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.

Showing 1-5 of 5 results.

A135901 Number of terms in row n of irregular triangle A135879 for n>=0.

Original entry on oeis.org

1, 2, 4, 7, 12, 18, 26, 36, 47, 60, 75, 92, 111, 132, 154, 178, 204, 232, 262, 294, 328, 364, 402, 442, 484, 527, 572, 619, 668, 719, 772, 827, 884, 943, 1004, 1067, 1132, 1199, 1268, 1339, 1412, 1487, 1564, 1643, 1724, 1807, 1892, 1978, 2066, 2156, 2248
Offset: 0

Views

Author

Paul D. Hanna, Dec 16 2007

Keywords

Examples

			A(x) = (1+x^2)/(1-x)^3 - [x+x^2+x^3+x^5+x^8+x^14+x^25+x^47+x^90+x^176+...]/(1-x)^2.
		

Crossrefs

Cf. A135879.

Programs

  • PARI
    {a(n)=polcoeff((1+x^2)/(1-x+x*O(x^n))^3 - sum(k=1,#binary(n)+2,x^((2^(k+1)+6*k+3+(-1)^k)/12))/(1-x +x*O(x^n) )^2,n)}

Formula

G.f.: A(x) = (1+x^2)/(1-x)^3 - [Sum_{n>=1} x^b(n)]/(1-x)^2, where b(n) = (2^(n+1) + 6*n + 3 + (-1)^n)/12.

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.

Original entry on oeis.org

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

Views

Author

Paul D. Hanna, Dec 15 2007

Keywords

Comments

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

Examples

			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.
		

Crossrefs

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).

Programs

  • PARI
    {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]}

Formula

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

Views

Author

Paul D. Hanna, Dec 15 2007

Keywords

Comments

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

Examples

			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.
		

Crossrefs

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

Programs

  • PARI
    /* 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])}
    
  • PARI
    /* 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]}

Extensions

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

A135878 Square array, read by antidiagonals, where row n+1 is generated from row n by first removing terms at positions [(m+3)^2/4 - 2] for m>=0 and then taking partial sums, starting with all 1's in row 0.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 6, 6, 3, 1, 25, 25, 12, 4, 1, 138, 138, 63, 19, 5, 1, 970, 970, 421, 113, 28, 6, 1, 8390, 8390, 3472, 832, 190, 38, 7, 1, 86796, 86796, 34380, 7420, 1560, 283, 50, 8, 1, 1049546, 1049546, 399463, 78406, 15250, 2502, 411, 63, 9, 1, 14563135, 14563135
Offset: 0

Views

Author

Paul D. Hanna, Dec 14 2007

Keywords

Comments

Column 0 is A135881 which equals column 0 of triangle A135879 and also equals column 0 of triangle A135880. Compare to triangle A135879, which is generated by a complementary process. An interesting variant is square array A135876, in which column 0 equals the double factorials (A001147).

Examples

			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),2,(3),4,(5),6,7,(8),9,10,(11),12,13,14,(15),16,17,18,(19),20,...;
(2),6,(12),19,(28),38,50,(63),77,93,(110),128,148,169,(191),214,...;
(6),25,(63),113,(190),283,411,(559),728,942,(1181),1446,1766,2116,...;
(25),138,(421),832,(1560),2502,3948,(5714),7830,10740,(14130),18036,...;
(138),970,(3472),7420,(15250),25990,44026,(67112),95918,138343,(189598),..;
(970),8390,(34380),78406,(174324),312667,(563287),897471,1329234,2003240,..;
(8390),86796,(399463),962750,(2291984),4295224,8168819,(13523882),20656067,.;
(86796),1049546,(5344770),13513589,(34169656),66534382,132787852,(227380975),.;
(1049546),14563135,(81097517),213885369,(570682050),1149537869,2395865161,..;
(14563135),228448504,(1377986373),3773851534,(10568874312),21945438536,...;
where terms in parenthesis are removed before taking partial sums.
For example, to generate row 2 from row 1, remove terms at positions
{[(m+3)^2/4-2], m>=0} = [0,2,4,7,10,14,18,23,28,34,...] to obtain:
[2, 4, 6,7, 9,10, 12,13,14, 16,17,18, 20,21,22,23, ...]
then take partial sums to get row 2:
[2, 6, 12,19, 28,38, 50,63,77, 93,110,128, 148,169,191,214, ...].
Repeating this process will generate all the rows of the triangle.
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.
		

Crossrefs

Programs

  • PARI
    {T(n, k)=local(A=0, b=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==floor((b+3)^2/4)-2, b+=1, A+=T(n-1, c); d+=1); c+=1)); A}

A135877 Triangle, read by rows of n(n+1)+1 terms, where row n+1 is generated from row n by inserting zeros at positions [(m+2)^2/4 - 1] for m=1..2n+2 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, 1, 3, 3, 3, 2, 2, 1, 1, 15, 15, 15, 12, 12, 9, 9, 6, 4, 4, 2, 1, 1, 105, 105, 105, 90, 90, 75, 75, 60, 48, 48, 36, 27, 27, 18, 12, 8, 8, 4, 2, 1, 1, 945, 945, 945, 840, 840, 735, 735, 630, 540, 540, 450, 375, 375, 300, 240, 192, 192, 144, 108, 81, 81, 54, 36, 24, 16, 16, 8
Offset: 0

Views

Author

Paul D. Hanna, Dec 14 2007

Keywords

Comments

Compare to square array A135876 which is generated by a complementary process. Compare to triangle A127452 which generates the factorials in column 0. A very interesting variant is triangle A135879.

Examples

			Triangle begins:
1;
1, 1, 1;
3, 3, 3, 2, 2, 1, 1;
15, 15, 15, 12, 12, 9, 9, 6, 4, 4, 2, 1, 1;
105, 105, 105, 90, 90, 75, 75, 60, 48, 48, 36, 27, 27, 18, 12, 8, 8, 4, 2, 1, 1;
945, 945, 945, 840, 840, 735, 735, 630, 540, 540, 450, 375, 375, 300, 240, 192, 192, 144, 108, 81, 81, 54, 36, 24, 16, 16, 8, 4, 2, 1, 1; ...
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+2)^2/4 - 1] for m=1..2n+2 and then
taking reverse partial sums (i.e., summing from right to left).
Start with row 0, insert 2 zeros in front of the '1':
[0,0,1];
take reverse partial sums to get row 1:
[1,1,1];
insert zeros at positions [0,1,3,5]:
[0,0,1,0,1,0,1];
take reverse partial sums to get row 2:
[3,3,3,2,2,1,1];
insert zeros at positions [0,1,3,5,8,11]:
[0,0,3,0,3,0,3,2,0,2,1,0,1];
take reverse partial sums to get row 3:
[15,15,15,12,12,9,9,6,4,4,2,1,1];
insert zeros at positions [0,1,3,5,8,11,15,19]:
[0,0,15,0,15,0,15,12,0,12,9,0,9,6,4,0,4,2,1,0,1];
take reverse partial sums to get row 4:
[105,105,105,90,90,75,75,60,48,48,36,27,27,18,12,8,8,4,2,1,1].
		

Crossrefs

Programs

  • PARI
    {T(n,k)=local(A=[1],B);if(n>0,for(i=1,n,m=1;B=[0]; 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])} /* for(n=0,8,for(k=0,n*(n+1),print1(T(n,k),","));print("")) */

Formula

Column 0 equals the double factorials A001147(n) = (2n)!/(n!*2^n).
Showing 1-5 of 5 results.