Nov 09 2023]
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A125791[n_]:=2^Binomial[n,n-3];Array[A125791,15] (* Paolo Xausa, Nov 05 2023 *)
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a(n)=if(n<1,0,2^(n*(n-1)*(n-2)/6))
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/* As determinant of n X n matrix: */
{a(n)=local(q=2,A=Mat(1), B); for(m=1, n, 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(matdet(matrix(n,n,r,c,(A^c)[r,1])))}
for(n=1,15,print1(a(n),", "))
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% This generates all 3SAT problem instances
test:-test(4).
test(Max):-
between(1,Max,N),
nl,
one_in_three_monotone_3sat(N,Pss),
write(N:Pss),nl,
fail
; nl.
% generates all one-in-three monotone 3SAT problems involving N variables
one_in_three_monotone_3sat(N,Pss):-
ints(1,N,Is),
findall(Xs,ksubset(3,Is,Xs),Xss),
subset_of(Xss,Pss).
% subset generator
subset_of([],[]).
subset_of([X|Xs],Zs):-
subset_of(Xs,Ys),
add_element(X,Ys,Zs).
add_element(_,Ys,Ys).
add_element(X,Ys,[X|Ys]).
% subsets of K elements
ksubset(0,_,[]).
ksubset(K,[X|Xs],[X|Rs]):-K>0,K1 is K-1,ksubset(K1,Xs,Rs).
ksubset(K,[_|Xs],Rs):-K>0,ksubset(K,Xs,Rs).
% list of integers in [From..To]
ints(From,To,Is):-findall(I,between(From,To,I),Is).
% Paul Tarau (paul.tarau(AT)gmail.com), Jan 25 2008
Name simplified; determinant formula moved out of name into formula section by
Paul D. Hanna, Oct 16 2013
A078536
Infinite lower triangular matrix, M, that satisfies [M^4](i,j) = M(i+1,j+1) for all i,j>=0 where [M^n](i,j) denotes the element at row i, column j, of the n-th power of matrix M, with M(0,k)=1 and M(k,k)=1 for all k>=0.
Original entry on oeis.org
1, 1, 1, 1, 4, 1, 1, 28, 16, 1, 1, 524, 496, 64, 1, 1, 29804, 41136, 8128, 256, 1, 1, 5423660, 10272816, 2755264, 130816, 1024, 1, 1, 3276048300, 8220685104, 2804672704, 178301696, 2096128, 4096, 1, 1, 6744720496300, 21934062166320, 9139625620672, 729250931456, 11442760704, 33550336, 16384, 1
Offset: 0
The 4th power of matrix is the same matrix excluding the first row and column:
[1,__0,__0,_0,0]^4=[____1,____0,___0,__0,0]
[1,__1,__0,_0,0]___[____4,____1,___0,__0,0]
[1,__4,__1,_0,0]___[___28,___16,___1,__0,0]
[1,_28,_16,_1,0]___[__524,__496,__64,__1,0]
[1,524,496,64,1]___[29804,41136,8128,256,1]
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dim = 9;
a[, 0] = 1; a[i, i_] = 1; a[i_, j_] /; j > i = 0;
M = Table[a[i, j], {i, 0, dim-1}, {j, 0, dim-1}];
M4 = MatrixPower[M, 4];
sol = Table[M4[[i, j]] == M[[i+1, j+1]], {i, 1, dim-1}, {j, 1, dim-1}] // Flatten // Solve;
Table[a[i, j], {i, 0, dim-1}, {j, 0, i}] /. sol // Flatten (* Jean-François Alcover, Oct 20 2019 *)
A098539
Lower triangular matrix T, read by rows, that shifts left one column under the matrix square of T, with T(n,0)=T(n,1) for n>0 and T(n,n)=1 for n>=0.
Original entry on oeis.org
1, 1, 1, 2, 2, 1, 6, 6, 4, 1, 26, 26, 20, 8, 1, 166, 166, 140, 72, 16, 1, 1626, 1626, 1460, 888, 272, 32, 1, 25510, 25510, 23884, 16392, 6256, 1056, 64, 1, 664666, 664666, 639156, 479736, 215696, 46816, 4160, 128, 1, 29559718, 29559718, 28895052
Offset: 0
Rows of T begin:
[1],
[1,1],
[2,2,1],
[6,6,4,1],
[26,26,20,8,1],
[166,166,140,72,16,1],
[1626,1626,1460,888,272,32,1],
[25510,25510,23884,16392,6256,1056,64,1],
[664666,664666,639156,479736,215696,46816,4160,128,1],...
Matrix square T^2 begins:
[1],
[2,1],
[6,4,1],
[26,20,8,1],
[166,140,72,16,1],
[1626,1460,888,272,32,1],...
showing that T^2 is the same as T after shifting left one column.
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T(n,k)=local(A,B,C,m);A=matrix(1,1);A[1,1]=1; for(m=2,n+1,B=A^2;C=matrix(m,m);for(i=1,m, for(j=1,i, if(i<3 || j==i || j>m-1,C[i,j]=1,if(j==1,C[i,j]=B[i-1,1],C[i,j]=B[i-1,j-1]));)); A=C);A[n+1,k+1]
A111825
Triangle P, read by rows, that satisfies [P^6](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(6*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, 6, 1, 1, 96, 36, 1, 1, 6306, 3816, 216, 1, 1, 1883076, 1625436, 139536, 1296, 1, 1, 2700393702, 3121837776, 360839016, 5036256, 7776, 1, 1, 19324893252552, 28794284803908, 4200503990976, 78293629296, 181382976, 46656, 1
Offset: 0
Let q=6; 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 - 4/2!*x^2 + 42/3!*x^3 + 7296/4!*x^4 +... (A111829).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(6*x) + m^3/3!*L(x)*L(6*x)*L(6^2*x) +
m^4/4!*L(x)*L(6*x)*L(6^2*x)*L(6^3*x) + ...
Triangle P begins:
1;
1,1;
1,6,1;
1,96,36,1;
1,6306,3816,216,1;
1,1883076,1625436,139536,1296,1;
1,2700393702,3121837776,360839016,5036256,7776,1; ...
where P^6 shifts columns left and up one place:
1;
6,1;
96,36,1;
6306,3816,216,1; ...
Original entry on oeis.org
1, 2, 9, 84, 1625, 64350, 5174449, 841185704, 275723872209, 181906966455026, 241258554545388985, 642662865556736504700, 3436011253857466940820073, 36852501476559726217536067974, 792571351187806816558255494473185
Offset: 0
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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^n)[n+1,c+1]))
Original entry on oeis.org
1, 4, 35, 656, 25509, 2026564, 326603719, 106355219008, 69808185542089, 92203545302072964, 244779396712068825067, 1305009502037405316440848, 13963029918525356899170492525, 299675759834305402824238609624548
Offset: 0
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a(n)=local(q=2,A=Mat(1), B); for(m=1, n+2, 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+1,(A^n)[n+2,c+1]))
A111820
Triangle P, read by rows, that satisfies [P^5](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(5*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, 5, 1, 1, 55, 25, 1, 1, 2055, 1525, 125, 1, 1, 291430, 311525, 38875, 625, 1, 1, 165397680, 239305275, 40338875, 975625, 3125, 1, 1, 390075741430, 735920617775, 157056792000, 5077475625, 24409375, 15625, 1
Offset: 0
Let q=5; 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 - 3/2!*x^2 + 16/3!*x^3 + 2814/4!*x^4 +... (A111824).
Thus the g.f. of column 0 of matrix power P^m is:
1 + m*L(x) + m^2/2!*L(x)*L(5*x) + m^3/3!*L(x)*L(5*x)*L(5^2*x) +
m^4/4!*L(x)*L(5*x)*L(5^2*x)*L(5^3*x) + ...
Triangle P begins:
1;
1,1;
1,5,1;
1,55,25,1;
1,2055,1525,125,1;
1,291430,311525,38875,625,1;
1,165397680,239305275,40338875,975625,3125,1; ...
where P^5 shifts columns left and up one place:
1;
5,1;
55,25,1;
2055,1525,125,1;
291430,311525,38875,625,1; ...
Comments