A135878 -id:A135878 - cp's OEIS frontend

A135880 Triangle P, read by rows, where column k of P^2 equals column 0 of P^(2k+2) such that column 0 of P^2 equals column 0 of P shift one place left, with P(0,0)=1.

1, 1, 1, 2, 2, 1, 6, 7, 3, 1, 25, 34, 15, 4, 1, 138, 215, 99, 26, 5, 1, 970, 1698, 814, 216, 40, 6, 1, 8390, 16220, 8057, 2171, 400, 57, 7, 1, 86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1, 1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9, 1, 14563135

Offset: 0

Paul D. Hanna, Dec 15 2007

Amazingly, column 0 (A135881) also equals column 0 of tables A135878 and A135879, both of which have unusual recurrences seemingly unrelated to this triangle.

			Triangle P begins:
1;
1, 1;
2, 2, 1;
6, 7, 3, 1;
25, 34, 15, 4, 1;
138, 215, 99, 26, 5, 1;
970, 1698, 814, 216, 40, 6, 1;
8390, 16220, 8057, 2171, 400, 57, 7, 1;
86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1;
1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9,
1;
14563135, 35219202, 19003467, 5352788, 1004176, 140908, 15855,
1504, 126, 10, 1;
where column k of P equals column 0 of R^(k+1) where R =
A135894.
Triangle Q = P^2 = A135885 begins:
1;
2, 1;
6, 4, 1;
25, 20, 6, 1;
138, 126, 42, 8, 1;
970, 980, 351, 72, 10, 1;
8390, 9186, 3470, 748, 110, 12, 1;
86796, 101492, 39968, 8936, 1365, 156, 14, 1;
1049546, 1296934, 528306, 121532, 19090, 2250, 210, 16, 1; ...
where column k of Q equals column 0 of Q^(k+1) for k>=0;
thus column k of P^2 equals column 0 of P^(2k+2).
Triangle R = A135894 begins:
1;
1, 1;
2, 3, 1;
6, 12, 5, 1;
25, 63, 30, 7, 1;
138, 421, 220, 56, 9, 1;
970, 3472, 1945, 525, 90, 11, 1;
8390, 34380, 20340, 5733, 1026, 132, 13, 1;
86796, 399463, 247066, 72030, 13305, 1771, 182, 15, 1; ...
where column k of R equals column 0 of P^(2k+1) for k>=0.
Surprisingly, column 0 is also found in triangle A135879:
1;
1,1;
2,2,1,1;
6,6,4,4,2,2,1;
25,25,19,19,13,13,9,5,5,3,1,1;
138,138,113,113,88,88,69,50,50,37,24,24,15,10,5,5,2,1; ...
and is generated by a process that seems completely unrelated.

Cf. columns: A135881, A135882, A135883, A135884.
Cf. related tables: A135885 (Q=P^2), A135894 (R).
Cf. A135888 (P^3), A135891 (P^4), A135892 (P^5), A135893 (P^6).
Cf. A135898 (P^-1*R), A135899 (P*R^-1*P), A135900 (R^-1*Q).
Cf. A135878, A135879.

{T(n,k)=local(P=Mat(1),R,PShR);if(n>0,for(n=0,n, PShR=matrix(#P,#P, r,c, if(r>=c,if(r==c,1,if(c==1,0,P[r-1,c-1]))));R=P*PShR; R=matrix(#P+1, #P+1, r,c, if(r>=c, if(r<#P+1,R[r,c], if(c==1,(P^2)[ #P,1],(P^(2*c-1))[r-c+1,1])))); P=matrix(#R, #R, r,c, if(r>=c, if(r<#R,P[r,c], (R^c)[r-c+1,1])))));P[n+1,k+1]}

Denote this triangle by P and define as follows.
Let [P^m]_k denote column k of matrix power P^m,
so that triangular matrix Q = A135885 may be defined by
[Q]_k = [P^(2k+2)]_0, for k>=0, such that
(1) Q = P^2 and (2) [Q]_0 = [P]_0 shifted left.
Define the dual triangular matrix R = A135894 by
[R]_k = [P^(2k+1)]_0, for k>=0.
Then columns of P may be formed from powers of R:
[P]_k = [R^(k+1)]_0, for k>=0.
Further, columns of powers of P, Q and R satisfy:
[R^(j+1)]_k = [P^(2k+1)]_j,
[Q^(j+1)]_k = [P^(2k+2)]_j,
[Q^(j+1)]_k = [Q^(k+1)]_j,
[P^(2j+2)]_k = [P^(2k+2)]_j, for all j>=0, k>=0.
Also, we have the column transformations:
R * [P]k = [P]{k+1},
Q * [Q]k = [Q]{k+1},
Q * [R]k = [R]{k+1},
P^2 * [Q]k = [Q]{k+1},
P^2 * [R]k = [R]{k+1}, for all k>=0.
Other identities include the matrix products:
P^-1*R (A135898) = P shifted right one column;
P*R^-1*P (A135899) = Q shifted down one row;
R^-1*Q (A135900) = R shifted down one row.

A135881 Column 0 of triangle A135880.

Original entry on oeis.org

1, 1, 2, 6, 25, 138, 970, 8390, 86796, 1049546, 14563135, 228448504, 4002300038, 77523038603, 1646131568618, 38043008887356, 950967024783228, 25573831547118764, 736404945614783668, 22611026430036582671

Offset: 0

Paul D. Hanna, Dec 15 2007

nonn

Amazingly, this sequence also equals column 0 of tables A135878 and A135879, which have unusual recurrences seemingly unrelated to triangle A135880.

			Equals column 0 of triangle P=A135880:
1;
1, 1;
2, 2, 1;
6, 7, 3, 1;
25, 34, 15, 4, 1;
138, 215, 99, 26, 5, 1;
970, 1698, 814, 216, 40, 6, 1;
8390, 16220, 8057, 2171, 400, 57, 7, 1; ...
where column k of P^2 equals column 0 of P^(2k+2)
such that column 0 of P^2 equals this sequence shift left.
Also equals column 0 of irregular triangle A135879:
1;
1,1;
2,2,1,1;
6,6,4,4,2,2,1;
25,25,19,19,13,13,9,5,5,3,1,1;
138,138,113,113,88,88,69,50,50,37,24,24,15,10,5,5,2,1; ...
which has a recurrence similar to that of triangle A135877
which generates the double factorials.

Paul D. Hanna, Table of n, a(n) for n = 0..100

Cf. A135880, A135879, A135878; other columns: A135882, A135883, A135884.

/* Generated as column 0 in triangle A135880: */ {a(n)=local(P=Mat(1),R,PShR);if(n==0,1,for(i=0,n, PShR=matrix(#P,#P, r,c, if(r>=c,if(r==c,1,if(c==1,0,P[r-1,c-1]))));R=P*PShR; R=matrix(#P+1, #P+1, r,c, if(r>=c, if(r<#P+1,R[r,c],if(c==1,(P^2)[ #P,1],(P^(2*c-1))[r-c+1,1])))); P=matrix(#R, #R, r,c, if(r>=c, if(r<#R,P[r,c], (R^c)[r-c+1,1]))));P[n+1,1])}

/* Generated as column 0 in triangle A135879 (faster): */ {a(n)=local(A=[1],B);if(n>0,for(i=1,n,m=1;B=[]; for(j=1,#A,if(j+m-1==floor((m+2)^2/4)-1,m+=1;B=concat(B,0));B=concat(B,A[ j])); A=Vec(Polrev(Vec(Pol(B)/(1-x+O(x^#B)))))));A[1]}

Typo in entries (false comma) corrected by N. J. A. Sloane, Jan 23 2008

A135876 Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms at positions [(m+2)^2/4 - 1] for m>=0 and then taking partial sums, starting with all 1's in row 0.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 15, 8, 3, 1, 105, 48, 15, 4, 1, 945, 384, 105, 23, 5, 1, 10395, 3840, 945, 176, 33, 6, 1, 135135, 46080, 10395, 1689, 279, 44, 7, 1, 2027025, 645120, 135135, 19524, 2895, 400, 57, 8, 1, 34459425, 10321920, 2027025, 264207, 35685, 4384, 561

Offset: 0

Paul D. Hanna, Dec 14 2007

This is the double factorial analog of Moessner's factorial array (A125714). Compare to triangle A135877 which is generated by a complementary process. A very interesting variant is square array A135878.

			Square array begins:
(1),(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,1,(1),1,1,1,(1),1,1,1,1,(1),...;
(1),(2),3,(4),5,(6),7,8,(9),10,11,(12),13,14,15,(16),17,18,19,(20),...;
(3),(8),15,(23),33,(44),57,71,(86),103,121,(140),161,183,206,(230),..;
(15),(48),105,(176),279,(400),561,744,(950),1206,1489,(1800),2171,..;
(105),(384),945,(1689),2895,(4384),6555,9129,(12139),16161,20763,..;
(945),(3840),10395,(19524),35685,(56448),89055,129072,(177331),245778,...;
(10395),(46080),135135,(264207),509985,(836352),1381905,2071215,(2924172),.;
(135135),(645120),2027025,(4098240),8294895,(14026752),24137505,...; ...
where terms in parenthesis are removed before taking partial sums.
For example, to generate row 2 from row 1, remove terms at positions
{[(m+2)^2/4-1], m>=0} = [0,1,3,5,8,11,15,19,24,29,35,...] to obtain:
[3, 5, 7,8, 10,11, 13,14,15, 17,18,19, 21,22,23,24, 25,26,27,28, ...]
then take partial sums to get row 2:
[3, 8, 15,23, 33,44, 57,71,86, 103,121,140, 161,183,206,230, ...].
Repeating this process will generate all the rows of the triangle,
where column 0 will be the odd double factorials (A001147)
and column 1 will be the even double factorials (A000165).

Cf. columns: A001147, A000165, A004041, A129890; variants: A135878, A125714.

{T(n, k)=local(A=0, b=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==floor((b+2)^2/4)-1, b+=1, A+=T(n-1, c); d+=1); c+=1)); A}

T(n,0) = (2n)!/n!/2^n; T(n,1) = 2^n*n!; T(n,2) = (2n+1)!/n!/2^n; T(n,3) = A004041(n) = (2n+1)!/n!/2^n * Sum_{k=0..n} 1/(2k+1). T(n,4) = A129890(n) = 2^(n+1)*(n+1)! - (2n+1)!/n!/2^n = T(n+1,1)-T(n+1,0).

A135879 Triangle, read by rows of A135901(n) terms, where row n+1 is generated from row n by inserting zeros at positions [(m+3)^2/4 - 2], as m=0,1,2,3,... and then taking partial sums from right to left, starting with a single 1 in row 0.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 1, 6, 6, 4, 4, 2, 2, 1, 25, 25, 19, 19, 13, 13, 9, 5, 5, 3, 1, 1, 138, 138, 113, 113, 88, 88, 69, 50, 50, 37, 24, 24, 15, 10, 5, 5, 2, 1, 970, 970, 832, 832, 694, 694, 581, 468, 468, 380, 292, 292, 223, 173, 123, 123, 86, 62, 38, 38, 23, 13, 8, 3, 3, 1, 8390

Offset: 0

Paul D. Hanna, Dec 14 2007

Column 0 is A135881 which equals column 0 of square array A135878 and also equals column 0 of triangle A135880. Compare to square array A135878, which is generated by a complementary process. An interesting variant is triangle A135877 in which column 0 equals the double factorials (A001147).

			Triangle begins:
1;
1, 1;
2, 2, 1, 1;
6, 6, 4, 4, 2, 2, 1;
25, 25, 19, 19, 13, 13, 9, 5, 5, 3, 1, 1;
138, 138, 113, 113, 88, 88, 69, 50, 50, 37, 24, 24, 15, 10, 5, 5, 2, 1;
970, 970, 832, 832, 694, 694, 581, 468, 468, 380, 292, 292, 223, 173, 123, 123, 86, 62, 38, 38, 23, 13, 8, 3, 3, 1;
8390, 8390, 7420, 7420, 6450, 6450, 5618, 4786, 4786, 4092, 3398, 3398, 2817, 2349, 1881, 1881, 1501, 1209, 917, 917, 694, 521, 398, 275, 275, 189, 127, 89, 51, 51, 28, 15, 7, 4, 1, 1;
There are A135901(n) number of terms in row n.
To generate the triangle, start with a single 1 in row 0,
and then obtain row n+1 from row n by inserting zeros at
positions {[(m+3)^2/4 - 2], m=0,1,2,...} and then
taking reverse partial sums (i.e., summing from right to left).
Start with row 0, insert a zero in front of the '1' at position 0:
[0,1];
take reverse partial sums to get row 1:
[1,1];
insert zeros at positions [0,2]:
[0,1,0,1];
take reverse partial sums to get row 2:
[2,2,1,1];
insert zeros at positions [0,2,4]:
[0,2,0,2,0,1,1];
take reverse partial sums to get row 3:
[6,6,4,4,2,2,1];
insert zeros at positions [0,2,4,7]:
[0,6,0,6,0,4,4,0,2,2,0,1];
take reverse partial sums to get row 4:
[25,25,19,19,13,13,9,5,5,3,1,1];
insert zeros at positions [0,2,4,7,10,14]:
[0,25,0,25,0,19,19,0,13,13,0,9,5,5,0,3,1,1];
take reverse partial sums to get row 5:
[138,138,113,113,88,88,69,50,50,37,24,24,15,10,5,5,2,1].
Triangle A135880 begins:
1;
1, 1;
2, 2, 1;
6, 7, 3, 1;
25, 34, 15, 4, 1;
138, 215, 99, 26, 5, 1;
970, 1698, 814, 216, 40, 6, 1; ...
and is generated by matrix powers of itself.

Cf. A135881, A135901, A135878, A135880; variants: A135877, A127452, A125781.

{T(n,k)=local(A=[1],B);if(n>0,for(i=1,n,m=1;B=[]; for(j=1,#A,if(j+m-1==floor((m+2)^2/4)-1,m+=1;B=concat(B,0));B=concat(B,A[j])); A=Vec(Polrev(Vec(Pol(B)/(1-x+O(x^#B)))))));if(k+1>#A,0,A[k+1])}

A152405 Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions {m*(m+1)/2, m>=0} and then taking partial sums, starting with all 1's in row 0.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 14, 8, 3, 1, 86, 45, 14, 4, 1, 645, 318, 86, 22, 5, 1, 5662, 2671, 645, 152, 31, 6, 1, 56632, 25805, 5662, 1251, 232, 41, 7, 1, 633545, 280609, 56632, 11869, 2026, 327, 53, 8, 1, 7820115, 3381993, 633545, 126987, 20143, 2991, 457, 66, 9, 1

Offset: 0

Paul D. Hanna, Dec 05 2008

			Table begins:
(1),(1),1,(1),1,1,(1),1,1,1,(1),1,1,1,1,(1),1,...;
(1),(2),3,(4),5,6,(7),8,9,10,(11),12,13,14,15,(16),...;
(3),(8),14,(22),31,41,(53),66,80,95,(112),130,149,169,190,...;
(14),(45),86,(152),232,327,(457),606,775,965,(1202),1464,1752,2067,...;
(86),(318),645,(1251),2026,2991,(4455),6207,8274,10684,(13934),17653,...;
(645),(2671),5662,(11869),20143,30827,(48480),70355,96990,128959,...;
(5662),(25805),56632,(126987),223977,352936,(582183),874664,1240239,...;
(56632),(280609),633545,(1508209),2748448,4438122,(7641111),11831184,...;
(633545),(3381993),7820115,(19651299),36837937,60743909,...; ...
where row n equals the partial sums of row n-1 after removing terms
at positions {m*(m+1)/2, m>=0} (marked by parenthesis in above table).
For example, to generate row 3 from row 2:
[3,8, 14, 22, 31,41, 53, 66,80,95, 112, 130,...]
remove terms at positions {0,1,3,6,10,...}, yielding:
[14, 31,41, 66,80,95, 130,149,169,190, ...]
then take partial sums to obtain row 3:
[14, 45,86, 152,232,327, 457,606,775,965, ...].
Continuing in this way generates all rows of this table.
RELATION TO POWERS OF A SPECIAL TRIANGULAR MATRIX.
Columns 0 and 1 are found in triangle T=A152400, which begins:
1;
1, 1;
3, 2, 1;
14, 8, 3, 1;
86, 45, 15, 4, 1;
645, 318, 99, 24, 5, 1;
5662, 2671, 794, 182, 35, 6, 1;
56632, 25805, 7414, 1636, 300, 48, 7, 1; ...
where column k of T = column 0 of matrix power T^(k+1) for k>=0.
Furthermore, matrix powers of triangle T=A152400 satisfy:
column k of T^(j+1) = column j of T^(k+1) for all j>=0, k>=0.
Column 3 of this square array = column 1 of T^2:
1;
2, 1;
8, 4, 1;
45, 22, 6, 1;
318, 152, 42, 8, 1;
2671, 1251, 345, 68, 10, 1;
25805, 11869, 3253, 648, 100, 12, 1; ...
RELATED TRIANGLE A127714 begins:
1;
1, 1, 1;
1, 2, 2, 3, 3, 3;
1, 3, 5, 5, 8, 11, 11, 14, 14, 14;
1, 4, 9, 14, 14, 22, 33, 44, 44, 58, 72, 72, 86, 86, 86;...
where right border = column 0 of this square array.

Cf. columns: A127715, A152401, A152404.
Cf. related triangles: A152400, A127714.
Cf. variants: A125781, A127054, A136212, A136217, A135876, A135878, A136730.

{T(n, k)=local(A=0, m=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==m*(m+1)/2, m+=1, A+=T(n-1, c); d+=1); c+=1)); A}

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A135880 Triangle P, read by rows, where column k of P^2 equals column 0 of P^(2k+2) such that column 0 of P^2 equals column 0 of P shift one place left, with P(0,0)=1.

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A135881 Column 0 of triangle A135880.

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A135876 Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms at positions [(m+2)^2/4 - 1] for m>=0 and then taking partial sums, starting with all 1's in row 0.

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A135879 Triangle, read by rows of A135901(n) terms, where row n+1 is generated from row n by inserting zeros at positions [(m+3)^2/4 - 2], as m=0,1,2,3,... and then taking partial sums from right to left, starting with a single 1 in row 0.

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A152405 Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions {m*(m+1)/2, m>=0} and then taking partial sums, starting with all 1's in row 0.

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