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 1-4 of 4 results.

A127497 Main diagonal of triangle A127496: a(n) = A127496(n,n) for n>=0.

Original entry on oeis.org

1, 1, 2, 7, 23, 83, 306, 1135, 4257, 16095, 61222, 233956, 897499, 3454211, 13331358, 51575877, 199954871, 776642435, 3021505570, 11772342667, 45927761857, 179393034333, 701465614758, 2745595628417, 10756232646630, 42174080381586
Offset: 0

Views

Author

Paul D. Hanna, Jan 16 2007

Keywords

Links

Crossrefs

Cf. A127496, A127498.

Formula

a(n) ~ c * 4^n / sqrt(n), where c = 0.1893687633123847717351285248043394631769854231497885277419320199497639583... - Vaclav Kotesovec, Jun 15 2018

A127498 Secondary diagonal of triangle A127496: a(n) = A127496(n+1,n) for n>=0.

Original entry on oeis.org

1, 2, 5, 16, 53, 186, 672, 2465, 9158, 34354, 129833, 493534, 1884955, 7227522, 27804584, 107268790, 414852381, 1607836478, 6243201267, 24282747900, 94587991975, 368940451638, 1440796181876, 5632827891526, 22043782012620
Offset: 0

Views

Author

Paul D. Hanna, Jan 16 2007

Keywords

Crossrefs

Cf. A127496, A127497.

A305601 G.f. A(x) satisfies: [x^k] (1+x)^(n*(n+1)/2) * A(x) = 0 for k = n*(n-1)/2 + 1 through k = n*(n+1)/2 for n >= 1.

Original entry on oeis.org

1, -1, 0, 2, -7, 21, -56, 125, -209, 154, 572, -3404, 11930, -35394, 99144, -274550, 757813, -2057867, 5392764, -13383194, 30829315, -63995085, 112304664, -133488537, -70063101, 1177164984, -5212740024, 17744267816, -53365305260, 149452146536, -401301906464, 1053638673004, -2741495192679, 7122821187935, -18514807074104, 48019064944442, -123571120120435, 313403811733896, -778001059367703, 1877334690759250, -4370190271978998
Offset: 0

Views

Author

Paul D. Hanna, Jun 14 2018

Keywords

Comments

The diagonals in the table of coefficients of x^k in (1+x)^n * A(x) forms the rows of irregular triangle A127496.
Equals the inverse binomial transform of A305605.

Examples

			G.f. A(x) = 1 - x + 2*x^3 - 7*x^4 + 21*x^5 - 56*x^6 + 125*x^7 - 209*x^8 + 154*x^9 + 572*x^10 - 3404*x^11 + 11930*x^12 - 35394*x^13 + 99144*x^14 - 274550*x^15 + 757813*x^16 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in (1+x)^n * A(x) begins:
n= 0: [1, -1,  0,   2,  -7,   21,  -56,  125, -209,  154, 572, -3404, ...];
n= 1: [1,  0, -1,   2,  -5,   14,  -35,   69,  -84,  -55, 726, -2832, ...];
n= 2: [1,  1, -1,   1,  -3,    9,  -21,   34,  -15, -139, 671, -2106, ...];
n= 3: [1,  2,  0,   0,  -2,    6,  -12,   13,   19, -154, 532, -1435, ...];
n= 4: [1,  3,  2,   0,  -2,    4,   -6,    1,   32, -135, 378,  -903, ...];
n= 5: [1,  4,  5,   2,  -2,    2,   -2,   -5,   33, -103, 243,  -525, ...];
n= 6: [1,  5,  9,   7,   0,    0,    0,   -7,   28,  -70, 140,  -282, ...];
n= 7: [1,  6, 14,  16,   7,    0,    0,   -7,   21,  -42,  70,  -142, ...];
n= 8: [1,  7, 20,  30,  23,    7,    0,   -7,   14,  -21,  28,   -72, ...];
n= 9: [1,  8, 27,  50,  53,   30,    7,   -7,    7,   -7,   7,   -44, ...];
n=10: [1,  9, 35,  77, 103,   83,   37,    0,    0,    0,   0,   -37, ...];
n=11: [1, 10, 44, 112, 180,  186,  120,   37,    0,    0,   0,   -37, ...];
n=12: [1, 11, 54, 156, 292,  366,  306,  157,   37,    0,   0,   -37, ...];
n=13: [1, 12, 65, 210, 448,  658,  672,  463,  194,   37,   0,   -37, ...];
n=14: [1, 13, 77, 275, 658, 1106, 1330, 1135,  657,  231,  37,   -37, ...];
n=15: [1, 14, 90, 352, 933, 1764, 2436, 2465, 1792,  888, 268, 0, 0, 0, 0, 0, -268, ...]; ...
which illustrates the occurrences of zeros in the table.
RELATED SEQUENCES.
Notice that [x^(n*(n-1)/2)] (1+x)^(n*(n+1)/2) * A(x) = A107877(n), which begins:
[1, 1, 2, 7, 37, 268, 2496, 28612, 391189, 6230646, 113521387, ...].
Also, note that the coefficient of x^(n*(n-1)/2) in (1+x)^(n*(n+1)/2) * A(x) yields -A107877(n).
Also, observe that [x^n] (1+x)^(2*n) * A(x) = A127497(n), which begins:
[1, 1, 2, 7, 23, 83, 306, 1135, 4257, 16095, 61222, 233956, ...].
The initial terms of the diagonals in the above table forms the rows of irregular triangle A127496:
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; ...
in which row n equals the partial sums of the prior row with the final term repeated n more times at the end.

Links

Crossrefs

Cf. A107877, A127497, A127496, A305605, A305600.

Programs

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

Formula

G.f. A(x) also satisfies:
(1) A(x) = 1/(1+x) - Sum_{n>=1} A107877(n) * (x/(1+x))^(n*(n+1)/2+1) / (1+x)^n.
(2) [x^(n*(n-1)/2)] (1+x)^(n*(n+1)/2) * A(x) = A107877(n) for n >= 0.
(3) [x^(n*(n+1)/2 + 1)] (1+x)^(n*(n+1)/2) * A(x) = -A107877(n) for n >= 0.
(4) [x^n] (1+x)^(2*n) * A(x) = A127497(n) = A127496(n,n) for n>=0.
(5) [x^k] (1+x)^(n+k) * A(x) = A127496(n,k) for k = 0..n*(n+1)/2, for n>=0.
(6) [x^n] (1+x)^n * A(x) = A305605(n) for n >= 0.

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

Views

Author

Paul D. Hanna, Jun 14 2018

Keywords

Examples

			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.

Links

Crossrefs

Cf. A305601, A107877, A127496.

Programs

  • PARI
    /* 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(""))

Formula

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.
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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